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| author | kinitrupti | 2017-05-12 18:40:35 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
| commit | 64d949698432e05f2a372d9edc859c5b9df1f438 (patch) | |
| tree | 012fd5b4ac9102cdcf5bc56305e49d6714fa5951 /Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics | |
| parent | 9c6ab8cbf3e1a84c780386abf4852d84cdd32d56 (diff) | |
| download | Python-Textbook-Companions-64d949698432e05f2a372d9edc859c5b9df1f438.tar.gz Python-Textbook-Companions-64d949698432e05f2a372d9edc859c5b9df1f438.tar.bz2 Python-Textbook-Companions-64d949698432e05f2a372d9edc859c5b9df1f438.zip | |
Revised list of TBCs
Diffstat (limited to 'Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics')
57 files changed, 0 insertions, 18900 deletions
diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01.ipynb deleted file mode 100755 index 1b47a56a..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01.ipynb +++ /dev/null @@ -1,165 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:973d427aa265e6008d444ff4e5a0026589bdadc695d5f549c16acb669fdf56e2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1. Fundamental Concepts of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 1.1, Page Number 7"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "Pi = 3.21e5 #Recommended tyre pressure, Pa\n",
- "Ti = -5.00 #Initial Tyre temperature, \u00b0C\n",
- "Tf = 28.00 #Final Tyre temperature, \u00b0C\n",
- "\n",
- "#Calculations\n",
- "Ti = 273.16 + Ti\n",
- "Tf = 273.16 + Tf\n",
- "pf = Pi*Tf/Ti #Final tyre pressure, Pa\n",
- "\n",
- "#Results\n",
- "print 'Final Tyre pressure is %6.2e Pa'%pf"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Final Tyre pressure is 3.61e+05 Pa\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.2, Page Number 8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "phe = 1.5 #Pressure in Helium chamber, bar\n",
- "vhe = 2.0 #Volume of Helium chamber, L\n",
- "pne = 2.5 #Pressure in Neon chamber, bar\n",
- "vne = 3.0 #Volume of Neon chamber, L\n",
- "pxe = 1.0 #Pressure in Xenon chamber, bar\n",
- "vxe = 1.0 #Volume of Xenon chamber, L\n",
- "R = 8.314e-2 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "#Calculations\n",
- "\n",
- "nhe = phe*vhe/(R*T) #Number of moles of Helium, mol\n",
- "nne = pne*vne/(R*T) #Number of moles of Neon, mol\n",
- "nxe = pxe*vxe/(R*T) #Number of moles of Xenon, mol\n",
- "n = nhe + nne + nxe #Total number of moles, mol\n",
- "V = vhe + vne + vxe #Total volume of system, L\n",
- "xhe = nhe/n\n",
- "xne = nne/n\n",
- "xxe = nxe/n\n",
- "P = n*R*T/(V)\n",
- "phe = P*xhe #Partial pressure of Helium, bar\n",
- "pne = P*xne #Partial pressure of Neon, bar\n",
- "pxe = P*xxe #Partial pressure of Xenon, bar\n",
- "\n",
- "#Results\n",
- "print 'Moles of He=%4.3f, Ne=%4.3f and, Xe=%4.3f in mol'%(nhe,nne,nxe) \n",
- "print 'Mole fraction of xHe=%4.3f, xNe=%4.3f and, xXe=%4.3f'%(xhe,xne,xxe)\n",
- "print 'Final pressure is %4.3f bar'%P\n",
- "print 'Partial pressure of pHe=%4.3f, pNe=%4.3f and, pXe=%4.3f in bar'%(phe,pne,pxe)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moles of He=0.121, Ne=0.303 and, Xe=0.040 in mol\n",
- "Mole fraction of xHe=0.261, xNe=0.652 and, xXe=0.087\n",
- "Final pressure is 1.917 bar\n",
- "Partial pressure of pHe=0.500, pNe=1.250 and, pXe=0.167 in bar\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.4, Page Number 10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "T = 300.0 #Nitrogen temperature, K\n",
- "v1 = 250.00 #Molar volume, L\n",
- "v2 = 0.1 #Molar volume, L\n",
- "a = 1.37 #Van der Waals parameter a, bar.dm6/mol2 \n",
- "b = 0.0387 #Van der Waals parameter b, dm3/mol\n",
- "R = 8.314e-2 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "n = 1.\n",
- "#Calculations\n",
- "\n",
- "p1 = n*R*T/v1 \n",
- "p2 = n*R*T/v2\n",
- "pv1 = n*R*T/(v1-n*b)- n**2*a/v1**2\n",
- "pv2 = n*R*T/(v2-n*b)- n**2*a/v2**2\n",
- "\n",
- "#Results\n",
- "print 'Pressure from ideal gas law = %4.2e bar nad from Van der Waals equation = %4.2e bar '%(p1, pv1)\n",
- "print 'Pressure from ideal gas law = %4.1f bar nad from Van der Waals equation = %4.1f bar '%(p2, pv2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure from ideal gas law = 9.98e-02 bar nad from Van der Waals equation = 9.98e-02 bar \n",
- "Pressure from ideal gas law = 249.4 bar nad from Van der Waals equation = 269.9 bar \n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01_1.ipynb deleted file mode 100755 index 1b47a56a..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01_1.ipynb +++ /dev/null @@ -1,165 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:973d427aa265e6008d444ff4e5a0026589bdadc695d5f549c16acb669fdf56e2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1. Fundamental Concepts of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 1.1, Page Number 7"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "Pi = 3.21e5 #Recommended tyre pressure, Pa\n",
- "Ti = -5.00 #Initial Tyre temperature, \u00b0C\n",
- "Tf = 28.00 #Final Tyre temperature, \u00b0C\n",
- "\n",
- "#Calculations\n",
- "Ti = 273.16 + Ti\n",
- "Tf = 273.16 + Tf\n",
- "pf = Pi*Tf/Ti #Final tyre pressure, Pa\n",
- "\n",
- "#Results\n",
- "print 'Final Tyre pressure is %6.2e Pa'%pf"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Final Tyre pressure is 3.61e+05 Pa\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.2, Page Number 8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "phe = 1.5 #Pressure in Helium chamber, bar\n",
- "vhe = 2.0 #Volume of Helium chamber, L\n",
- "pne = 2.5 #Pressure in Neon chamber, bar\n",
- "vne = 3.0 #Volume of Neon chamber, L\n",
- "pxe = 1.0 #Pressure in Xenon chamber, bar\n",
- "vxe = 1.0 #Volume of Xenon chamber, L\n",
- "R = 8.314e-2 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "#Calculations\n",
- "\n",
- "nhe = phe*vhe/(R*T) #Number of moles of Helium, mol\n",
- "nne = pne*vne/(R*T) #Number of moles of Neon, mol\n",
- "nxe = pxe*vxe/(R*T) #Number of moles of Xenon, mol\n",
- "n = nhe + nne + nxe #Total number of moles, mol\n",
- "V = vhe + vne + vxe #Total volume of system, L\n",
- "xhe = nhe/n\n",
- "xne = nne/n\n",
- "xxe = nxe/n\n",
- "P = n*R*T/(V)\n",
- "phe = P*xhe #Partial pressure of Helium, bar\n",
- "pne = P*xne #Partial pressure of Neon, bar\n",
- "pxe = P*xxe #Partial pressure of Xenon, bar\n",
- "\n",
- "#Results\n",
- "print 'Moles of He=%4.3f, Ne=%4.3f and, Xe=%4.3f in mol'%(nhe,nne,nxe) \n",
- "print 'Mole fraction of xHe=%4.3f, xNe=%4.3f and, xXe=%4.3f'%(xhe,xne,xxe)\n",
- "print 'Final pressure is %4.3f bar'%P\n",
- "print 'Partial pressure of pHe=%4.3f, pNe=%4.3f and, pXe=%4.3f in bar'%(phe,pne,pxe)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moles of He=0.121, Ne=0.303 and, Xe=0.040 in mol\n",
- "Mole fraction of xHe=0.261, xNe=0.652 and, xXe=0.087\n",
- "Final pressure is 1.917 bar\n",
- "Partial pressure of pHe=0.500, pNe=1.250 and, pXe=0.167 in bar\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.4, Page Number 10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "T = 300.0 #Nitrogen temperature, K\n",
- "v1 = 250.00 #Molar volume, L\n",
- "v2 = 0.1 #Molar volume, L\n",
- "a = 1.37 #Van der Waals parameter a, bar.dm6/mol2 \n",
- "b = 0.0387 #Van der Waals parameter b, dm3/mol\n",
- "R = 8.314e-2 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "n = 1.\n",
- "#Calculations\n",
- "\n",
- "p1 = n*R*T/v1 \n",
- "p2 = n*R*T/v2\n",
- "pv1 = n*R*T/(v1-n*b)- n**2*a/v1**2\n",
- "pv2 = n*R*T/(v2-n*b)- n**2*a/v2**2\n",
- "\n",
- "#Results\n",
- "print 'Pressure from ideal gas law = %4.2e bar nad from Van der Waals equation = %4.2e bar '%(p1, pv1)\n",
- "print 'Pressure from ideal gas law = %4.1f bar nad from Van der Waals equation = %4.1f bar '%(p2, pv2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure from ideal gas law = 9.98e-02 bar nad from Van der Waals equation = 9.98e-02 bar \n",
- "Pressure from ideal gas law = 249.4 bar nad from Van der Waals equation = 269.9 bar \n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01_2.ipynb deleted file mode 100755 index 1b47a56a..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter01_2.ipynb +++ /dev/null @@ -1,165 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:973d427aa265e6008d444ff4e5a0026589bdadc695d5f549c16acb669fdf56e2"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1. Fundamental Concepts of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 1.1, Page Number 7"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "Pi = 3.21e5 #Recommended tyre pressure, Pa\n",
- "Ti = -5.00 #Initial Tyre temperature, \u00b0C\n",
- "Tf = 28.00 #Final Tyre temperature, \u00b0C\n",
- "\n",
- "#Calculations\n",
- "Ti = 273.16 + Ti\n",
- "Tf = 273.16 + Tf\n",
- "pf = Pi*Tf/Ti #Final tyre pressure, Pa\n",
- "\n",
- "#Results\n",
- "print 'Final Tyre pressure is %6.2e Pa'%pf"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Final Tyre pressure is 3.61e+05 Pa\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.2, Page Number 8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "phe = 1.5 #Pressure in Helium chamber, bar\n",
- "vhe = 2.0 #Volume of Helium chamber, L\n",
- "pne = 2.5 #Pressure in Neon chamber, bar\n",
- "vne = 3.0 #Volume of Neon chamber, L\n",
- "pxe = 1.0 #Pressure in Xenon chamber, bar\n",
- "vxe = 1.0 #Volume of Xenon chamber, L\n",
- "R = 8.314e-2 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "#Calculations\n",
- "\n",
- "nhe = phe*vhe/(R*T) #Number of moles of Helium, mol\n",
- "nne = pne*vne/(R*T) #Number of moles of Neon, mol\n",
- "nxe = pxe*vxe/(R*T) #Number of moles of Xenon, mol\n",
- "n = nhe + nne + nxe #Total number of moles, mol\n",
- "V = vhe + vne + vxe #Total volume of system, L\n",
- "xhe = nhe/n\n",
- "xne = nne/n\n",
- "xxe = nxe/n\n",
- "P = n*R*T/(V)\n",
- "phe = P*xhe #Partial pressure of Helium, bar\n",
- "pne = P*xne #Partial pressure of Neon, bar\n",
- "pxe = P*xxe #Partial pressure of Xenon, bar\n",
- "\n",
- "#Results\n",
- "print 'Moles of He=%4.3f, Ne=%4.3f and, Xe=%4.3f in mol'%(nhe,nne,nxe) \n",
- "print 'Mole fraction of xHe=%4.3f, xNe=%4.3f and, xXe=%4.3f'%(xhe,xne,xxe)\n",
- "print 'Final pressure is %4.3f bar'%P\n",
- "print 'Partial pressure of pHe=%4.3f, pNe=%4.3f and, pXe=%4.3f in bar'%(phe,pne,pxe)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moles of He=0.121, Ne=0.303 and, Xe=0.040 in mol\n",
- "Mole fraction of xHe=0.261, xNe=0.652 and, xXe=0.087\n",
- "Final pressure is 1.917 bar\n",
- "Partial pressure of pHe=0.500, pNe=1.250 and, pXe=0.167 in bar\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1.4, Page Number 10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "T = 300.0 #Nitrogen temperature, K\n",
- "v1 = 250.00 #Molar volume, L\n",
- "v2 = 0.1 #Molar volume, L\n",
- "a = 1.37 #Van der Waals parameter a, bar.dm6/mol2 \n",
- "b = 0.0387 #Van der Waals parameter b, dm3/mol\n",
- "R = 8.314e-2 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "n = 1.\n",
- "#Calculations\n",
- "\n",
- "p1 = n*R*T/v1 \n",
- "p2 = n*R*T/v2\n",
- "pv1 = n*R*T/(v1-n*b)- n**2*a/v1**2\n",
- "pv2 = n*R*T/(v2-n*b)- n**2*a/v2**2\n",
- "\n",
- "#Results\n",
- "print 'Pressure from ideal gas law = %4.2e bar nad from Van der Waals equation = %4.2e bar '%(p1, pv1)\n",
- "print 'Pressure from ideal gas law = %4.1f bar nad from Van der Waals equation = %4.1f bar '%(p2, pv2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure from ideal gas law = 9.98e-02 bar nad from Van der Waals equation = 9.98e-02 bar \n",
- "Pressure from ideal gas law = 249.4 bar nad from Van der Waals equation = 269.9 bar \n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02.ipynb deleted file mode 100755 index 432f8e79..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02.ipynb +++ /dev/null @@ -1,420 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:79d03e9c97c3d8ac9e7b819d4383e36696e6a2ea64bb801edb77a95675cdc73c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 02: Heat, Work, Internal Energy, Enthalpy, and The First Law of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.1, Page 18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration Part a\n",
- "vi = 20.0 #Initial volume of ideal gas, L\n",
- "vf = 85.0 #final volume of ideal gas, L\n",
- "Pext = 2.5 #External Pressure against which work is done, bar\n",
- "\n",
- "#Calculations\n",
- "w = -Pext*1e5*(vf-vi)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'Part a: Work done in expansion is %6.1f kJ'%(w/1000)\n",
- "\n",
- "#Variable Declaration Part b\n",
- "ri = 1.00 #Initial diameter of bubble, cm\n",
- "rf = 3.25 #final diameter of bubble, cm\n",
- "sigm = 71.99 #Surface tension, N/m\n",
- "\n",
- "from math import pi\n",
- "#Calculations\n",
- "w = -2*sigm*4*pi*(rf**2-ri**2)*1e-4\n",
- "\n",
- "#Results\n",
- "print 'Part b: Work done in expansion of bubble is %4.2f J'%w\n",
- "\n",
- "#Variable Declaration Part c\n",
- "i = 3.20 #Current through heating coil, A \n",
- "v = 14.5 #fVoltage applied across coil, volts\n",
- "t = 30.0 #time for which current is applied,s\n",
- "\n",
- "from math import pi\n",
- "#Calculations\n",
- "w = v*i*t\n",
- "\n",
- "#Results\n",
- "print 'Part c: Work done in paasing the cuurent through coil is %4.2f kJ'%(w/1000)\n",
- "\n",
- "#Variable Declaration Part d\n",
- "k = 100.0 #Constant in F = -kx, N/cm \n",
- "dl = -0.15 #stretch , cm\n",
- "\n",
- "from math import pi\n",
- "#Calculations\n",
- "w = -k*(dl**2-0)/2\n",
- "\n",
- "#Results\n",
- "print 'Part d: Work done stretching th fiber is %4.2f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Part a: Work done in expansion is -16.2 kJ\n",
- "Part b: Work done in expansion of bubble is -1.73 J\n",
- "Part c: Work done in paasing the cuurent through coil is 1.39 kJ\n",
- "Part c: Work done stretching th fiber is -1.12 J\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.2, Page Number 20"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration \n",
- "m = 100.0 #Mass of water, g \n",
- "T = 100.0 #Temperature of water, \u00b0C\n",
- "Pext = 1.0 #External Pressure on assembly, bar\n",
- "x = 10.0 #percent of water vaporised at 1 bar,-\n",
- "i = 2.00 #current through heating coil, A\n",
- "v = 12.0 #Voltage applied, v\n",
- "t = 1.0e3 #time for which current applied, s \n",
- "rhol = 997 #Density of liquid, kg/m3\n",
- "rhog = 0.59 #Density of vapor, kg/m3\n",
- "\n",
- "#Calculations\n",
- "q = i*v*t\n",
- "vi = m/(rhol*100)*1e-3\n",
- "vf = m*(100-x)*1e-3/(rhol*100) + m*x*1e-3/(rhog*100)\n",
- "w = -Pext*(vf-vi)*1e5\n",
- "#Results\n",
- "print 'Heat added to the water %4.2f kJ'%(q/1000)\n",
- "print 'Work done in vaporizing liquid is %4.2f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat added to the water 24.00 kJ\n",
- "Work done in vaporizing liquid is -1703.84 J\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.3, Page Number 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration Part d\n",
- "m = 1.5 #mass of water in surrounding, kg \n",
- "dT = 14.2 #Change in temperature of water, \u00b0C or K\n",
- "cp = 4.18 #Specific heat of water at constant pressure, J/(g.K)\n",
- "\n",
- "#Calculations\n",
- "qp = m*cp*dT\n",
- "\n",
- "#Results\n",
- "print 'Heat removed by water at constant pressure %4.2f kJ'%qp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat removed by water at constant pressure 89.03 kJ\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.4, Page Number 28"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "#Variable declaration\n",
- "n = 2.0 #moles of ideal gas\n",
- "R = 8.314 #Ideal gas constant, bar.L/(mol.K)\n",
- "#For reverssible Isothermal expansion \n",
- "Pi1 = 25.0 #Initial Pressure of ideal gas, bar\n",
- "Vi1 = 4.50 #Initial volume of ideal gas, L\n",
- "Pf1 = 4.50 #Fianl Pressure of ideal gas, bar\n",
- "Pext = 4.50 #External pressure, bar \n",
- "Pint = 11.0 #Intermediate pressure, bar\n",
- "\n",
- "#Calcualtions reverssible Isothermal expansion \n",
- "T1 = Pi1*Vi1/(n*R)\n",
- "Vf1 = n*R*T1/Pf1\n",
- "w = -n*R*T1*log(Vf1/Vi1)\n",
- "\n",
- "#Results\n",
- "print 'For reverssible Isothermal expansion'\n",
- "print 'Work done = %4.2e J'%w\n",
- "\n",
- "#Calcualtions Single step irreverssible expansion \n",
- "\n",
- "w = -Pext*1e5*(Vf1-Vi1)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'For Single step reverssible expansion'\n",
- "print 'Work done = %4.2e J'%w\n",
- "\n",
- "#Calcualtions Two step irreverssible expansion \n",
- "Vint = n*R*T1/(Pint)\n",
- "w = -Pint*1e5*(Vint-Vi1)*1e-3 - Pf1*1e5*(Vf1-Vint)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'For Two step reverssible expansion'\n",
- "print 'Work done = %4.2e J'%w\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For reverssible Isothermal expansion\n",
- "Work done = -1.93e+02 J\n",
- "For Single step reverssible expansion\n",
- "Work done = -9.22e+03 J\n",
- "For Twi step reverssible expansion\n",
- "Work done = -1.29e+04 J\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.5, Page Number 32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable declaration\n",
- "n = 2.5 #moles of ideal gas\n",
- "R = 0.08314 #Ideal gas constant, bar.L/(mol.K)\n",
- "cvm = 20.79 #Heat Capacity at constant volume, J/(mol.K)\n",
- "\n",
- "p1 = 16.6 #Pressure at point 1, bar\n",
- "v1 = 1.00 #Volume at point 1, L\n",
- "p2 = 16.6 #Pressure at point 2, bar\n",
- "v2 = 25.0 #Volume at point 2, L \n",
- "v3 = 25.0 #Volume at point 3, L\n",
- "\n",
- "#Calculations\n",
- "T1 = p1*v1/(n*R)\n",
- "T2 = p2*v2/(n*R)\n",
- "T3 = T1 #from problem statement\n",
- " #for path 1-2\n",
- "DU12 = n*cvm*(T2-T1)\n",
- "w12 = -p1*1e5*(v2-v1)*1e-3\n",
- "q12 = DU12 - w12\n",
- "DH12 = DU12 + n*R*(T2-T1)*1e2\n",
- "\n",
- " #for path 2-3\n",
- "w23 = 0.0\n",
- "DU23 = q23 = n*cvm*(T3-T2)\n",
- "DH23 = -DH12\n",
- "\n",
- "\n",
- " #for path 3-1\n",
- "DU31 = 0.0 #Isothemal process\n",
- "DH31 = 0.0\n",
- "w31 = -n*R*1e2*T1*log(v1/v3)\n",
- "q31 = -w31\n",
- "\n",
- "DU = DU12+DU23+DU31\n",
- "w = w12+w23+w31\n",
- "q = q12+q23+q31\n",
- "DH = DH12+DH23+DH31\n",
- "\n",
- "#Results\n",
- "print 'For Path q w DU DH '\n",
- "print '1-2 %7.2f %7.2f %7.2f %7.2f'%(q12,w12,DU12,DH12)\n",
- "print '2-3 %7.2f %7.2f %7.2f %7.2f'%(q23,w23,DU23,DH23)\n",
- "print '3-1 %7.2f %7.2f %7.2f %7.2f'%(q31,w31,DU31,DH31)\n",
- "print 'Overall %7.2f %7.2f %7.2f %7.2f'%(q,w,DU,DH)\n",
- "print 'all values are in J'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For Path q w DU DH \n",
- "1-2 139463.96 -39840.00 99623.96 139463.96\n",
- "2-3 -99623.96 0.00 -99623.96 -139463.96\n",
- "3-1 -5343.33 5343.33 0.00 0.00\n",
- "Overall 34496.67 -34496.67 0.00 0.00\n",
- "all values are in J\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.6, Page Number 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "n = 2.5 #moles of ideal gas\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "cvm = 12.47 #Heat Capacity at constant volume, J/(mol.K)\n",
- "\n",
- "pext = 1.00 #External Pressure, bar\n",
- "Ti = 325. #Initial Temeprature, K\n",
- "pi = 2.50 #Initial Pressure, bar\n",
- "pf = 1.25 #Final pressure, bar \n",
- "\n",
- "#Calculations Adiabatic process q = 0; DU = w\n",
- "q = 0.0 \n",
- "Tf = Ti*(cvm + R*pext/pi)/(cvm + R*pext/pf )\n",
- "DU = w = n*cvm*(Tf-Ti)\n",
- "DH = DU + n*R*(Tf-Ti)\n",
- "\n",
- "#Results\n",
- "print 'The final temperature at end of adiabatic procees is %4.1f K'%Tf\n",
- "print 'The enthalpy change of adiabatic procees is %4.1f J'%DH\n",
- "print 'The Internal energy change of adiabatic procees is %4.1f J'%DU\n",
- "print 'The work done in expansion of adiabatic procees is %4.1f J'%w\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final temperature at end of adiabatic procees is 268.5 K\n",
- "The enthalpy change of adiabatic procees is -2937.0 J\n",
- "The Internal energy change of adiabatic procees is -1762.2 J\n",
- "The work done in expansion of adiabatic procees is -1762.2 J\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.7, Page Number 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration Part d\n",
- "h1 = 1000.0 #initial Altitude of cloud, m \n",
- "hf = 3500.0 #Final Altitude of cloud, m \n",
- "p1 = 0.802 #Pressure at h1, atm \n",
- "pf = 0.602 #Pressure at hf, atm\n",
- "T1 = 288.0 #Initial temperature of cloud, K\n",
- "cp = 28.86 #Specific heat of air, J/mol.K\n",
- "R = 8.314 #Gas constant, J/mol.K\n",
- "\n",
- "from math import log, exp\n",
- "\n",
- "#Calculations\n",
- "Tf = exp(-(cp/(cp-R)-1)/(cp/(cp-R))*log(p1/pf))*T1\n",
- "#Results\n",
- "print 'Final temperature of cloud %4.1f K'%Tf\n",
- "if Tf < 273:\n",
- " print 'You can expect cloud'\n",
- "else:\n",
- " print 'You can not expect cloud'\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Final temperature of cloud 265.2 K\n",
- "You can expect cloud\n"
- ]
- }
- ],
- "prompt_number": 34
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02_1.ipynb deleted file mode 100755 index 432f8e79..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02_1.ipynb +++ /dev/null @@ -1,420 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:79d03e9c97c3d8ac9e7b819d4383e36696e6a2ea64bb801edb77a95675cdc73c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 02: Heat, Work, Internal Energy, Enthalpy, and The First Law of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.1, Page 18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration Part a\n",
- "vi = 20.0 #Initial volume of ideal gas, L\n",
- "vf = 85.0 #final volume of ideal gas, L\n",
- "Pext = 2.5 #External Pressure against which work is done, bar\n",
- "\n",
- "#Calculations\n",
- "w = -Pext*1e5*(vf-vi)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'Part a: Work done in expansion is %6.1f kJ'%(w/1000)\n",
- "\n",
- "#Variable Declaration Part b\n",
- "ri = 1.00 #Initial diameter of bubble, cm\n",
- "rf = 3.25 #final diameter of bubble, cm\n",
- "sigm = 71.99 #Surface tension, N/m\n",
- "\n",
- "from math import pi\n",
- "#Calculations\n",
- "w = -2*sigm*4*pi*(rf**2-ri**2)*1e-4\n",
- "\n",
- "#Results\n",
- "print 'Part b: Work done in expansion of bubble is %4.2f J'%w\n",
- "\n",
- "#Variable Declaration Part c\n",
- "i = 3.20 #Current through heating coil, A \n",
- "v = 14.5 #fVoltage applied across coil, volts\n",
- "t = 30.0 #time for which current is applied,s\n",
- "\n",
- "from math import pi\n",
- "#Calculations\n",
- "w = v*i*t\n",
- "\n",
- "#Results\n",
- "print 'Part c: Work done in paasing the cuurent through coil is %4.2f kJ'%(w/1000)\n",
- "\n",
- "#Variable Declaration Part d\n",
- "k = 100.0 #Constant in F = -kx, N/cm \n",
- "dl = -0.15 #stretch , cm\n",
- "\n",
- "from math import pi\n",
- "#Calculations\n",
- "w = -k*(dl**2-0)/2\n",
- "\n",
- "#Results\n",
- "print 'Part d: Work done stretching th fiber is %4.2f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Part a: Work done in expansion is -16.2 kJ\n",
- "Part b: Work done in expansion of bubble is -1.73 J\n",
- "Part c: Work done in paasing the cuurent through coil is 1.39 kJ\n",
- "Part c: Work done stretching th fiber is -1.12 J\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.2, Page Number 20"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration \n",
- "m = 100.0 #Mass of water, g \n",
- "T = 100.0 #Temperature of water, \u00b0C\n",
- "Pext = 1.0 #External Pressure on assembly, bar\n",
- "x = 10.0 #percent of water vaporised at 1 bar,-\n",
- "i = 2.00 #current through heating coil, A\n",
- "v = 12.0 #Voltage applied, v\n",
- "t = 1.0e3 #time for which current applied, s \n",
- "rhol = 997 #Density of liquid, kg/m3\n",
- "rhog = 0.59 #Density of vapor, kg/m3\n",
- "\n",
- "#Calculations\n",
- "q = i*v*t\n",
- "vi = m/(rhol*100)*1e-3\n",
- "vf = m*(100-x)*1e-3/(rhol*100) + m*x*1e-3/(rhog*100)\n",
- "w = -Pext*(vf-vi)*1e5\n",
- "#Results\n",
- "print 'Heat added to the water %4.2f kJ'%(q/1000)\n",
- "print 'Work done in vaporizing liquid is %4.2f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat added to the water 24.00 kJ\n",
- "Work done in vaporizing liquid is -1703.84 J\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.3, Page Number 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration Part d\n",
- "m = 1.5 #mass of water in surrounding, kg \n",
- "dT = 14.2 #Change in temperature of water, \u00b0C or K\n",
- "cp = 4.18 #Specific heat of water at constant pressure, J/(g.K)\n",
- "\n",
- "#Calculations\n",
- "qp = m*cp*dT\n",
- "\n",
- "#Results\n",
- "print 'Heat removed by water at constant pressure %4.2f kJ'%qp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat removed by water at constant pressure 89.03 kJ\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.4, Page Number 28"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "#Variable declaration\n",
- "n = 2.0 #moles of ideal gas\n",
- "R = 8.314 #Ideal gas constant, bar.L/(mol.K)\n",
- "#For reverssible Isothermal expansion \n",
- "Pi1 = 25.0 #Initial Pressure of ideal gas, bar\n",
- "Vi1 = 4.50 #Initial volume of ideal gas, L\n",
- "Pf1 = 4.50 #Fianl Pressure of ideal gas, bar\n",
- "Pext = 4.50 #External pressure, bar \n",
- "Pint = 11.0 #Intermediate pressure, bar\n",
- "\n",
- "#Calcualtions reverssible Isothermal expansion \n",
- "T1 = Pi1*Vi1/(n*R)\n",
- "Vf1 = n*R*T1/Pf1\n",
- "w = -n*R*T1*log(Vf1/Vi1)\n",
- "\n",
- "#Results\n",
- "print 'For reverssible Isothermal expansion'\n",
- "print 'Work done = %4.2e J'%w\n",
- "\n",
- "#Calcualtions Single step irreverssible expansion \n",
- "\n",
- "w = -Pext*1e5*(Vf1-Vi1)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'For Single step reverssible expansion'\n",
- "print 'Work done = %4.2e J'%w\n",
- "\n",
- "#Calcualtions Two step irreverssible expansion \n",
- "Vint = n*R*T1/(Pint)\n",
- "w = -Pint*1e5*(Vint-Vi1)*1e-3 - Pf1*1e5*(Vf1-Vint)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'For Two step reverssible expansion'\n",
- "print 'Work done = %4.2e J'%w\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For reverssible Isothermal expansion\n",
- "Work done = -1.93e+02 J\n",
- "For Single step reverssible expansion\n",
- "Work done = -9.22e+03 J\n",
- "For Twi step reverssible expansion\n",
- "Work done = -1.29e+04 J\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.5, Page Number 32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable declaration\n",
- "n = 2.5 #moles of ideal gas\n",
- "R = 0.08314 #Ideal gas constant, bar.L/(mol.K)\n",
- "cvm = 20.79 #Heat Capacity at constant volume, J/(mol.K)\n",
- "\n",
- "p1 = 16.6 #Pressure at point 1, bar\n",
- "v1 = 1.00 #Volume at point 1, L\n",
- "p2 = 16.6 #Pressure at point 2, bar\n",
- "v2 = 25.0 #Volume at point 2, L \n",
- "v3 = 25.0 #Volume at point 3, L\n",
- "\n",
- "#Calculations\n",
- "T1 = p1*v1/(n*R)\n",
- "T2 = p2*v2/(n*R)\n",
- "T3 = T1 #from problem statement\n",
- " #for path 1-2\n",
- "DU12 = n*cvm*(T2-T1)\n",
- "w12 = -p1*1e5*(v2-v1)*1e-3\n",
- "q12 = DU12 - w12\n",
- "DH12 = DU12 + n*R*(T2-T1)*1e2\n",
- "\n",
- " #for path 2-3\n",
- "w23 = 0.0\n",
- "DU23 = q23 = n*cvm*(T3-T2)\n",
- "DH23 = -DH12\n",
- "\n",
- "\n",
- " #for path 3-1\n",
- "DU31 = 0.0 #Isothemal process\n",
- "DH31 = 0.0\n",
- "w31 = -n*R*1e2*T1*log(v1/v3)\n",
- "q31 = -w31\n",
- "\n",
- "DU = DU12+DU23+DU31\n",
- "w = w12+w23+w31\n",
- "q = q12+q23+q31\n",
- "DH = DH12+DH23+DH31\n",
- "\n",
- "#Results\n",
- "print 'For Path q w DU DH '\n",
- "print '1-2 %7.2f %7.2f %7.2f %7.2f'%(q12,w12,DU12,DH12)\n",
- "print '2-3 %7.2f %7.2f %7.2f %7.2f'%(q23,w23,DU23,DH23)\n",
- "print '3-1 %7.2f %7.2f %7.2f %7.2f'%(q31,w31,DU31,DH31)\n",
- "print 'Overall %7.2f %7.2f %7.2f %7.2f'%(q,w,DU,DH)\n",
- "print 'all values are in J'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For Path q w DU DH \n",
- "1-2 139463.96 -39840.00 99623.96 139463.96\n",
- "2-3 -99623.96 0.00 -99623.96 -139463.96\n",
- "3-1 -5343.33 5343.33 0.00 0.00\n",
- "Overall 34496.67 -34496.67 0.00 0.00\n",
- "all values are in J\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.6, Page Number 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "n = 2.5 #moles of ideal gas\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "cvm = 12.47 #Heat Capacity at constant volume, J/(mol.K)\n",
- "\n",
- "pext = 1.00 #External Pressure, bar\n",
- "Ti = 325. #Initial Temeprature, K\n",
- "pi = 2.50 #Initial Pressure, bar\n",
- "pf = 1.25 #Final pressure, bar \n",
- "\n",
- "#Calculations Adiabatic process q = 0; DU = w\n",
- "q = 0.0 \n",
- "Tf = Ti*(cvm + R*pext/pi)/(cvm + R*pext/pf )\n",
- "DU = w = n*cvm*(Tf-Ti)\n",
- "DH = DU + n*R*(Tf-Ti)\n",
- "\n",
- "#Results\n",
- "print 'The final temperature at end of adiabatic procees is %4.1f K'%Tf\n",
- "print 'The enthalpy change of adiabatic procees is %4.1f J'%DH\n",
- "print 'The Internal energy change of adiabatic procees is %4.1f J'%DU\n",
- "print 'The work done in expansion of adiabatic procees is %4.1f J'%w\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final temperature at end of adiabatic procees is 268.5 K\n",
- "The enthalpy change of adiabatic procees is -2937.0 J\n",
- "The Internal energy change of adiabatic procees is -1762.2 J\n",
- "The work done in expansion of adiabatic procees is -1762.2 J\n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.7, Page Number 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration Part d\n",
- "h1 = 1000.0 #initial Altitude of cloud, m \n",
- "hf = 3500.0 #Final Altitude of cloud, m \n",
- "p1 = 0.802 #Pressure at h1, atm \n",
- "pf = 0.602 #Pressure at hf, atm\n",
- "T1 = 288.0 #Initial temperature of cloud, K\n",
- "cp = 28.86 #Specific heat of air, J/mol.K\n",
- "R = 8.314 #Gas constant, J/mol.K\n",
- "\n",
- "from math import log, exp\n",
- "\n",
- "#Calculations\n",
- "Tf = exp(-(cp/(cp-R)-1)/(cp/(cp-R))*log(p1/pf))*T1\n",
- "#Results\n",
- "print 'Final temperature of cloud %4.1f K'%Tf\n",
- "if Tf < 273:\n",
- " print 'You can expect cloud'\n",
- "else:\n",
- " print 'You can not expect cloud'\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Final temperature of cloud 265.2 K\n",
- "You can expect cloud\n"
- ]
- }
- ],
- "prompt_number": 34
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02_2.ipynb deleted file mode 100755 index 33af2f86..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter02_2.ipynb +++ /dev/null @@ -1,422 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:fd4fe45c4daf974ea5ee4870185f4ea6168b1d380170021a0ef0ab1146d13121"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 02: Heat, Work, Internal Energy, Enthalpy, and The First Law of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.1, Page 18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi\n",
- "\n",
- "#Variable Declaration Part a\n",
- "vi = 20.0 #Initial volume of ideal gas, L\n",
- "vf = 85.0 #final volume of ideal gas, L\n",
- "Pext = 2.5 #External Pressure against which work is done, bar\n",
- "\n",
- "#Calculations\n",
- "w = -Pext*1e5*(vf-vi)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'Part a: Work done in expansion is %6.1f kJ'%(w/1000)\n",
- "\n",
- "#Variable Declaration Part b\n",
- "ri = 1.00 #Initial diameter of bubble, cm\n",
- "rf = 3.25 #final diameter of bubble, cm\n",
- "sigm = 71.99 #Surface tension, N/m\n",
- "\n",
- "#Calculations\n",
- "w = -2*sigm*4*pi*(rf**2-ri**2)*1e-4\n",
- "\n",
- "#Results\n",
- "print 'Part b: Work done in expansion of bubble is %4.2f J'%w\n",
- "\n",
- "#Variable Declaration Part c\n",
- "i = 3.20 #Current through heating coil, A \n",
- "v = 14.5 #fVoltage applied across coil, volts\n",
- "t = 30.0 #time for which current is applied,s\n",
- "\n",
- "from math import pi\n",
- "#Calculations\n",
- "w = v*i*t\n",
- "\n",
- "#Results\n",
- "print 'Part c: Work done in paasing the cuurent through coil is %4.2f kJ'%(w/1000)\n",
- "\n",
- "#Variable Declaration Part d\n",
- "k = 100.0 #Constant in F = -kx, N/cm \n",
- "dl = -0.15 #stretch , cm\n",
- "\n",
- "from math import pi\n",
- "#Calculations\n",
- "w = -k*(dl**2-0)/2\n",
- "\n",
- "#Results\n",
- "print 'Part d: Work done stretching th fiber is %4.2f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Part a: Work done in expansion is -16.2 kJ\n",
- "Part b: Work done in expansion of bubble is -1.73 J\n",
- "Part c: Work done in paasing the cuurent through coil is 1.39 kJ\n",
- "Part d: Work done stretching th fiber is -1.12 J\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.2, Page Number 20"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration \n",
- "m = 100.0 #Mass of water, g \n",
- "T = 100.0 #Temperature of water, \u00b0C\n",
- "Pext = 1.0 #External Pressure on assembly, bar\n",
- "x = 10.0 #percent of water vaporised at 1 bar,-\n",
- "i = 2.00 #current through heating coil, A\n",
- "v = 12.0 #Voltage applied, v\n",
- "t = 1.0e3 #time for which current applied, s \n",
- "rhol = 997 #Density of liquid, kg/m3\n",
- "rhog = 0.59 #Density of vapor, kg/m3\n",
- "\n",
- "#Calculations\n",
- "q = i*v*t\n",
- "vi = m/(rhol*100)*1e-3\n",
- "vf = m*(100-x)*1e-3/(rhol*100) + m*x*1e-3/(rhog*100)\n",
- "w = -Pext*(vf-vi)*1e5\n",
- "#Results\n",
- "print 'Heat added to the water %4.2f kJ'%(q/1000)\n",
- "print 'Work done in vaporizing liquid is %4.2f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat added to the water 24.00 kJ\n",
- "Work done in vaporizing liquid is -1703.84 J\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.3, Page Number 22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration Part d\n",
- "m = 1.5 #mass of water in surrounding, kg \n",
- "dT = 14.2 #Change in temperature of water, \u00b0C or K\n",
- "cp = 4.18 #Specific heat of water at constant pressure, J/(g.K)\n",
- "\n",
- "#Calculations\n",
- "qp = m*cp*dT\n",
- "\n",
- "#Results\n",
- "print 'Heat removed by water at constant pressure %4.2f kJ'%qp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat removed by water at constant pressure 89.03 kJ\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.4, Page Number 28"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable declaration\n",
- "n = 2.0 #moles of ideal gas\n",
- "R = 8.314 #Ideal gas constant, bar.L/(mol.K)\n",
- "#For reverssible Isothermal expansion \n",
- "Pi1 = 25.0 #Initial Pressure of ideal gas, bar\n",
- "Vi1 = 4.50 #Initial volume of ideal gas, L\n",
- "Pf1 = 4.50 #Fianl Pressure of ideal gas, bar\n",
- "Pext = 4.50 #External pressure, bar \n",
- "Pint = 11.0 #Intermediate pressure, bar\n",
- "\n",
- "#Calcualtions reverssible Isothermal expansion \n",
- "T1 = Pi1*Vi1/(n*R)\n",
- "Vf1 = n*R*T1/Pf1\n",
- "w = -n*R*T1*log(Vf1/Vi1)\n",
- "\n",
- "#Results\n",
- "print 'For reverssible Isothermal expansion'\n",
- "print 'Work done = %4.2e J'%w\n",
- "\n",
- "#Calcualtions Single step irreverssible expansion \n",
- "\n",
- "w = -Pext*1e5*(Vf1-Vi1)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'For Single step reverssible expansion'\n",
- "print 'Work done = %4.2e J'%w\n",
- "\n",
- "#Calcualtions Two step irreverssible expansion \n",
- "Vint = n*R*T1/(Pint)\n",
- "w = -Pint*1e5*(Vint-Vi1)*1e-3 - Pf1*1e5*(Vf1-Vint)*1e-3\n",
- "\n",
- "#Results\n",
- "print 'For Two step reverssible expansion'\n",
- "print 'Work done = %4.2e J'%w\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For reverssible Isothermal expansion\n",
- "Work done = -1.93e+02 J\n",
- "For Single step reverssible expansion\n",
- "Work done = -9.22e+03 J\n",
- "For Two step reverssible expansion\n",
- "Work done = -1.29e+04 J\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.5, Page Number 32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable declaration\n",
- "n = 2.5 #moles of ideal gas\n",
- "R = 0.08314 #Ideal gas constant, bar.L/(mol.K)\n",
- "cvm = 20.79 #Heat Capacity at constant volume, J/(mol.K)\n",
- "\n",
- "p1 = 16.6 #Pressure at point 1, bar\n",
- "v1 = 1.00 #Volume at point 1, L\n",
- "p2 = 16.6 #Pressure at point 2, bar\n",
- "v2 = 25.0 #Volume at point 2, L \n",
- "v3 = 25.0 #Volume at point 3, L\n",
- "\n",
- "#Calculations\n",
- "T1 = p1*v1/(n*R)\n",
- "T2 = p2*v2/(n*R)\n",
- "T3 = T1 #from problem statement\n",
- " #for path 1-2\n",
- "DU12 = n*cvm*(T2-T1)\n",
- "w12 = -p1*1e5*(v2-v1)*1e-3\n",
- "q12 = DU12 - w12\n",
- "DH12 = DU12 + n*R*(T2-T1)*1e2\n",
- "\n",
- " #for path 2-3\n",
- "w23 = 0.0\n",
- "DU23 = q23 = n*cvm*(T3-T2)\n",
- "DH23 = -DH12\n",
- "\n",
- "\n",
- " #for path 3-1\n",
- "DU31 = 0.0 #Isothemal process\n",
- "DH31 = 0.0\n",
- "w31 = -n*R*1e2*T1*log(v1/v3)\n",
- "q31 = -w31\n",
- "\n",
- "DU = DU12+DU23+DU31\n",
- "w = w12+w23+w31\n",
- "q = q12+q23+q31\n",
- "DH = DH12+DH23+DH31\n",
- "\n",
- "#Results\n",
- "print 'For Path q w DU DH '\n",
- "print '1-2 %7.2f %7.2f %7.2f %7.2f'%(q12,w12,DU12,DH12)\n",
- "print '2-3 %7.2f %7.2f %7.2f %7.2f'%(q23,w23,DU23,DH23)\n",
- "print '3-1 %7.2f %7.2f %7.2f %7.2f'%(q31,w31,DU31,DH31)\n",
- "print 'Overall %7.2f %7.2f %7.2f %7.2f'%(q,w,DU,DH)\n",
- "print 'all values are in J'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For Path q w DU DH \n",
- "1-2 139463.96 -39840.00 99623.96 139463.96\n",
- "2-3 -99623.96 0.00 -99623.96 -139463.96\n",
- "3-1 -5343.33 5343.33 0.00 0.00\n",
- "Overall 34496.67 -34496.67 0.00 0.00\n",
- "all values are in J\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.6, Page Number 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration Part d\n",
- "n = 2.5 #moles of ideal gas\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "cvm = 12.47 #Heat Capacity at constant volume, J/(mol.K)\n",
- "\n",
- "pext = 1.00 #External Pressure, bar\n",
- "Ti = 325. #Initial Temeprature, K\n",
- "pi = 2.50 #Initial Pressure, bar\n",
- "pf = 1.25 #Final pressure, bar \n",
- "\n",
- "#Calculations Adiabatic process q = 0; DU = w\n",
- "q = 0.0 \n",
- "Tf = Ti*(cvm + R*pext/pi)/(cvm + R*pext/pf )\n",
- "DU = w = n*cvm*(Tf-Ti)\n",
- "DH = DU + n*R*(Tf-Ti)\n",
- "\n",
- "#Results\n",
- "print 'The final temperature at end of adiabatic procees is %4.1f K'%Tf\n",
- "print 'The enthalpy change of adiabatic procees is %4.1f J'%DH\n",
- "print 'The Internal energy change of adiabatic procees is %4.1f J'%DU\n",
- "print 'The work done in expansion of adiabatic procees is %4.1f J'%w\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The final temperature at end of adiabatic procees is 268.5 K\n",
- "The enthalpy change of adiabatic procees is -2937.0 J\n",
- "The Internal energy change of adiabatic procees is -1762.2 J\n",
- "The work done in expansion of adiabatic procees is -1762.2 J\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 2.7, Page Number 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, exp\n",
- "\n",
- "#Variable Declaration Part d\n",
- "h1 = 1000.0 #initial Altitude of cloud, m \n",
- "hf = 3500.0 #Final Altitude of cloud, m \n",
- "p1 = 0.802 #Pressure at h1, atm \n",
- "pf = 0.602 #Pressure at hf, atm\n",
- "T1 = 288.0 #Initial temperature of cloud, K\n",
- "cp = 28.86 #Specific heat of air, J/mol.K\n",
- "R = 8.314 #Gas constant, J/mol.K\n",
- "\n",
- "#Calculations\n",
- "Tf = exp(-(cp/(cp-R)-1)/(cp/(cp-R))*log(p1/pf))*T1\n",
- "#Results\n",
- "print 'Final temperature of cloud %4.1f K'%Tf\n",
- "if Tf < 273:\n",
- " print 'You can expect cloud'\n",
- "else:\n",
- " print 'You can not expect cloud'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Final temperature of cloud 265.2 K\n",
- "You can expect cloud\n"
- ]
- }
- ],
- "prompt_number": 7
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03.ipynb deleted file mode 100755 index 55805cb9..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03.ipynb +++ /dev/null @@ -1,239 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ff90ca36306843dda7b5d7f4afff0c2bcd17928b8d2090ef8ae2b28da2c7c555"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3: Importance of State Functions: Internal Energy and Enthalpy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.2, Page Number 45"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log`\n",
- "#Variable Declaration\n",
- "betaOH = 11.2e-4 #Thermal exapnasion coefficient of ethanol, \u00b0C\n",
- "betagl = 2.00e-5 #Thermal exapnasion coefficient of glass, \u00b0C\n",
- "kOH = 11.0e-5 #Isothermal compressibility of ethanol, /bar\n",
- "dT = 10.0 #Increase in Temperature, \u00b0C\n",
- "\n",
- "#Calcualtions\n",
- "vfbyvi = (1+ betagl*dT)\n",
- "dP = betaOH*dT/kOH-(1./kOH)*log(vfbyvi)\n",
- "\n",
- "#Results\n",
- "print 'Pressure increase in capillary %4.1f bar'%dP"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure increase in capillary 100.0\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.4, Page Number 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "cpsubysy = 1000 #Specific heat ration of surrounding and system\n",
- "Tpreci = 0.006 #Precision in Temperature measurement, \u00b0C\n",
- "\n",
- "#Calcualtions\n",
- "dtgas = -cpsubysy*(-Tpreci)\n",
- "\n",
- "#Results\n",
- "print 'Minimum detectable temperature change of gas +-%4.1f \u00b0C'%dtgas"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Minimum detectable temperature change of gas +- 6.0 \u00b0C\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.6, Page Number 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #number of mole of N2, mol \n",
- "Ti = 200.0 #Intial Temperature, K\n",
- "Pi = 5.00 #Initial pressure, bar\n",
- "Tf = 400.0 #Intial Temperature, K\n",
- "Pf = 20.0 #Initial pressure, bar\n",
- "a = 0.137 #van der Waals constant a, Pa.m3/(mol2)\n",
- "b = 3.87e-5 #van der Waals constant b, m3/(mol)\n",
- "A, B, C, D = 22.5, -1.187e-2,2.3968e-5, -1.0176e-8\n",
- " #Constants in Cvm equation J, K and mol\n",
- "vi = 3.28e-3 #initial volume, m3/mol\n",
- "vf = 7.88e-3 #Final volume, m3/mol\n",
- "\n",
- "#Calculations\n",
- "T = symbols('T')\n",
- "dUT = n**2*a*(1./vi-1./vf)\n",
- "dUV = integrate( A + B*T + C*T**2 + D*T**3, (T,Ti,Tf))\n",
- "\n",
- "#Results\n",
- "print 'dUT = %4.1f J: This is wrongly reported in book'%dUT\n",
- "print 'dUV = %4.1f J'%dUV"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dUT = 24.4 J: This is wrongly reported in book\n",
- "dUV = 4174.1 J\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.7, Page Number 53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "m = 143.0 #Mass of graphite, g \n",
- "Ti = 300.0 #Intial Temperature, K\n",
- "Tf = 600.0 #Intial Temperature, K\n",
- "A, B, C, D, E = -12.19,0.1126,-1.947e-4,1.919e-7,-7.8e-11\n",
- " #Constants in Cvm equation J, K and mol\n",
- "M = 12.01\n",
- "\n",
- "#Calculations\n",
- "\n",
- "T = symbols('T')\n",
- "dH = (m/M)*integrate( A + B*T + C*T**2 + D*T**3 + E*T**4, (T,Ti,Tf))\n",
- "expr = A + B*T + C*T**2 + D*T**3 + E*T**4\n",
- "cpm = expr.subs(T,300.)\n",
- "qp = (m/M)*cpm*(Tf-Ti)\n",
- "err = abs(dH-qp)/dH\n",
- "#Results\n",
- "print 'dH = %6.1f kJ'%(dH/1000)\n",
- "print 'qp = %6.1f kJ'%(qp/1000)\n",
- "print 'Error in calculations %4.1f'%(err*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dH = 46.8 kJ\n",
- "qp = 30.8 kJ\n",
- "Error in calculations 34.3\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.9, Page Number 56"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 124.0 #Mass of liquid methanol, g\n",
- "Pi = 1.0 #Initial Pressure, bar\n",
- "Ti = 298.0 #Intial Temperature, K\n",
- "Pf = 2.5 #Final Pressure, bar\n",
- "Tf = 425.0 #Intial Temperature, K\n",
- "rho = 0.791 #Density, g/cc\n",
- "Cpm = 81.1 #Specifi heat, J/(K.mol)\n",
- "M = 32.04\n",
- "\n",
- "#Calculations\n",
- "n = m/M\n",
- "DH = n*Cpm*(Tf-Ti)+ m*(Pf-Pi)*1e-6/rho\n",
- "\n",
- "#Results\n",
- "print 'Enthalpy change for change in state of methanol is %4.1f kJ'%(DH/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy change for change in state of methanol is 39.9 kJ\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03_1.ipynb deleted file mode 100755 index 55805cb9..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03_1.ipynb +++ /dev/null @@ -1,239 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ff90ca36306843dda7b5d7f4afff0c2bcd17928b8d2090ef8ae2b28da2c7c555"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3: Importance of State Functions: Internal Energy and Enthalpy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.2, Page Number 45"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log`\n",
- "#Variable Declaration\n",
- "betaOH = 11.2e-4 #Thermal exapnasion coefficient of ethanol, \u00b0C\n",
- "betagl = 2.00e-5 #Thermal exapnasion coefficient of glass, \u00b0C\n",
- "kOH = 11.0e-5 #Isothermal compressibility of ethanol, /bar\n",
- "dT = 10.0 #Increase in Temperature, \u00b0C\n",
- "\n",
- "#Calcualtions\n",
- "vfbyvi = (1+ betagl*dT)\n",
- "dP = betaOH*dT/kOH-(1./kOH)*log(vfbyvi)\n",
- "\n",
- "#Results\n",
- "print 'Pressure increase in capillary %4.1f bar'%dP"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure increase in capillary 100.0\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.4, Page Number 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "cpsubysy = 1000 #Specific heat ration of surrounding and system\n",
- "Tpreci = 0.006 #Precision in Temperature measurement, \u00b0C\n",
- "\n",
- "#Calcualtions\n",
- "dtgas = -cpsubysy*(-Tpreci)\n",
- "\n",
- "#Results\n",
- "print 'Minimum detectable temperature change of gas +-%4.1f \u00b0C'%dtgas"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Minimum detectable temperature change of gas +- 6.0 \u00b0C\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.6, Page Number 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #number of mole of N2, mol \n",
- "Ti = 200.0 #Intial Temperature, K\n",
- "Pi = 5.00 #Initial pressure, bar\n",
- "Tf = 400.0 #Intial Temperature, K\n",
- "Pf = 20.0 #Initial pressure, bar\n",
- "a = 0.137 #van der Waals constant a, Pa.m3/(mol2)\n",
- "b = 3.87e-5 #van der Waals constant b, m3/(mol)\n",
- "A, B, C, D = 22.5, -1.187e-2,2.3968e-5, -1.0176e-8\n",
- " #Constants in Cvm equation J, K and mol\n",
- "vi = 3.28e-3 #initial volume, m3/mol\n",
- "vf = 7.88e-3 #Final volume, m3/mol\n",
- "\n",
- "#Calculations\n",
- "T = symbols('T')\n",
- "dUT = n**2*a*(1./vi-1./vf)\n",
- "dUV = integrate( A + B*T + C*T**2 + D*T**3, (T,Ti,Tf))\n",
- "\n",
- "#Results\n",
- "print 'dUT = %4.1f J: This is wrongly reported in book'%dUT\n",
- "print 'dUV = %4.1f J'%dUV"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dUT = 24.4 J: This is wrongly reported in book\n",
- "dUV = 4174.1 J\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.7, Page Number 53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "m = 143.0 #Mass of graphite, g \n",
- "Ti = 300.0 #Intial Temperature, K\n",
- "Tf = 600.0 #Intial Temperature, K\n",
- "A, B, C, D, E = -12.19,0.1126,-1.947e-4,1.919e-7,-7.8e-11\n",
- " #Constants in Cvm equation J, K and mol\n",
- "M = 12.01\n",
- "\n",
- "#Calculations\n",
- "\n",
- "T = symbols('T')\n",
- "dH = (m/M)*integrate( A + B*T + C*T**2 + D*T**3 + E*T**4, (T,Ti,Tf))\n",
- "expr = A + B*T + C*T**2 + D*T**3 + E*T**4\n",
- "cpm = expr.subs(T,300.)\n",
- "qp = (m/M)*cpm*(Tf-Ti)\n",
- "err = abs(dH-qp)/dH\n",
- "#Results\n",
- "print 'dH = %6.1f kJ'%(dH/1000)\n",
- "print 'qp = %6.1f kJ'%(qp/1000)\n",
- "print 'Error in calculations %4.1f'%(err*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dH = 46.8 kJ\n",
- "qp = 30.8 kJ\n",
- "Error in calculations 34.3\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.9, Page Number 56"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 124.0 #Mass of liquid methanol, g\n",
- "Pi = 1.0 #Initial Pressure, bar\n",
- "Ti = 298.0 #Intial Temperature, K\n",
- "Pf = 2.5 #Final Pressure, bar\n",
- "Tf = 425.0 #Intial Temperature, K\n",
- "rho = 0.791 #Density, g/cc\n",
- "Cpm = 81.1 #Specifi heat, J/(K.mol)\n",
- "M = 32.04\n",
- "\n",
- "#Calculations\n",
- "n = m/M\n",
- "DH = n*Cpm*(Tf-Ti)+ m*(Pf-Pi)*1e-6/rho\n",
- "\n",
- "#Results\n",
- "print 'Enthalpy change for change in state of methanol is %4.1f kJ'%(DH/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy change for change in state of methanol is 39.9 kJ\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03_2.ipynb deleted file mode 100755 index 0b7fcd00..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter03_2.ipynb +++ /dev/null @@ -1,239 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f533717d02de2958b8edaf7535383653f756ad9b4060abf4f556950fcb3e683e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3: Importance of State Functions: Internal Energy and Enthalpy"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.2, Page Number 45"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "#Variable Declaration\n",
- "betaOH = 11.2e-4 #Thermal exapnasion coefficient of ethanol, \u00b0C\n",
- "betagl = 2.00e-5 #Thermal exapnasion coefficient of glass, \u00b0C\n",
- "kOH = 11.0e-5 #Isothermal compressibility of ethanol, /bar\n",
- "dT = 10.0 #Increase in Temperature, \u00b0C\n",
- "\n",
- "#Calcualtions\n",
- "vfbyvi = (1+ betagl*dT)\n",
- "dP = betaOH*dT/kOH-(1./kOH)*log(vfbyvi)\n",
- "\n",
- "#Results\n",
- "print 'Pressure increase in capillary %4.1f bar'%dP"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure increase in capillary 100.0 bar\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.4, Page Number 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "cpsubysy = 1000 #Specific heat ration of surrounding and system\n",
- "Tpreci = 0.006 #Precision in Temperature measurement, \u00b0C\n",
- "\n",
- "#Calcualtions\n",
- "dtgas = -cpsubysy*(-Tpreci)\n",
- "\n",
- "#Results\n",
- "print 'Minimum detectable temperature change of gas +-%4.1f \u00b0C'%dtgas"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Minimum detectable temperature change of gas +- 6.0 \u00b0C\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.6, Page Number 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import symbols, integrate\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #number of mole of N2, mol \n",
- "Ti = 200.0 #Intial Temperature, K\n",
- "Pi = 5.00 #Initial pressure, bar\n",
- "Tf = 400.0 #Intial Temperature, K\n",
- "Pf = 20.0 #Initial pressure, bar\n",
- "a = 0.137 #van der Waals constant a, Pa.m3/(mol2)\n",
- "b = 3.87e-5 #van der Waals constant b, m3/(mol)\n",
- "A, B, C, D = 22.5, -1.187e-2,2.3968e-5, -1.0176e-8\n",
- " #Constants in Cvm equation J, K and mol\n",
- "vi = 3.28e-3 #initial volume, m3/mol\n",
- "vf = 7.88e-3 #Final volume, m3/mol\n",
- "\n",
- "#Calculations\n",
- "T = symbols('T')\n",
- "dUT = n**2*a*(1./vi-1./vf)\n",
- "dUV = integrate( A + B*T + C*T**2 + D*T**3, (T,Ti,Tf))\n",
- "\n",
- "#Results\n",
- "print 'dUT = %4.1f J: This is wrongly reported in book'%dUT\n",
- "print 'dUV = %4.1f J'%dUV"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dUT = 24.4 J: This is wrongly reported in book\n",
- "dUV = 4174.1 J\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.7, Page Number 53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import symbols, integrate\n",
- "\n",
- "#Variable Declaration\n",
- "m = 143.0 #Mass of graphite, g \n",
- "Ti = 300.0 #Intial Temperature, K\n",
- "Tf = 600.0 #Intial Temperature, K\n",
- "A, B, C, D, E = -12.19,0.1126,-1.947e-4,1.919e-7,-7.8e-11\n",
- " #Constants in Cvm equation J, K and mol\n",
- "M = 12.01\n",
- "\n",
- "#Calculations\n",
- "\n",
- "T = symbols('T')\n",
- "dH = (m/M)*integrate( A + B*T + C*T**2 + D*T**3 + E*T**4, (T,Ti,Tf))\n",
- "expr = A + B*T + C*T**2 + D*T**3 + E*T**4\n",
- "cpm = expr.subs(T,300.)\n",
- "qp = (m/M)*cpm*(Tf-Ti)\n",
- "err = abs(dH-qp)/dH\n",
- "#Results\n",
- "print 'dH = %6.1f kJ'%(dH/1000)\n",
- "print 'qp = %6.1f kJ'%(qp/1000)\n",
- "print 'Error in calculations %4.1f'%(err*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dH = 46.8 kJ\n",
- "qp = 30.8 kJ\n",
- "Error in calculations 34.3\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 3.9, Page Number 56"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 124.0 #Mass of liquid methanol, g\n",
- "Pi = 1.0 #Initial Pressure, bar\n",
- "Ti = 298.0 #Intial Temperature, K\n",
- "Pf = 2.5 #Final Pressure, bar\n",
- "Tf = 425.0 #Intial Temperature, K\n",
- "rho = 0.791 #Density, g/cc\n",
- "Cpm = 81.1 #Specifi heat, J/(K.mol)\n",
- "M = 32.04\n",
- "\n",
- "#Calculations\n",
- "n = m/M\n",
- "DH = n*Cpm*(Tf-Ti)+ m*(Pf-Pi)*1e-6/rho\n",
- "\n",
- "#Results\n",
- "print 'Enthalpy change for change in state of methanol is %4.1f kJ'%(DH/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy change for change in state of methanol is 39.9 kJ\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04.ipynb deleted file mode 100755 index c9094478..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04.ipynb +++ /dev/null @@ -1,204 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c9abd3a4addb064e928970d9b53877d65c2d3bdaa244a23d35c0b5616eab425e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 04: Thermochemistry"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 4.1, Page Number 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varialble Declaration\n",
- "DH0_H2O = 241.8 #Std Enthalpy of reaxtion of Water Fomation backward rxn, kJ/mol\n",
- "DH0_2H = 2*218.0 #Std Enthalpy of formation of Hydrogen atom, kJ/mol\n",
- "DH0_O = 249.2 #Std Enthalpy of formation of Oxygen atom, kJ/mol\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "Dn = 2.0\n",
- "T = 298.15 #Std. Temperature, K\n",
- "#Calculation\n",
- "DH0_2HO = DH0_H2O + DH0_2H + DH0_O\n",
- "DU0 = (DH0_2HO - Dn*R*T*1e-3)/2\n",
- "\n",
- "#Results\n",
- "print 'Avergae Enthalpy change required for breaking both OH bonds %4.1f kJ/mol'%DH0_2HO\n",
- "print 'Average bond energy required for breaking both OH bonds %4.1f kJ/mol'%DU0"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Avergae Enthalpy change required for breaking both OH bonds 927.0 kJ/mol\n",
- "Average bond energy required for breaking both OH bonds 461.0 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.2, Page Number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import numpy as np\n",
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "a = ([29.064, 31.695, 28.165]) #Constant 'a' in Heat capacity equation, J/(mol.K)\n",
- "b = ([-0.8363e-3, 10.143e-3, 1.809e-3]) #Constant 'b' in Heat capacity equation, J/(mol.K)\n",
- "c = ([20.111e-7, -40.373e-7, 15.464e-7]) #Constant 'a' in Heat capacity equation, J/(mol.K)\n",
- "delHf0HCl = -92.3 #Std. Heat of formation of HCl, kJ/mol\n",
- "T1, T2 = 298.15, 1450 #Std and final temperature, K\n",
- "\n",
- "#Calculations\n",
- "T = symbols('T')\n",
- "DA = a[2]-(a[0]+a[1])/2\n",
- "DB = b[2]-(b[0]+b[1])/2\n",
- "DC = c[2]-(c[0]+c[1])/2\n",
- "\n",
- "expr = integrate( DA + DB*T + DC*T**2, (T,T1,T2))\n",
- "DHR1450= expr/1000 + delHf0HCl\n",
- "\n",
- "#Results\n",
- "print 'Heat of reaction for HCl formation is %4.1f kJ/mol'%DHR1450"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat of reaction for HCl formation is -95.1 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.3, Page Number 72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varialble Declaration\n",
- "ms1 = 0.972 #Mass of cyclohexane, g\n",
- "DT1 = 2.98 #Change in temperature for bath, \u00b0C\n",
- "DUR1 = -3913e3 #Std Internal energy change, J/mol\n",
- "mw = 1.812e3 #Mass of water, g\n",
- "ms2 = 0.857 #Mass of benzene, g\n",
- "Ms1 = 84.16\n",
- "Ms2 = 78.12\n",
- "DT2 = 2.36 #Change in temperature for bath, \u00b0C\n",
- "Mw = 18.02\n",
- "Cpw = 75.3 \n",
- "\n",
- "#Calculation\n",
- "\n",
- "Ccal = ((-ms1/Ms1)*DUR1-(mw/Mw)*Cpw*DT1)/DT1\n",
- "DUR2 = (-Ms2/ms2)*((mw/Mw)*Cpw*DT2+Ccal*DT2)\n",
- "\n",
- "#Results\n",
- "print 'Calorimeter constant %4.2e J/\u00b0C'%Ccal\n",
- "print 'Enthalpy of rection for benzene %4.2e J/mol'%DUR2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat capacity of calorimeter 7.59e+03 J/\u00b0C\n",
- "Enthalpy of rection for benzene -3.26e+06 J/mol\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4, Page Number 73"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varialble Declaration\n",
- "ms = 1.423 #Mass of Na2SO4, g\n",
- "mw = 100.34 #Mass of Na2SO4, g\n",
- "DT = 0.037 #Change in temperature for solution, K\n",
- "Mw = 18.02 #Molecular wt of Water\n",
- "Ms = 142.04 #Molecular wt of ms Na2SO4\n",
- "Ccal = 342.5 #Calorimeter constant, J/K\n",
- "#Data\n",
- "DHfNa = -240.1\n",
- "DHfSO4 = -909.3\n",
- "DHfNa2SO4 = -1387.1\n",
- "\n",
- "#Calculation\n",
- "DHs = (-Ms/ms)*((mw/Mw)*Cpw*DT+Ccal*DT)\n",
- "DHsolD = 2*DHfNa + DHfSO4 - DHfNa2SO4\n",
- "\n",
- "#Results\n",
- "print 'Enthalpy of solution for Na2SO4 %4.2e J/mol'%DHs\n",
- "print 'Enthalpy of solution for Na2SO4 from Data %4.2e J/mol'%DHsolD"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy of solution for Na2SO4 -2.81e+03 J/mol\n",
- "Enthalpy of solution for Na2SO4 from Data -2.40e+00 J/mol\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04_1.ipynb deleted file mode 100755 index c9094478..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04_1.ipynb +++ /dev/null @@ -1,204 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c9abd3a4addb064e928970d9b53877d65c2d3bdaa244a23d35c0b5616eab425e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 04: Thermochemistry"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 4.1, Page Number 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varialble Declaration\n",
- "DH0_H2O = 241.8 #Std Enthalpy of reaxtion of Water Fomation backward rxn, kJ/mol\n",
- "DH0_2H = 2*218.0 #Std Enthalpy of formation of Hydrogen atom, kJ/mol\n",
- "DH0_O = 249.2 #Std Enthalpy of formation of Oxygen atom, kJ/mol\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "Dn = 2.0\n",
- "T = 298.15 #Std. Temperature, K\n",
- "#Calculation\n",
- "DH0_2HO = DH0_H2O + DH0_2H + DH0_O\n",
- "DU0 = (DH0_2HO - Dn*R*T*1e-3)/2\n",
- "\n",
- "#Results\n",
- "print 'Avergae Enthalpy change required for breaking both OH bonds %4.1f kJ/mol'%DH0_2HO\n",
- "print 'Average bond energy required for breaking both OH bonds %4.1f kJ/mol'%DU0"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Avergae Enthalpy change required for breaking both OH bonds 927.0 kJ/mol\n",
- "Average bond energy required for breaking both OH bonds 461.0 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.2, Page Number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import numpy as np\n",
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "a = ([29.064, 31.695, 28.165]) #Constant 'a' in Heat capacity equation, J/(mol.K)\n",
- "b = ([-0.8363e-3, 10.143e-3, 1.809e-3]) #Constant 'b' in Heat capacity equation, J/(mol.K)\n",
- "c = ([20.111e-7, -40.373e-7, 15.464e-7]) #Constant 'a' in Heat capacity equation, J/(mol.K)\n",
- "delHf0HCl = -92.3 #Std. Heat of formation of HCl, kJ/mol\n",
- "T1, T2 = 298.15, 1450 #Std and final temperature, K\n",
- "\n",
- "#Calculations\n",
- "T = symbols('T')\n",
- "DA = a[2]-(a[0]+a[1])/2\n",
- "DB = b[2]-(b[0]+b[1])/2\n",
- "DC = c[2]-(c[0]+c[1])/2\n",
- "\n",
- "expr = integrate( DA + DB*T + DC*T**2, (T,T1,T2))\n",
- "DHR1450= expr/1000 + delHf0HCl\n",
- "\n",
- "#Results\n",
- "print 'Heat of reaction for HCl formation is %4.1f kJ/mol'%DHR1450"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat of reaction for HCl formation is -95.1 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.3, Page Number 72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varialble Declaration\n",
- "ms1 = 0.972 #Mass of cyclohexane, g\n",
- "DT1 = 2.98 #Change in temperature for bath, \u00b0C\n",
- "DUR1 = -3913e3 #Std Internal energy change, J/mol\n",
- "mw = 1.812e3 #Mass of water, g\n",
- "ms2 = 0.857 #Mass of benzene, g\n",
- "Ms1 = 84.16\n",
- "Ms2 = 78.12\n",
- "DT2 = 2.36 #Change in temperature for bath, \u00b0C\n",
- "Mw = 18.02\n",
- "Cpw = 75.3 \n",
- "\n",
- "#Calculation\n",
- "\n",
- "Ccal = ((-ms1/Ms1)*DUR1-(mw/Mw)*Cpw*DT1)/DT1\n",
- "DUR2 = (-Ms2/ms2)*((mw/Mw)*Cpw*DT2+Ccal*DT2)\n",
- "\n",
- "#Results\n",
- "print 'Calorimeter constant %4.2e J/\u00b0C'%Ccal\n",
- "print 'Enthalpy of rection for benzene %4.2e J/mol'%DUR2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat capacity of calorimeter 7.59e+03 J/\u00b0C\n",
- "Enthalpy of rection for benzene -3.26e+06 J/mol\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4, Page Number 73"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varialble Declaration\n",
- "ms = 1.423 #Mass of Na2SO4, g\n",
- "mw = 100.34 #Mass of Na2SO4, g\n",
- "DT = 0.037 #Change in temperature for solution, K\n",
- "Mw = 18.02 #Molecular wt of Water\n",
- "Ms = 142.04 #Molecular wt of ms Na2SO4\n",
- "Ccal = 342.5 #Calorimeter constant, J/K\n",
- "#Data\n",
- "DHfNa = -240.1\n",
- "DHfSO4 = -909.3\n",
- "DHfNa2SO4 = -1387.1\n",
- "\n",
- "#Calculation\n",
- "DHs = (-Ms/ms)*((mw/Mw)*Cpw*DT+Ccal*DT)\n",
- "DHsolD = 2*DHfNa + DHfSO4 - DHfNa2SO4\n",
- "\n",
- "#Results\n",
- "print 'Enthalpy of solution for Na2SO4 %4.2e J/mol'%DHs\n",
- "print 'Enthalpy of solution for Na2SO4 from Data %4.2e J/mol'%DHsolD"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy of solution for Na2SO4 -2.81e+03 J/mol\n",
- "Enthalpy of solution for Na2SO4 from Data -2.40e+00 J/mol\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04_2.ipynb deleted file mode 100755 index 8490aa32..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter04_2.ipynb +++ /dev/null @@ -1,206 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:333f059bd9f5c0f0e37eb3d999bc59b67dae36f47a0afa092d85ee79643d8c58"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 04: Thermochemistry"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 4.1, Page Number 68"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varialble Declaration\n",
- "DH0_H2O = 241.8 #Std Enthalpy of reaxtion of Water Fomation backward rxn, kJ/mol\n",
- "DH0_2H = 2*218.0 #Std Enthalpy of formation of Hydrogen atom, kJ/mol\n",
- "DH0_O = 249.2 #Std Enthalpy of formation of Oxygen atom, kJ/mol\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "Dn = 2.0\n",
- "T = 298.15 #Std. Temperature, K\n",
- "#Calculation\n",
- "DH0_2HO = DH0_H2O + DH0_2H + DH0_O\n",
- "DU0 = (DH0_2HO - Dn*R*T*1e-3)/2\n",
- "\n",
- "#Results\n",
- "print 'Avergae Enthalpy change required for breaking both OH bonds %4.1f kJ/mol'%DH0_2HO\n",
- "print 'Average bond energy required for breaking both OH bonds %4.1f kJ/mol'%DU0"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Avergae Enthalpy change required for breaking both OH bonds 927.0 kJ/mol\n",
- "Average bond energy required for breaking both OH bonds 461.0 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.2, Page Number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import numpy as np\n",
- "from sympy import symbols, integrate\n",
- "\n",
- "#Variable Declaration\n",
- "a = ([29.064, 31.695, 28.165]) #Constant 'a' in Heat capacity equation, J/(mol.K)\n",
- "b = ([-0.8363e-3, 10.143e-3, 1.809e-3]) #Constant 'b' in Heat capacity equation, J/(mol.K)\n",
- "c = ([20.111e-7, -40.373e-7, 15.464e-7]) #Constant 'a' in Heat capacity equation, J/(mol.K)\n",
- "delHf0HCl = -92.3 #Std. Heat of formation of HCl, kJ/mol\n",
- "T1, T2 = 298.15, 1450 #Std and final temperature, K\n",
- "\n",
- "#Calculations\n",
- "T = symbols('T')\n",
- "DA = a[2]-(a[0]+a[1])/2\n",
- "DB = b[2]-(b[0]+b[1])/2\n",
- "DC = c[2]-(c[0]+c[1])/2\n",
- "\n",
- "expr = integrate( DA + DB*T + DC*T**2, (T,T1,T2))\n",
- "DHR1450= expr/1000 + delHf0HCl\n",
- "\n",
- "#Results\n",
- "print 'Heat of reaction for HCl formation is %4.1f kJ/mol'%DHR1450"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat of reaction for HCl formation is -95.1 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.3, Page Number 72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Varialble Declaration\n",
- "ms1 = 0.972 #Mass of cyclohexane, g\n",
- "DT1 = 2.98 #Change in temperature for bath, \u00b0C\n",
- "DUR1 = -3913e3 #Std Internal energy change, J/mol\n",
- "mw = 1.812e3 #Mass of water, g\n",
- "ms2 = 0.857 #Mass of benzene, g\n",
- "Ms1 = 84.16\n",
- "Ms2 = 78.12\n",
- "DT2 = 2.36 #Change in temperature for bath, \u00b0C\n",
- "Mw = 18.02\n",
- "Cpw = 75.3 \n",
- "\n",
- "#Calculation\n",
- "\n",
- "Ccal = ((-ms1/Ms1)*DUR1-(mw/Mw)*Cpw*DT1)/DT1\n",
- "DUR2 = (-Ms2/ms2)*((mw/Mw)*Cpw*DT2+Ccal*DT2)\n",
- "\n",
- "#Results\n",
- "print 'Calorimeter constant %4.2e J/\u00b0C'%Ccal\n",
- "print 'Enthalpy of rection for benzene %4.2e J/mol'%DUR2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Calorimeter constant 7.59e+03 J/\u00b0C\n",
- "Enthalpy of rection for benzene -3.26e+06 J/mol\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4.4, Page Number 73"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Varialble Declaration\n",
- "ms = 1.423 #Mass of Na2SO4, g\n",
- "mw = 100.34 #Mass of Na2SO4, g\n",
- "DT = 0.037 #Change in temperature for solution, K\n",
- "Mw = 18.02 #Molecular wt of Water\n",
- "Ms = 142.04 #Molecular wt of ms Na2SO4\n",
- "Ccal = 342.5 #Calorimeter constant, J/K\n",
- "#Data\n",
- "DHfNa = -240.1\n",
- "DHfSO4 = -909.3\n",
- "DHfNa2SO4 = -1387.1\n",
- "\n",
- "#Calculation\n",
- "DHs = (-Ms/ms)*((mw/Mw)*Cpw*DT+Ccal*DT)\n",
- "DHsolD = 2*DHfNa + DHfSO4 - DHfNa2SO4\n",
- "\n",
- "#Results\n",
- "print 'Enthalpy of solution for Na2SO4 %4.2e J/mol'%DHs\n",
- "print 'Enthalpy of solution for Na2SO4 from Data %4.2e J/mol'%DHsolD"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy of solution for Na2SO4 -2.81e+03 J/mol\n",
- "Enthalpy of solution for Na2SO4 from Data -2.40e+00 J/mol\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05.ipynb deleted file mode 100755 index ce8818c7..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05.ipynb +++ /dev/null @@ -1,425 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:2c9e6fedd22fa6b2ae65a19697803d5aa951ae2982e5a733c3ba66fccdc2f728"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5: Enthalpy and the Second and Third Laws of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.1, Page Number 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "Th, Tc = 500.,200. #Temeperatures IN Which reversible heat engine works, K\n",
- "q = 1000. #Heat absorbed by heat engine, J\n",
- "\n",
- "#Calcualtions\n",
- "eps = 1.-Tc/Th\n",
- "w = eps*q\n",
- "\n",
- "#Results\n",
- "print 'Efficiency of heat engine is %4.3f'%eps\n",
- "print 'Work done by heat engine is %4.1f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Efficiency of heat engine is 0.600\n",
- "Work done by heat engine is 600.0 J\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.4, Page Number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "Ti, Tf = 320.,650. #Initial and final state Temeperatures of CO2, K\n",
- "vi, vf = 80.,120. #Initial and final state volume of CO2, K\n",
- "A, B, C, D = 31.08,-0.01452,3.1415e-5,-1.4973e-8\n",
- " #Constants in constant volume Heat capacity equation in J, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (A + B*T + C*T**2 + D*T**3)/T, (T,Ti,Tf)) \n",
- "dS2 = n*R*log(vf/vi)\n",
- "dS = dS1 + dS2\n",
- "#Results\n",
- "print 'Entropy change of process is %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 24.43 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.5, Page Number 88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 2.5 #Number of moles of CO2\n",
- "Ti, Tf = 450.,800. #Initial and final state Temeperatures of CO2, K\n",
- "pi, pf = 1.35,3.45 #Initial and final state pressure of CO2, K\n",
- "A, B, C, D = 18.86,7.937e-2,-6.7834e-5,2.4426e-8\n",
- " #Constants in constant pressure Heat capacity equation in J, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (A + B*T + C*T**2 + D*T**3)/T, (T,Ti,Tf)) \n",
- "dS2 = n*R*log(pf/pi)\n",
- "dS = dS1 - dS2\n",
- "#Results\n",
- "print 'Entropy change of process is %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 48.55 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.6, Page Number 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 3.0 #Number of moles of CO2\n",
- "Ti, Tf = 300.,600. #Initial and final state Temeperatures of CO2, K\n",
- "pi, pf = 1.00,3.00 #Initial and final state pressure of CO2, K\n",
- "cpm = 27.98 #Specific heat of mercury, J/(mol.K)\n",
- "M = 200.59 #Molecualr wt of mercury, g/(mol)\n",
- "beta = 1.81e-4 #per K\n",
- "rho = 13.54 #Density of mercury, g/cm3\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "dS1 = n*cpm*log(Tf/Ti)\n",
- "dS2 = n*(M/(rho*1e6))*beta*(pf-pi)*1e5\n",
- "dS = dS1 - dS2\n",
- "\n",
- "#Results\n",
- "print 'Entropy change of process is %4.1f J/(mol.K)'%dS\n",
- "print 'Ratio of pressure to temperature dependent term %3.1e\\nhence effect of pressure dependent term isvery less'%(dS2/dS1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 58.2 J/(mol.K)\n",
- "Ratio of pressure to temperature dependent term 2.8e-05\n",
- "hence effect of pressure dependent term isvery less\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.7, Page Number 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "T = 300.0 #Temeperatures of Water bath, K\n",
- "vi, vf = 25.0,10.0 #Initial and final state Volume of Ideal Gas, L\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "qrev = n*R*T*log(vf/vi)\n",
- "w = -qrev\n",
- "dSsys = qrev/T\n",
- "dSsur = -dSsys\n",
- "dS = dSsys + dSsur\n",
- "\n",
- "#Results\n",
- "print 'Entropy change of surrounding is %4.1f J/(mol.K)'%dSsur\n",
- "print 'Entropy change of system is %4.1f J/(mol.K)'%dSsys\n",
- "print 'Total Entropy changeis %4.1f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of surrounding is 7.6 J/(mol.K)\n",
- "Entropy change of system is -7.6 J/(mol.K)\n",
- "Total Entropy changeis 0.0 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.8, Page Number 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "T = 300.0 #Temeperatures of Water bath, K\n",
- "vi, vf = 25.0,10.0 #Initial and final state Volume of Ideal Gas, L\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "pext = n*R*T/(vf/1e3)\n",
- "pi = n*R*T/(vi/1e3)\n",
- "q = pext*(vf-vi)/1e3\n",
- "qrev = n*R*T*log(vf/vi)\n",
- "w = -q\n",
- "dSsur = -q/T\n",
- "dSsys = qrev/T\n",
- "dS = dSsys + dSsur\n",
- "\n",
- "#Results\n",
- "print 'Constant external pressure and initial pressure are %4.3e J,and %4.3e J respectively'%(pext,pi)\n",
- "print 'Heat in reverssible and irreversible processes are %4.1f J,and %4.1f J respectively'%(qrev,q)\n",
- "print 'Entropy change of system is %4.1f J/(mol.K)'%dSsys\n",
- "print 'Entropy change of surrounding is %4.2f J/(mol.K)'%dSsur\n",
- "print 'Total Entropy changeis %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Constant external pressure and initial pressure are 2.494e+05 J,and 9.977e+04 J respectively\n",
- "Heat in reverssible and irreversible processes are -2285.4 J,and -3741.3 J respectively\n",
- "Entropy change of system is -7.6 J/(mol.K)\n",
- "Entropy change of surrounding is 12.47 J/(mol.K)\n",
- "Total Entropy changeis 4.85 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.9, Page Number 96"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "pi, pf = 1.35,3.45 #Initial and final state pressure of CO2, K\n",
- "D1 = 2.11e-3 #Constants in constant pressure Heat capacity equation for K<T<12.97K, in J, mol, K units\n",
- "A2, B2, C2, D2 = -5.666,0.6927,-5.191e-3,9.943e-4\n",
- " #Constants in constant pressure Heat capacity equation for 12.97<T<23.66, J, mol, K units\n",
- "A3, B3, C3, D3 = 31.70,-2.038,0.08384,-6.685e-4\n",
- " #Constants in constant pressure Heat capacity equation for 23.66<T<43.76, J, mol, K units\n",
- "A4 = 46.094 #Constants in constant pressure Heat capacity equation for 43.76<T<54.39, J/(mol.K)\n",
- "A5, B5, C5, D5 = 81.268,-1.1467,0.01516,-6.407e-5\n",
- " #Constants in constant pressure Heat capacity equation for 54.39<T<90.20K, J, mol, K units\n",
- "A6, B6, C6, D6 = 32.71,-0.04093,1.545e-4,-1.819e-7\n",
- " #Constants in constant pressure Heat capacity equation for 90.20<T<298.15 KJ, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "Ltrans1 = 93.80 #Entalpy of transition at 23.66K, J/mol\n",
- "Ltrans2 = 743.0 #Entalpy of transition at 43.76K, J/mol\n",
- "Ltrans3 = 445.0 #Entalpy of transition at 54.39K, J/mol\n",
- "Ltrans4 = 6815. #Entalpy of transition at 90.20K, J/mol\n",
- "T1 = 12.97 #Maximum applicabliltiy temeprature for first heat capacity equation, K\n",
- "T12 = 23.66 #Phase Change temperature from Solid III--II, K\n",
- "T23 = 43.76 #Phase Change temperature from Solid II--I, K\n",
- "T34 = 54.39 #Phase Change temperature from Solid I--liquid, K\n",
- "T45 = 90.20 #Phase Change temperature from liquid--gas, K\n",
- "Ts = 298.15 #Std. Temeprature, K\n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (D1*T**3)/T, (T,0,T1)) \n",
- "dS2 = n*integrate( (A2 + B2*T + C2*T**2 + D2*T**3)/T, (T,T1,T12)) \n",
- "dS21 = Ltrans1/T12\n",
- "dS3 = n*integrate( (A3 + B3*T + C3*T**2 + D3*T**3)/T, (T,T12,T23)) \n",
- "dS31 = Ltrans2/T23\n",
- "dS4 = n*integrate( (A4)/T, (T,T23,T34)) \n",
- "dS41 = Ltrans3/T34\n",
- "dS5 = n*integrate( (A5 + B5*T + C5*T**2 + D5*T**3)/T, (T,T34,T45)) \n",
- "dS51 = Ltrans4/T45\n",
- "dS6 = n*integrate( (A6 + B6*T + C6*T**2 + D6*T**3)/T, (T,T45,Ts))\n",
- "#print dS1+dS2,dS21\n",
- "#print dS3, dS31\n",
- "#print dS4, dS41\n",
- "#print dS5, dS51\n",
- "#print dS6\n",
- "dS = dS1+dS2+dS21+dS3+dS31+dS4+dS41+dS5+dS51+dS6\n",
- "\n",
- "#Results\n",
- "print 'Entropy change Sm0 for O2 is %4.1f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change Sm0 for O2 is 204.8 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.10, Page Number 99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2 formed, mol\n",
- "p = 1. #Pressure of CO2, K\n",
- "\n",
- "A1, B1, C1, D1 = 18.86,7.937e-2,-6.7834e-5,2.4426e-8\n",
- " #Constants in constant pressure Heat capacity equation for CO2, J/(mol.K)\n",
- "A2, B2, C2, D2 = 30.81,-1.187e-2,2.3968e-5, 0.0\n",
- " #Constants in constant pressure Heat capacity equation for O2, J/(mol.K)\n",
- "A3, B3, C3, D3 = 31.08,-1.452e-2,3.1415e-5 ,-1.4793e-8 \n",
- " #Constants in constant pressure Heat capacity equation for CO, J/(mol.K)\n",
- "DSr298CO = 197.67 #Std. Entropy change for CO, J/(mol.K)\n",
- "DSr298CO2 = 213.74 #Std. Entropy change for CO, J/(mol.K)\n",
- "DSr298O2 = 205.138 #Std. Entropy change for CO, J/(mol.K)\n",
- "Tr = 475. #Reaction temperature, K\n",
- "Ts = 298.15 #Std. temperature, K\n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "v1,v2,v3 = 1.,1./2,1.\n",
- "DSr = DSr298CO2*v1 - DSr298CO*v1 - DSr298O2*v2\n",
- "DA = v1*A1-v2*A2-v3*A3\n",
- "DB = v1*B1-v2*B2-v3*B3\n",
- "DC = v1*C1-v2*C2-v3*C3\n",
- "DD = v1*D1-v2*D2-v3*D3\n",
- "dS = DSr + n*integrate( (DA + DB*T + DC*T**2 + DD*T**3)/T, (T,Ts,Tr)) \n",
- "\n",
- "#Results\n",
- "print 'Entropy change for reaction at %4d K is %4.2f J/(mol.K)'%(Tr,dS)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change Sm0 for O2 is 204.83 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05_1.ipynb deleted file mode 100755 index ce8818c7..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05_1.ipynb +++ /dev/null @@ -1,425 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:2c9e6fedd22fa6b2ae65a19697803d5aa951ae2982e5a733c3ba66fccdc2f728"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5: Enthalpy and the Second and Third Laws of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.1, Page Number 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "Th, Tc = 500.,200. #Temeperatures IN Which reversible heat engine works, K\n",
- "q = 1000. #Heat absorbed by heat engine, J\n",
- "\n",
- "#Calcualtions\n",
- "eps = 1.-Tc/Th\n",
- "w = eps*q\n",
- "\n",
- "#Results\n",
- "print 'Efficiency of heat engine is %4.3f'%eps\n",
- "print 'Work done by heat engine is %4.1f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Efficiency of heat engine is 0.600\n",
- "Work done by heat engine is 600.0 J\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.4, Page Number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "Ti, Tf = 320.,650. #Initial and final state Temeperatures of CO2, K\n",
- "vi, vf = 80.,120. #Initial and final state volume of CO2, K\n",
- "A, B, C, D = 31.08,-0.01452,3.1415e-5,-1.4973e-8\n",
- " #Constants in constant volume Heat capacity equation in J, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (A + B*T + C*T**2 + D*T**3)/T, (T,Ti,Tf)) \n",
- "dS2 = n*R*log(vf/vi)\n",
- "dS = dS1 + dS2\n",
- "#Results\n",
- "print 'Entropy change of process is %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 24.43 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.5, Page Number 88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 2.5 #Number of moles of CO2\n",
- "Ti, Tf = 450.,800. #Initial and final state Temeperatures of CO2, K\n",
- "pi, pf = 1.35,3.45 #Initial and final state pressure of CO2, K\n",
- "A, B, C, D = 18.86,7.937e-2,-6.7834e-5,2.4426e-8\n",
- " #Constants in constant pressure Heat capacity equation in J, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (A + B*T + C*T**2 + D*T**3)/T, (T,Ti,Tf)) \n",
- "dS2 = n*R*log(pf/pi)\n",
- "dS = dS1 - dS2\n",
- "#Results\n",
- "print 'Entropy change of process is %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 48.55 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.6, Page Number 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 3.0 #Number of moles of CO2\n",
- "Ti, Tf = 300.,600. #Initial and final state Temeperatures of CO2, K\n",
- "pi, pf = 1.00,3.00 #Initial and final state pressure of CO2, K\n",
- "cpm = 27.98 #Specific heat of mercury, J/(mol.K)\n",
- "M = 200.59 #Molecualr wt of mercury, g/(mol)\n",
- "beta = 1.81e-4 #per K\n",
- "rho = 13.54 #Density of mercury, g/cm3\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "dS1 = n*cpm*log(Tf/Ti)\n",
- "dS2 = n*(M/(rho*1e6))*beta*(pf-pi)*1e5\n",
- "dS = dS1 - dS2\n",
- "\n",
- "#Results\n",
- "print 'Entropy change of process is %4.1f J/(mol.K)'%dS\n",
- "print 'Ratio of pressure to temperature dependent term %3.1e\\nhence effect of pressure dependent term isvery less'%(dS2/dS1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 58.2 J/(mol.K)\n",
- "Ratio of pressure to temperature dependent term 2.8e-05\n",
- "hence effect of pressure dependent term isvery less\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.7, Page Number 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "T = 300.0 #Temeperatures of Water bath, K\n",
- "vi, vf = 25.0,10.0 #Initial and final state Volume of Ideal Gas, L\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "qrev = n*R*T*log(vf/vi)\n",
- "w = -qrev\n",
- "dSsys = qrev/T\n",
- "dSsur = -dSsys\n",
- "dS = dSsys + dSsur\n",
- "\n",
- "#Results\n",
- "print 'Entropy change of surrounding is %4.1f J/(mol.K)'%dSsur\n",
- "print 'Entropy change of system is %4.1f J/(mol.K)'%dSsys\n",
- "print 'Total Entropy changeis %4.1f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of surrounding is 7.6 J/(mol.K)\n",
- "Entropy change of system is -7.6 J/(mol.K)\n",
- "Total Entropy changeis 0.0 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.8, Page Number 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "T = 300.0 #Temeperatures of Water bath, K\n",
- "vi, vf = 25.0,10.0 #Initial and final state Volume of Ideal Gas, L\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "pext = n*R*T/(vf/1e3)\n",
- "pi = n*R*T/(vi/1e3)\n",
- "q = pext*(vf-vi)/1e3\n",
- "qrev = n*R*T*log(vf/vi)\n",
- "w = -q\n",
- "dSsur = -q/T\n",
- "dSsys = qrev/T\n",
- "dS = dSsys + dSsur\n",
- "\n",
- "#Results\n",
- "print 'Constant external pressure and initial pressure are %4.3e J,and %4.3e J respectively'%(pext,pi)\n",
- "print 'Heat in reverssible and irreversible processes are %4.1f J,and %4.1f J respectively'%(qrev,q)\n",
- "print 'Entropy change of system is %4.1f J/(mol.K)'%dSsys\n",
- "print 'Entropy change of surrounding is %4.2f J/(mol.K)'%dSsur\n",
- "print 'Total Entropy changeis %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Constant external pressure and initial pressure are 2.494e+05 J,and 9.977e+04 J respectively\n",
- "Heat in reverssible and irreversible processes are -2285.4 J,and -3741.3 J respectively\n",
- "Entropy change of system is -7.6 J/(mol.K)\n",
- "Entropy change of surrounding is 12.47 J/(mol.K)\n",
- "Total Entropy changeis 4.85 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.9, Page Number 96"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "pi, pf = 1.35,3.45 #Initial and final state pressure of CO2, K\n",
- "D1 = 2.11e-3 #Constants in constant pressure Heat capacity equation for K<T<12.97K, in J, mol, K units\n",
- "A2, B2, C2, D2 = -5.666,0.6927,-5.191e-3,9.943e-4\n",
- " #Constants in constant pressure Heat capacity equation for 12.97<T<23.66, J, mol, K units\n",
- "A3, B3, C3, D3 = 31.70,-2.038,0.08384,-6.685e-4\n",
- " #Constants in constant pressure Heat capacity equation for 23.66<T<43.76, J, mol, K units\n",
- "A4 = 46.094 #Constants in constant pressure Heat capacity equation for 43.76<T<54.39, J/(mol.K)\n",
- "A5, B5, C5, D5 = 81.268,-1.1467,0.01516,-6.407e-5\n",
- " #Constants in constant pressure Heat capacity equation for 54.39<T<90.20K, J, mol, K units\n",
- "A6, B6, C6, D6 = 32.71,-0.04093,1.545e-4,-1.819e-7\n",
- " #Constants in constant pressure Heat capacity equation for 90.20<T<298.15 KJ, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "Ltrans1 = 93.80 #Entalpy of transition at 23.66K, J/mol\n",
- "Ltrans2 = 743.0 #Entalpy of transition at 43.76K, J/mol\n",
- "Ltrans3 = 445.0 #Entalpy of transition at 54.39K, J/mol\n",
- "Ltrans4 = 6815. #Entalpy of transition at 90.20K, J/mol\n",
- "T1 = 12.97 #Maximum applicabliltiy temeprature for first heat capacity equation, K\n",
- "T12 = 23.66 #Phase Change temperature from Solid III--II, K\n",
- "T23 = 43.76 #Phase Change temperature from Solid II--I, K\n",
- "T34 = 54.39 #Phase Change temperature from Solid I--liquid, K\n",
- "T45 = 90.20 #Phase Change temperature from liquid--gas, K\n",
- "Ts = 298.15 #Std. Temeprature, K\n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (D1*T**3)/T, (T,0,T1)) \n",
- "dS2 = n*integrate( (A2 + B2*T + C2*T**2 + D2*T**3)/T, (T,T1,T12)) \n",
- "dS21 = Ltrans1/T12\n",
- "dS3 = n*integrate( (A3 + B3*T + C3*T**2 + D3*T**3)/T, (T,T12,T23)) \n",
- "dS31 = Ltrans2/T23\n",
- "dS4 = n*integrate( (A4)/T, (T,T23,T34)) \n",
- "dS41 = Ltrans3/T34\n",
- "dS5 = n*integrate( (A5 + B5*T + C5*T**2 + D5*T**3)/T, (T,T34,T45)) \n",
- "dS51 = Ltrans4/T45\n",
- "dS6 = n*integrate( (A6 + B6*T + C6*T**2 + D6*T**3)/T, (T,T45,Ts))\n",
- "#print dS1+dS2,dS21\n",
- "#print dS3, dS31\n",
- "#print dS4, dS41\n",
- "#print dS5, dS51\n",
- "#print dS6\n",
- "dS = dS1+dS2+dS21+dS3+dS31+dS4+dS41+dS5+dS51+dS6\n",
- "\n",
- "#Results\n",
- "print 'Entropy change Sm0 for O2 is %4.1f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change Sm0 for O2 is 204.8 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.10, Page Number 99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2 formed, mol\n",
- "p = 1. #Pressure of CO2, K\n",
- "\n",
- "A1, B1, C1, D1 = 18.86,7.937e-2,-6.7834e-5,2.4426e-8\n",
- " #Constants in constant pressure Heat capacity equation for CO2, J/(mol.K)\n",
- "A2, B2, C2, D2 = 30.81,-1.187e-2,2.3968e-5, 0.0\n",
- " #Constants in constant pressure Heat capacity equation for O2, J/(mol.K)\n",
- "A3, B3, C3, D3 = 31.08,-1.452e-2,3.1415e-5 ,-1.4793e-8 \n",
- " #Constants in constant pressure Heat capacity equation for CO, J/(mol.K)\n",
- "DSr298CO = 197.67 #Std. Entropy change for CO, J/(mol.K)\n",
- "DSr298CO2 = 213.74 #Std. Entropy change for CO, J/(mol.K)\n",
- "DSr298O2 = 205.138 #Std. Entropy change for CO, J/(mol.K)\n",
- "Tr = 475. #Reaction temperature, K\n",
- "Ts = 298.15 #Std. temperature, K\n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "v1,v2,v3 = 1.,1./2,1.\n",
- "DSr = DSr298CO2*v1 - DSr298CO*v1 - DSr298O2*v2\n",
- "DA = v1*A1-v2*A2-v3*A3\n",
- "DB = v1*B1-v2*B2-v3*B3\n",
- "DC = v1*C1-v2*C2-v3*C3\n",
- "DD = v1*D1-v2*D2-v3*D3\n",
- "dS = DSr + n*integrate( (DA + DB*T + DC*T**2 + DD*T**3)/T, (T,Ts,Tr)) \n",
- "\n",
- "#Results\n",
- "print 'Entropy change for reaction at %4d K is %4.2f J/(mol.K)'%(Tr,dS)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change Sm0 for O2 is 204.83 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05_2.ipynb deleted file mode 100755 index 21cc530e..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter05_2.ipynb +++ /dev/null @@ -1,426 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:fb3a9fa8ccc4e6b6189db2b7b007068058fcf43a7754989741838789e7fd6c7a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5: Enthalpy and the Second and Third Laws of Thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.1, Page Number 84"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "Th, Tc = 500.,200. #Temeperatures IN Which reversible heat engine works, K\n",
- "q = 1000. #Heat absorbed by heat engine, J\n",
- "\n",
- "#Calcualtions\n",
- "eps = 1.-Tc/Th\n",
- "w = eps*q\n",
- "\n",
- "#Results\n",
- "print 'Efficiency of heat engine is %4.3f'%eps\n",
- "print 'Work done by heat engine is %4.1f J'%w"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Efficiency of heat engine is 0.600\n",
- "Work done by heat engine is 600.0 J\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.4, Page Number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "Ti, Tf = 320.,650. #Initial and final state Temeperatures of CO2, K\n",
- "vi, vf = 80.,120. #Initial and final state volume of CO2, K\n",
- "A, B, C, D = 31.08,-0.01452,3.1415e-5,-1.4973e-8\n",
- " #Constants in constant volume Heat capacity equation in J, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (A + B*T + C*T**2 + D*T**3)/T, (T,Ti,Tf)) \n",
- "dS2 = n*R*log(vf/vi)\n",
- "dS = dS1 + dS2\n",
- "#Results\n",
- "print 'Entropy change of process is %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 24.43 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.5, Page Number 88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 2.5 #Number of moles of CO2\n",
- "Ti, Tf = 450.,800. #Initial and final state Temeperatures of CO2, K\n",
- "pi, pf = 1.35,3.45 #Initial and final state pressure of CO2, K\n",
- "A, B, C, D = 18.86,7.937e-2,-6.7834e-5,2.4426e-8\n",
- " #Constants in constant pressure Heat capacity equation in J, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (A + B*T + C*T**2 + D*T**3)/T, (T,Ti,Tf)) \n",
- "dS2 = n*R*log(pf/pi)\n",
- "dS = dS1 - dS2\n",
- "#Results\n",
- "print 'Entropy change of process is %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 48.55 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.6, Page Number 89"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 3.0 #Number of moles of CO2\n",
- "Ti, Tf = 300.,600. #Initial and final state Temeperatures of CO2, K\n",
- "pi, pf = 1.00,3.00 #Initial and final state pressure of CO2, K\n",
- "cpm = 27.98 #Specific heat of mercury, J/(mol.K)\n",
- "M = 200.59 #Molecualr wt of mercury, g/(mol)\n",
- "beta = 1.81e-4 #per K\n",
- "rho = 13.54 #Density of mercury, g/cm3\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "dS1 = n*cpm*log(Tf/Ti)\n",
- "dS2 = n*(M/(rho*1e6))*beta*(pf-pi)*1e5\n",
- "dS = dS1 - dS2\n",
- "\n",
- "#Results\n",
- "print 'Entropy change of process is %4.1f J/(mol.K)'%dS\n",
- "print 'Ratio of pressure to temperature dependent term %3.1e\\nhence effect of pressure dependent term isvery less'%(dS2/dS1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of process is 58.2 J/(mol.K)\n",
- "Ratio of pressure to temperature dependent term 2.8e-05\n",
- "hence effect of pressure dependent term isvery less\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.7, Page Number 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "T = 300.0 #Temeperatures of Water bath, K\n",
- "vi, vf = 25.0,10.0 #Initial and final state Volume of Ideal Gas, L\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "qrev = n*R*T*log(vf/vi)\n",
- "w = -qrev\n",
- "dSsys = qrev/T\n",
- "dSsur = -dSsys\n",
- "dS = dSsys + dSsur\n",
- "\n",
- "#Results\n",
- "print 'Entropy change of surrounding is %4.1f J/(mol.K)'%dSsur\n",
- "print 'Entropy change of system is %4.1f J/(mol.K)'%dSsys\n",
- "print 'Total Entropy changeis %4.1f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change of surrounding is 7.6 J/(mol.K)\n",
- "Entropy change of system is -7.6 J/(mol.K)\n",
- "Total Entropy changeis 0.0 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.8, Page Number 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "T = 300.0 #Temeperatures of Water bath, K\n",
- "vi, vf = 25.0,10.0 #Initial and final state Volume of Ideal Gas, L\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "\n",
- "#Calcualtions\n",
- "pext = n*R*T/(vf/1e3)\n",
- "pi = n*R*T/(vi/1e3)\n",
- "q = pext*(vf-vi)/1e3\n",
- "qrev = n*R*T*log(vf/vi)\n",
- "w = -q\n",
- "dSsur = -q/T\n",
- "dSsys = qrev/T\n",
- "dS = dSsys + dSsur\n",
- "\n",
- "#Results\n",
- "print 'Constant external pressure and initial pressure are %4.3e J,and %4.3e J respectively'%(pext,pi)\n",
- "print 'Heat in reverssible and irreversible processes are %4.1f J,and %4.1f J respectively'%(qrev,q)\n",
- "print 'Entropy change of system is %4.1f J/(mol.K)'%dSsys\n",
- "print 'Entropy change of surrounding is %4.2f J/(mol.K)'%dSsur\n",
- "print 'Total Entropy changeis %4.2f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Constant external pressure and initial pressure are 2.494e+05 J,and 9.977e+04 J respectively\n",
- "Heat in reverssible and irreversible processes are -2285.4 J,and -3741.3 J respectively\n",
- "Entropy change of system is -7.6 J/(mol.K)\n",
- "Entropy change of surrounding is 12.47 J/(mol.K)\n",
- "Total Entropy changeis 4.85 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.9, Page Number 96"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2\n",
- "pi, pf = 1.35,3.45 #Initial and final state pressure of CO2, K\n",
- "D1 = 2.11e-3 #Constants in constant pressure Heat capacity equation for K<T<12.97K, in J, mol, K units\n",
- "A2, B2, C2, D2 = -5.666,0.6927,-5.191e-3,9.943e-4\n",
- " #Constants in constant pressure Heat capacity equation for 12.97<T<23.66, J, mol, K units\n",
- "A3, B3, C3, D3 = 31.70,-2.038,0.08384,-6.685e-4\n",
- " #Constants in constant pressure Heat capacity equation for 23.66<T<43.76, J, mol, K units\n",
- "A4 = 46.094 #Constants in constant pressure Heat capacity equation for 43.76<T<54.39, J/(mol.K)\n",
- "A5, B5, C5, D5 = 81.268,-1.1467,0.01516,-6.407e-5\n",
- " #Constants in constant pressure Heat capacity equation for 54.39<T<90.20K, J, mol, K units\n",
- "A6, B6, C6, D6 = 32.71,-0.04093,1.545e-4,-1.819e-7\n",
- " #Constants in constant pressure Heat capacity equation for 90.20<T<298.15 KJ, mol, K units\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K) \n",
- "Ltrans1 = 93.80 #Entalpy of transition at 23.66K, J/mol\n",
- "Ltrans2 = 743.0 #Entalpy of transition at 43.76K, J/mol\n",
- "Ltrans3 = 445.0 #Entalpy of transition at 54.39K, J/mol\n",
- "Ltrans4 = 6815. #Entalpy of transition at 90.20K, J/mol\n",
- "T1 = 12.97 #Maximum applicabliltiy temeprature for first heat capacity equation, K\n",
- "T12 = 23.66 #Phase Change temperature from Solid III--II, K\n",
- "T23 = 43.76 #Phase Change temperature from Solid II--I, K\n",
- "T34 = 54.39 #Phase Change temperature from Solid I--liquid, K\n",
- "T45 = 90.20 #Phase Change temperature from liquid--gas, K\n",
- "Ts = 298.15 #Std. Temeprature, K\n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "dS1 = n*integrate( (D1*T**3)/T, (T,0,T1)) \n",
- "dS2 = n*integrate( (A2 + B2*T + C2*T**2 + D2*T**3)/T, (T,T1,T12)) \n",
- "dS21 = Ltrans1/T12\n",
- "dS3 = n*integrate( (A3 + B3*T + C3*T**2 + D3*T**3)/T, (T,T12,T23)) \n",
- "dS31 = Ltrans2/T23\n",
- "dS4 = n*integrate( (A4)/T, (T,T23,T34)) \n",
- "dS41 = Ltrans3/T34\n",
- "dS5 = n*integrate( (A5 + B5*T + C5*T**2 + D5*T**3)/T, (T,T34,T45)) \n",
- "dS51 = Ltrans4/T45\n",
- "dS6 = n*integrate( (A6 + B6*T + C6*T**2 + D6*T**3)/T, (T,T45,Ts))\n",
- "#print dS1+dS2,dS21\n",
- "#print dS3, dS31\n",
- "#print dS4, dS41\n",
- "#print dS5, dS51\n",
- "#print dS6\n",
- "dS = dS1+dS2+dS21+dS3+dS31+dS4+dS41+dS5+dS51+dS6\n",
- "\n",
- "#Results\n",
- "print 'Entropy change Sm0 for O2 is %4.1f J/(mol.K)'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change Sm0 for O2 is 204.8 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 5.10, Page Number 99"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import integrate, symbols\n",
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "n = 1.0 #Number of moles of CO2 formed, mol\n",
- "p = 1. #Pressure of CO2, K\n",
- "\n",
- "A1, B1, C1, D1 = 18.86,7.937e-2,-6.7834e-5,2.4426e-8\n",
- " #Constants in constant pressure Heat capacity equation for CO2, J/(mol.K)\n",
- "A2, B2, C2, D2 = 30.81,-1.187e-2,2.3968e-5, 0.0\n",
- " #Constants in constant pressure Heat capacity equation for O2, J/(mol.K)\n",
- "A3, B3, C3, D3 = 31.08,-1.452e-2,3.1415e-5 ,-1.4793e-8 \n",
- " #Constants in constant pressure Heat capacity equation for CO, J/(mol.K)\n",
- "DSr298CO = 197.67 #Std. Entropy change for CO, J/(mol.K)\n",
- "DSr298CO2 = 213.74 #Std. Entropy change for CO, J/(mol.K)\n",
- "DSr298O2 = 205.138 #Std. Entropy change for CO, J/(mol.K)\n",
- "Tr = 475. #Reaction temperature, K\n",
- "Ts = 298.15 #Std. temperature, K\n",
- "#Calcualtions\n",
- "T = symbols('T')\n",
- "v1,v2,v3 = 1.,1./2,1.\n",
- "DSr = DSr298CO2*v1 - DSr298CO*v1 - DSr298O2*v2\n",
- "DA = v1*A1-v2*A2-v3*A3\n",
- "DB = v1*B1-v2*B2-v3*B3\n",
- "DC = v1*C1-v2*C2-v3*C3\n",
- "DD = v1*D1-v2*D2-v3*D3\n",
- "dS = DSr + n*integrate( (DA + DB*T + DC*T**2 + DD*T**3)/T, (T,Ts,Tr)) \n",
- "\n",
- "#Results\n",
- "print 'Entropy change for reaction at %4d K is %4.2f J/(mol.K)'%(Tr,dS)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change for reaction at 475 K is -88.26 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06.ipynb deleted file mode 100755 index c6258811..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06.ipynb +++ /dev/null @@ -1,709 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:399beb745330e541e567baa0f45fde7ecc89dabcc65d8db71d8e5921a036072a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 06: Chemical Equilibrium"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.1, Page Number 117"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dHcCH4 = -891.0 #Std. heat of combustion for CH4, kJ/mol\n",
- "dHcC8H18 = -5471.0 #Std. heat of combustion for C8H18, kJ/mol\n",
- "\n",
- "T = 298.15\n",
- "SmCO2, SmCH4, SmH2O, SmO2, SmC8H18 = 213.8,186.3,70.0,205.2, 316.1\n",
- "dnCH4 = -2.\n",
- "dnC8H18 = 4.5\n",
- "R = 8.314\n",
- "#Calculations\n",
- "dACH4 = dHcCH4*1e3 - dnCH4*R*T - T*(SmCO2 + 2*SmH2O - SmCH4 - 2*SmO2)\n",
- "dAC8H18 = dHcC8H18*1e3 - dnC8H18*R*T - T*(8*SmCO2 + 9*SmH2O - SmC8H18 - 25.*SmO2/2) \n",
- "#Results \n",
- "print 'Maximum Available work through combustion of CH4 %4.1f kJ/mol'%(dACH4/1000)\n",
- "print 'Maximum Available work through combustion of C8H18 %4.1f kJ/mol'%(dAC8H18/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Available work through combustion of CH4 -813.6 kJ/mol\n",
- "Maximum Available work through combustion of C8H18 -5320.9 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.2, Page Number 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dHcCH4 = -891.0 #Std. heat of combustion for CH4, kJ/mol\n",
- "dHcC8H18 = -5471.0 #Std. heat of combustion for C8H18, kJ/mol\n",
- "\n",
- "T = 298.15\n",
- "SmCO2, SmCH4, SmH2O, SmO2, SmC8H18 = 213.8,186.3,70.0,205.2, 316.1\n",
- "dnCH4 = -2.\n",
- "dnC8H18 = 4.5\n",
- "R = 8.314\n",
- "#Calculations\n",
- "dGCH4 = dHcCH4*1e3 - T*(SmCO2 + 2*SmH2O - SmCH4 - 2*SmO2)\n",
- "dGC8H18 = dHcC8H18*1e3 - T*(8*SmCO2 + 9*SmH2O - SmC8H18 - 25.*SmO2/2) \n",
- "#Results \n",
- "print 'Maximum nonexapnasion work through combustion of CH4 %4.1f kJ/mol'%(dGCH4/1000)\n",
- "print 'Maximum nonexapnasion work through combustion of C8H18 %4.1f kJ/mol'%(dGC8H18/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum nonexapnasion work through combustion of CH4 -818.6 kJ/mol\n",
- "Maximum nonexapnasion work through combustion of C8H18 -5309.8 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.4, Page Number 123"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGf298 = 370.7 #Std. free energy of formation for Fe (g), kJ/mol\n",
- "dHf298 = 416.3 #Std. Enthalpy of formation for Fe (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "T = 400. #Temperature in K\n",
- "R = 8.314\n",
- "\n",
- "#Calculations\n",
- "\n",
- "dGf = T*(dGf298*1e3/T0 + dHf298*1e3*(1./T - 1./T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. free energy of formation for Fe(g) at 400 K is %4.1f kJ/mol'%(dGf/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. free energy of formation for Fe(g) at 400 K is 355.1 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.5, Page Number 127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "#Variable Declaration\n",
- "nHe = 1.0 #Number of moles of He\n",
- "nNe = 3.0 #Number of moles of Ne\n",
- "nAr = 2.0 #Number of moles of Ar\n",
- "nXe = 2.5 #Number of moles of Xe\n",
- "T = 298.15 #Temperature in K\n",
- "P = 1.0 #Pressure, bar\n",
- "R = 8.314\n",
- "\n",
- "#Calculations\n",
- "n = nHe + nNe + nAr + nXe\n",
- "dGmix = n*R*T*((nHe/n)*log(nHe/n) + (nNe/n)*log(nNe/n) +(nAr/n)*log(nAr/n) + (nXe/n)*log(nXe/n))\n",
- "dSmix = n*R*((nHe/n)*log(nHe/n) + (nNe/n)*log(nNe/n) +(nAr/n)*log(nAr/n) + (nXe/n)*log(nXe/n))\n",
- "\n",
- "#Results \n",
- "print 'Std. free energy Change on mixing is %3.1e J'%(dGmix)\n",
- "print 'Std. entropy Change on mixing is %4.1f J'%(dSmix)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. free energy Change on mixing is -2.8e+04 J\n",
- "Std. entropy Change on mixing is -93.3 J\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.6, Page Number 128"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGfFe = 0.0 #Std. Gibbs energy of formation for Fe (S), kJ/mol\n",
- "dGfH2O = -237.1 #Std. Gibbs energy of formation for Water (g), kJ/mol\n",
- "dGfFe2O3 = -1015.4 #Std. Gibbs energy of formation for Fe2O3 (s), kJ/mol\n",
- "dGfH2 = 0.0 #Std. Gibbs energy of formation for Hydrogen (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nFe, nH2, nFe2O3, nH2O = 3,-4,-1,4\n",
- "\n",
- "#Calculations\n",
- "dGR = nFe*dGfFe + nH2O*dGfH2O + nFe2O3*dGfFe2O3 + nH2*dGfH2 \n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %4.2f kJ/mol'%(dGR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 67.00 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.7, Page Number 128"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGR = 67.0 #Std. Gibbs energy of formation for reaction, kJ, from previous problem\n",
- "dHfFe = 0.0 #Enthalpy of formation for Fe (S), kJ/mol\n",
- "dHfH2O = -285.8 #Enthalpy of formation for Water (g), kJ/mol\n",
- "dHfFe2O3 = -1118.4 #Enthalpy of formation for Fe2O3 (s), kJ/mol\n",
- "dHfH2 = 0.0 #Enthalpy of formation for Hydrogen (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "T = 525. #Temperature in K\n",
- "R = 8.314\n",
- "nFe, nH2, nFe2O3, nH2O = 3,-4,-1,4\n",
- "\n",
- "#Calculations\n",
- "dHR = nFe*dHfFe + nH2O*dHfH2O + nFe2O3*dHfFe2O3 + nH2*dHfH2 \n",
- "dGR2 = T*(dGR*1e3/T0 + dHR*1e3*(1./T - 1./T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. Enthalpy change for reactionat %4.1f is %4.2f kJ/mol'%(T, dHR)\n",
- "print 'Std. Gibbs energy change for reactionat %4.1f is %4.0f kJ/mol'%(T, dGR2/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Enthalpy change for reactionat 525.0 is -24.80 kJ/mol\n",
- "Std. Gibbs energy change for reactionat 525.0 is 137 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.8, Page Number 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "#Variable Declaration\n",
- "dGfNO2 = 51.3 #Std. Gibbs energy of formation for NO2 (g), kJ/mol\n",
- "dGfN2O4 = 99.8 #Std. Gibbs energy of formation for N2O4 (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "pNO2 = 0.350 #Partial pressure of NO2, bar\n",
- "pN2O4 = 0.650 #Partial pressure of N2O4, bar\n",
- "R = 8.314\n",
- "nNO2, nN2O4 = -2, 1 #Stoichiomentric coeff of NO2 and N2O4 respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nN2O4*dGfN2O4*1e3 + nNO2*dGfNO2*1e3 + R*T0*log(pN2O4/(pNO2)**2)\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 1.337 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.9, Page Number 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCO2 = -394.4 #Std. Gibbs energy of formation for CO2 (g), kJ/mol\n",
- "dGfH2 = 0.0 #Std. Gibbs energy of formation for H2 (g), kJ/mol\n",
- "dGfCO = 237.1 #Std. Gibbs energy of formation for CO (g), kJ/mol\n",
- "dGfH2O = 137.2 #Std. Gibbs energy of formation for H24 (l), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCO2, nH2, nCO, nH2O = 1,1,1,1 #Stoichiomentric coeff of CO2,H2,CO,H2O respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nCO2*dGfCO2 + nH2*dGfH2 + nCO*dGfCO + nH2O*dGfH2O\n",
- "Kp = exp(-dGR*1e3/(R*T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR/1e3)\n",
- "print 'Equilibrium constant for reaction is %5.3f '%(Kp)\n",
- "if Kp > 1: print 'Kp >> 1. hence, mixture will consists of product CO2 and H2'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is -0.020 kJ/mol\n",
- "Equilibrium constant for reaction is 3323.254 \n",
- "Kp >> 1. hence, mixture will consists of product CO2 and H2\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.11, Page Number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCl2 = 0.0 #Std. Gibbs energy of formation for CO2 (g), kJ/mol\n",
- "dGfCl = 105.7 #Std. Gibbs energy of formation for H2 (g), kJ/mol\n",
- "dHfCl2 = 0.0 #Std. Gibbs energy of formation for CO (g), kJ/mol\n",
- "dHfCl = 121.3 #Std. Gibbs energy of formation for H24 (l), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCl2, nCl= -1,2 #Stoichiomentric coeff of Cl2,Cl respectively in reaction\n",
- "PbyP0 = 0.01\n",
- "#Calculations\n",
- "dGR = nCl*dGfCl + nCl2*dGfCl2 \n",
- "dHR = nCl*dHfCl + nCl2*dHfCl2 \n",
- "func = lambda T: exp(-dGR*1e3/(R*T0) - dHR*1e3*(1./T - 1./T0)/R)\n",
- "Kp8 = func(800)\n",
- "Kp15 = func(1500)\n",
- "Kp20 = func(2000)\n",
- "DDiss = lambda K: sqrt(K/(K+4*PbyP0))\n",
- "alp8 = DDiss(Kp8)\n",
- "alp15 = DDiss(Kp15)\n",
- "alp20 = DDiss(Kp20)\n",
- "\n",
- "#Results \n",
- "print 'Part A'\n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR)\n",
- "print 'Std. Enthalpy change for reaction is %5.3f kJ/mol'%(dHR)\n",
- "print 'Equilibrium constants at 800, 1500, and 2000 K are %4.3e, %4.3e, and %4.3e'%(Kp8,Kp15,Kp20)\n",
- "\n",
- "print 'Part B'\n",
- "print 'Degree of dissociation at 800, 1500, and 2000 K are %4.3e, %4.3e, and %4.3e'%(alp8,alp15,alp20)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Part A\n",
- "Std. Gibbs energy change for reaction is 211.400 kJ/mol\n",
- "Std. Enthalpy change for reaction is 242.600 kJ/mol\n",
- "Equilibrium constants at 800, 1500, and 2000 K are 4.223e-11, 1.042e-03, and 1.349e-01\n",
- "Part B\n",
- "Degree of dissociation at 800, 1500, and 2000 K are 3.249e-05, 1.593e-01, and 8.782e-01\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.12, Page Number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "#Variable Declaration\n",
- "dGfCaCO3 = -1128.8 #Std. Gibbs energy of formation for CaCO3 (s), kJ/mol\n",
- "dGfCaO = -603.3 #Std. Gibbs energy of formation for CaO (s), kJ/mol\n",
- "dGfCO2 = -394.4 #Std. Gibbs energy of formation for O2 (g), kJ/mol\n",
- "dHfCaCO3 = -1206.9 #Std. Enthalpy Change of formation for CaCO3 (s), kJ/mol\n",
- "dHfCaO = -634.9 #Std. Enthalpy Change of formation for CaO (s), kJ/mol\n",
- "dHfCO2 = -393.5 #Std. Enthalpy Change of formation for O2 (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCaCO3, nCaO, nO2 = -1,1,1 #Stoichiomentric coeff of CaCO3, CaO, O2 respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nCaO*dGfCaO + nO2*dGfCO2 + nCaCO3*dGfCaCO3\n",
- "dHR = nCaO*dHfCaO + nO2*dHfCO2 + nCaCO3*dHfCaCO3\n",
- "\n",
- "func = lambda T: exp(-dGR*1e3/(R*T0) - dHR*1e3*(1./T - 1./T0)/R)\n",
- "\n",
- "Kp10 = func(1000)\n",
- "Kp11 = func(1100)\n",
- "Kp12 = func(1200)\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %4.1f kJ/mol'%(dGR)\n",
- "print 'Std. Enthalpy change for reaction is %4.1f kJ/mol'%(dHR)\n",
- "print 'Equilibrium constants at 1000, 1100, and 1200 K are %4.4f, %4.3fe, and %4.3f'%(Kp10,Kp11,Kp12)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 131.1 kJ/mol\n",
- "Std. Enthalpy change for reaction is 178.5 kJ/mol\n",
- "Equilibrium constants at 1000, 1100, and 1200 K are 0.0956, 0.673e, and 3.423\n"
- ]
- }
- ],
- "prompt_number": 88
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.13, Page Number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCG = 0.0 #Std. Gibbs energy of formation for CaCO3 (s), kJ/mol\n",
- "dGfCD = 2.90 #Std. Gibbs energy of formation for CaO (s), kJ/mol\n",
- "rhoG = 2.25e3 #Density of Graphite, kg/m3\n",
- "rhoD = 3.52e3 #Density of dimond, kg/m3\n",
- "T0 = 298.15 #Std. Temperature, K\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K) \n",
- "P0 = 1.0 #Pressure, bar\n",
- "M = 12.01 #Molceular wt of Carbon\n",
- "#Calculations\n",
- "P = P0*1e5 + dGfCD*1e3/((1./rhoG-1./rhoD)*M*1e-3)\n",
- "\n",
- "#Results \n",
- "print 'Pressure at which graphite and dimond will be in equilibrium is %4.2e bar'%(P/1e5)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure at which graphite and dimond will be in equilibrium is 1.51e+04 bar\n"
- ]
- }
- ],
- "prompt_number": 98
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.14, Page Number 143"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "beta = 2.04e-4 #Thermal exapansion coefficient, /K\n",
- "kapa = 45.9e-6 #Isothermal compressibility, /bar\n",
- "T = 298.15 #Std. Temperature, K\n",
- "R = 8.206e-2 #Ideal gas constant, atm.L/(mol.K) \n",
- "T1 = 320.0 #Temperature, K\n",
- "Pi = 1.0 #Initial Pressure, bar\n",
- "V = 1.00 #Volume, m3\n",
- "a = 1.35 #van der Waals constant a for nitrogen, atm.L2/mol2\n",
- "\n",
- "#Calculations\n",
- "dUbydV = Pf = (beta*T1-kapa*P0)/kapa\n",
- "dVT = V*kapa*(Pf-Pi)\n",
- "dVbyV = dVT*100/V\n",
- "Vm = Pi/(R*T1)\n",
- "dUbydVm = a/(Vm**2)\n",
- "\n",
- "#Results \n",
- "print 'dUbydV = %4.2e bar'%(dUbydV)\n",
- "print 'dVbyV = %4.3f percent'%(dVbyV)\n",
- "print 'dUbydVm = %4.0e atm'%(dUbydVm)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dUbydV = 1.42e+03 bar\n",
- "dVbyV = 6.519 percent\n",
- "dUbydVm = 9e+02 atm\n"
- ]
- }
- ],
- "prompt_number": 113
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.15, Page Number 144"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp, log\n",
- "#Variable Declaration\n",
- "m = 1000.0 #mass of mercury, g\n",
- "Pi, Ti = 1.00, 300.0 #Intial pressure and temperature, bar, K\n",
- "Pf, Tf = 300., 600.0 #Final pressure and temperature, bar, K\n",
- "rho = 13534. #Density of mercury, kg/m3\n",
- "beta = 18.1e-4 #Thermal exapansion coefficient for Hg, /K \n",
- "kapa = 3.91e-6 #Isothermal compressibility for Hg, /Pa\n",
- "Cpm = 27.98 #Molar Specific heat at constant pressure, J/(mol.K) \n",
- "M = 200.59 #Molecular wt of Hg, g/mol\n",
- "\n",
- "#Calculations\n",
- "Vi = m*1e-3/rho\n",
- "Vf = Vi*exp(-kapa*(Pf-Pi))\n",
- "Ut = m*Cpm*(Tf-Ti)/M \n",
- "Up = (beta*Ti/kapa-Pi)*1e5*(Vf-Vi) + (Vi-Vf+Vf*log(Vf/Vi))*1e5/kapa\n",
- "dU = Ut + Up\n",
- "Ht = m*Cpm*(Tf-Ti)/M\n",
- "Hp = ((1 + beta*(Tf-Ti))*Vi*exp(-kapa*Pi)/kapa)*(exp(-kapa*Pi)-exp(-kapa*Pf))\n",
- "dH = Ht + Hp\n",
- "#Results\n",
- "print 'Internal energy change is %6.2e J/mol in which \\ncontribution of temeprature dependent term %6.4f percent'%(dU,Ut*100/dH)\n",
- "print 'Enthalpy change is %4.3e J/mol in which \\ncontribution of temeprature dependent term %4.1f percent'%(dH,Ht*100/dH)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Internal energy change is 4.06e+04 J/mol in which \n",
- "contribution of temeprature dependent term 99.9999 percent\n",
- "Enthalpy change is 4.185e+04 J/mol in which \n",
- "contribution of temeprature dependent term 100.0 percent\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.16, Page Number 145"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "T = 300.0 #Temperature of Hg, K \n",
- "beta = 18.1e-4 #Thermal exapansion coefficient for Hg, /K \n",
- "kapa = 3.91e-6 #Isothermal compressibility for Hg, /Pa\n",
- "M = 0.20059 #Molecular wt of Hg, kg/mol \n",
- "rho = 13534 #Density of mercury, kg/m3\n",
- "Cpm = 27.98 #Experimental Molar specif heat at const pressure for mercury, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "Vm = M/rho\n",
- "DCpmCv = T*Vm*beta**2/kapa\n",
- "Cvm = Cpm - DCpmCv\n",
- "#Results\n",
- "print 'Difference in molar specific heats \\nat constant volume and constant pressure %4.2e J/(mol.K)'%DCpmCv\n",
- "print 'Molar Specific heat of Hg at const. volume is %4.2f J/(mol.K)'%Cvm"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference in molar specific heats \n",
- "at constant volume and constant pressure 3.73e-03 J/(mol.K)\n",
- "Molar Specific heat of Hg at const. volume is 27.98 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 149
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.17, Page Number 147"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "T = 298.15 #Std. Temperature, K \n",
- "P = 1.0 #Initial Pressure, bar\n",
- "Hm0, Sm0 = 0.0,154.8 #Std. molar enthalpy and entropy of Ar(g), kJ, mol, K units\n",
- "Sm0H2, Sm0O2 = 130.7,205.2 #Std. molar entropy of O2 and H2 (g), kJ/(mol.K)\n",
- "dGfH2O = -237.1 #Gibbs energy of formation for H2O(l), kJ/mol \n",
- "nH2, nO2 = 1, 1./2 #Stoichiomentric coefficients for H2 and O2 in water formation reaction \n",
- "\n",
- "#Calculations\n",
- "Gm0 = Hm0 - T*Sm0\n",
- "dGmH2O = dGfH2O*1000 - T*(nH2*Sm0H2 + nO2*Sm0O2)\n",
- "#Results\n",
- "print 'Molar Gibbs energy of Ar %4.3f kJ/mol'%(Gm0/1e3)\n",
- "print 'Molar Gibbs energy of Water %4.3f kJ/mol'%(dGmH2O/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molar Gibbs energy of Ar -46.154 kJ/mol\n",
- "Molar Gibbs energy of Water -306.658 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06_1.ipynb deleted file mode 100755 index c6258811..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06_1.ipynb +++ /dev/null @@ -1,709 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:399beb745330e541e567baa0f45fde7ecc89dabcc65d8db71d8e5921a036072a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 06: Chemical Equilibrium"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.1, Page Number 117"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dHcCH4 = -891.0 #Std. heat of combustion for CH4, kJ/mol\n",
- "dHcC8H18 = -5471.0 #Std. heat of combustion for C8H18, kJ/mol\n",
- "\n",
- "T = 298.15\n",
- "SmCO2, SmCH4, SmH2O, SmO2, SmC8H18 = 213.8,186.3,70.0,205.2, 316.1\n",
- "dnCH4 = -2.\n",
- "dnC8H18 = 4.5\n",
- "R = 8.314\n",
- "#Calculations\n",
- "dACH4 = dHcCH4*1e3 - dnCH4*R*T - T*(SmCO2 + 2*SmH2O - SmCH4 - 2*SmO2)\n",
- "dAC8H18 = dHcC8H18*1e3 - dnC8H18*R*T - T*(8*SmCO2 + 9*SmH2O - SmC8H18 - 25.*SmO2/2) \n",
- "#Results \n",
- "print 'Maximum Available work through combustion of CH4 %4.1f kJ/mol'%(dACH4/1000)\n",
- "print 'Maximum Available work through combustion of C8H18 %4.1f kJ/mol'%(dAC8H18/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Available work through combustion of CH4 -813.6 kJ/mol\n",
- "Maximum Available work through combustion of C8H18 -5320.9 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.2, Page Number 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dHcCH4 = -891.0 #Std. heat of combustion for CH4, kJ/mol\n",
- "dHcC8H18 = -5471.0 #Std. heat of combustion for C8H18, kJ/mol\n",
- "\n",
- "T = 298.15\n",
- "SmCO2, SmCH4, SmH2O, SmO2, SmC8H18 = 213.8,186.3,70.0,205.2, 316.1\n",
- "dnCH4 = -2.\n",
- "dnC8H18 = 4.5\n",
- "R = 8.314\n",
- "#Calculations\n",
- "dGCH4 = dHcCH4*1e3 - T*(SmCO2 + 2*SmH2O - SmCH4 - 2*SmO2)\n",
- "dGC8H18 = dHcC8H18*1e3 - T*(8*SmCO2 + 9*SmH2O - SmC8H18 - 25.*SmO2/2) \n",
- "#Results \n",
- "print 'Maximum nonexapnasion work through combustion of CH4 %4.1f kJ/mol'%(dGCH4/1000)\n",
- "print 'Maximum nonexapnasion work through combustion of C8H18 %4.1f kJ/mol'%(dGC8H18/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum nonexapnasion work through combustion of CH4 -818.6 kJ/mol\n",
- "Maximum nonexapnasion work through combustion of C8H18 -5309.8 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.4, Page Number 123"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGf298 = 370.7 #Std. free energy of formation for Fe (g), kJ/mol\n",
- "dHf298 = 416.3 #Std. Enthalpy of formation for Fe (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "T = 400. #Temperature in K\n",
- "R = 8.314\n",
- "\n",
- "#Calculations\n",
- "\n",
- "dGf = T*(dGf298*1e3/T0 + dHf298*1e3*(1./T - 1./T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. free energy of formation for Fe(g) at 400 K is %4.1f kJ/mol'%(dGf/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. free energy of formation for Fe(g) at 400 K is 355.1 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.5, Page Number 127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "#Variable Declaration\n",
- "nHe = 1.0 #Number of moles of He\n",
- "nNe = 3.0 #Number of moles of Ne\n",
- "nAr = 2.0 #Number of moles of Ar\n",
- "nXe = 2.5 #Number of moles of Xe\n",
- "T = 298.15 #Temperature in K\n",
- "P = 1.0 #Pressure, bar\n",
- "R = 8.314\n",
- "\n",
- "#Calculations\n",
- "n = nHe + nNe + nAr + nXe\n",
- "dGmix = n*R*T*((nHe/n)*log(nHe/n) + (nNe/n)*log(nNe/n) +(nAr/n)*log(nAr/n) + (nXe/n)*log(nXe/n))\n",
- "dSmix = n*R*((nHe/n)*log(nHe/n) + (nNe/n)*log(nNe/n) +(nAr/n)*log(nAr/n) + (nXe/n)*log(nXe/n))\n",
- "\n",
- "#Results \n",
- "print 'Std. free energy Change on mixing is %3.1e J'%(dGmix)\n",
- "print 'Std. entropy Change on mixing is %4.1f J'%(dSmix)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. free energy Change on mixing is -2.8e+04 J\n",
- "Std. entropy Change on mixing is -93.3 J\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.6, Page Number 128"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGfFe = 0.0 #Std. Gibbs energy of formation for Fe (S), kJ/mol\n",
- "dGfH2O = -237.1 #Std. Gibbs energy of formation for Water (g), kJ/mol\n",
- "dGfFe2O3 = -1015.4 #Std. Gibbs energy of formation for Fe2O3 (s), kJ/mol\n",
- "dGfH2 = 0.0 #Std. Gibbs energy of formation for Hydrogen (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nFe, nH2, nFe2O3, nH2O = 3,-4,-1,4\n",
- "\n",
- "#Calculations\n",
- "dGR = nFe*dGfFe + nH2O*dGfH2O + nFe2O3*dGfFe2O3 + nH2*dGfH2 \n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %4.2f kJ/mol'%(dGR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 67.00 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.7, Page Number 128"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGR = 67.0 #Std. Gibbs energy of formation for reaction, kJ, from previous problem\n",
- "dHfFe = 0.0 #Enthalpy of formation for Fe (S), kJ/mol\n",
- "dHfH2O = -285.8 #Enthalpy of formation for Water (g), kJ/mol\n",
- "dHfFe2O3 = -1118.4 #Enthalpy of formation for Fe2O3 (s), kJ/mol\n",
- "dHfH2 = 0.0 #Enthalpy of formation for Hydrogen (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "T = 525. #Temperature in K\n",
- "R = 8.314\n",
- "nFe, nH2, nFe2O3, nH2O = 3,-4,-1,4\n",
- "\n",
- "#Calculations\n",
- "dHR = nFe*dHfFe + nH2O*dHfH2O + nFe2O3*dHfFe2O3 + nH2*dHfH2 \n",
- "dGR2 = T*(dGR*1e3/T0 + dHR*1e3*(1./T - 1./T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. Enthalpy change for reactionat %4.1f is %4.2f kJ/mol'%(T, dHR)\n",
- "print 'Std. Gibbs energy change for reactionat %4.1f is %4.0f kJ/mol'%(T, dGR2/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Enthalpy change for reactionat 525.0 is -24.80 kJ/mol\n",
- "Std. Gibbs energy change for reactionat 525.0 is 137 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.8, Page Number 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "#Variable Declaration\n",
- "dGfNO2 = 51.3 #Std. Gibbs energy of formation for NO2 (g), kJ/mol\n",
- "dGfN2O4 = 99.8 #Std. Gibbs energy of formation for N2O4 (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "pNO2 = 0.350 #Partial pressure of NO2, bar\n",
- "pN2O4 = 0.650 #Partial pressure of N2O4, bar\n",
- "R = 8.314\n",
- "nNO2, nN2O4 = -2, 1 #Stoichiomentric coeff of NO2 and N2O4 respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nN2O4*dGfN2O4*1e3 + nNO2*dGfNO2*1e3 + R*T0*log(pN2O4/(pNO2)**2)\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 1.337 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.9, Page Number 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCO2 = -394.4 #Std. Gibbs energy of formation for CO2 (g), kJ/mol\n",
- "dGfH2 = 0.0 #Std. Gibbs energy of formation for H2 (g), kJ/mol\n",
- "dGfCO = 237.1 #Std. Gibbs energy of formation for CO (g), kJ/mol\n",
- "dGfH2O = 137.2 #Std. Gibbs energy of formation for H24 (l), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCO2, nH2, nCO, nH2O = 1,1,1,1 #Stoichiomentric coeff of CO2,H2,CO,H2O respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nCO2*dGfCO2 + nH2*dGfH2 + nCO*dGfCO + nH2O*dGfH2O\n",
- "Kp = exp(-dGR*1e3/(R*T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR/1e3)\n",
- "print 'Equilibrium constant for reaction is %5.3f '%(Kp)\n",
- "if Kp > 1: print 'Kp >> 1. hence, mixture will consists of product CO2 and H2'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is -0.020 kJ/mol\n",
- "Equilibrium constant for reaction is 3323.254 \n",
- "Kp >> 1. hence, mixture will consists of product CO2 and H2\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.11, Page Number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCl2 = 0.0 #Std. Gibbs energy of formation for CO2 (g), kJ/mol\n",
- "dGfCl = 105.7 #Std. Gibbs energy of formation for H2 (g), kJ/mol\n",
- "dHfCl2 = 0.0 #Std. Gibbs energy of formation for CO (g), kJ/mol\n",
- "dHfCl = 121.3 #Std. Gibbs energy of formation for H24 (l), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCl2, nCl= -1,2 #Stoichiomentric coeff of Cl2,Cl respectively in reaction\n",
- "PbyP0 = 0.01\n",
- "#Calculations\n",
- "dGR = nCl*dGfCl + nCl2*dGfCl2 \n",
- "dHR = nCl*dHfCl + nCl2*dHfCl2 \n",
- "func = lambda T: exp(-dGR*1e3/(R*T0) - dHR*1e3*(1./T - 1./T0)/R)\n",
- "Kp8 = func(800)\n",
- "Kp15 = func(1500)\n",
- "Kp20 = func(2000)\n",
- "DDiss = lambda K: sqrt(K/(K+4*PbyP0))\n",
- "alp8 = DDiss(Kp8)\n",
- "alp15 = DDiss(Kp15)\n",
- "alp20 = DDiss(Kp20)\n",
- "\n",
- "#Results \n",
- "print 'Part A'\n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR)\n",
- "print 'Std. Enthalpy change for reaction is %5.3f kJ/mol'%(dHR)\n",
- "print 'Equilibrium constants at 800, 1500, and 2000 K are %4.3e, %4.3e, and %4.3e'%(Kp8,Kp15,Kp20)\n",
- "\n",
- "print 'Part B'\n",
- "print 'Degree of dissociation at 800, 1500, and 2000 K are %4.3e, %4.3e, and %4.3e'%(alp8,alp15,alp20)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Part A\n",
- "Std. Gibbs energy change for reaction is 211.400 kJ/mol\n",
- "Std. Enthalpy change for reaction is 242.600 kJ/mol\n",
- "Equilibrium constants at 800, 1500, and 2000 K are 4.223e-11, 1.042e-03, and 1.349e-01\n",
- "Part B\n",
- "Degree of dissociation at 800, 1500, and 2000 K are 3.249e-05, 1.593e-01, and 8.782e-01\n"
- ]
- }
- ],
- "prompt_number": 71
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.12, Page Number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "#Variable Declaration\n",
- "dGfCaCO3 = -1128.8 #Std. Gibbs energy of formation for CaCO3 (s), kJ/mol\n",
- "dGfCaO = -603.3 #Std. Gibbs energy of formation for CaO (s), kJ/mol\n",
- "dGfCO2 = -394.4 #Std. Gibbs energy of formation for O2 (g), kJ/mol\n",
- "dHfCaCO3 = -1206.9 #Std. Enthalpy Change of formation for CaCO3 (s), kJ/mol\n",
- "dHfCaO = -634.9 #Std. Enthalpy Change of formation for CaO (s), kJ/mol\n",
- "dHfCO2 = -393.5 #Std. Enthalpy Change of formation for O2 (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCaCO3, nCaO, nO2 = -1,1,1 #Stoichiomentric coeff of CaCO3, CaO, O2 respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nCaO*dGfCaO + nO2*dGfCO2 + nCaCO3*dGfCaCO3\n",
- "dHR = nCaO*dHfCaO + nO2*dHfCO2 + nCaCO3*dHfCaCO3\n",
- "\n",
- "func = lambda T: exp(-dGR*1e3/(R*T0) - dHR*1e3*(1./T - 1./T0)/R)\n",
- "\n",
- "Kp10 = func(1000)\n",
- "Kp11 = func(1100)\n",
- "Kp12 = func(1200)\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %4.1f kJ/mol'%(dGR)\n",
- "print 'Std. Enthalpy change for reaction is %4.1f kJ/mol'%(dHR)\n",
- "print 'Equilibrium constants at 1000, 1100, and 1200 K are %4.4f, %4.3fe, and %4.3f'%(Kp10,Kp11,Kp12)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 131.1 kJ/mol\n",
- "Std. Enthalpy change for reaction is 178.5 kJ/mol\n",
- "Equilibrium constants at 1000, 1100, and 1200 K are 0.0956, 0.673e, and 3.423\n"
- ]
- }
- ],
- "prompt_number": 88
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.13, Page Number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCG = 0.0 #Std. Gibbs energy of formation for CaCO3 (s), kJ/mol\n",
- "dGfCD = 2.90 #Std. Gibbs energy of formation for CaO (s), kJ/mol\n",
- "rhoG = 2.25e3 #Density of Graphite, kg/m3\n",
- "rhoD = 3.52e3 #Density of dimond, kg/m3\n",
- "T0 = 298.15 #Std. Temperature, K\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K) \n",
- "P0 = 1.0 #Pressure, bar\n",
- "M = 12.01 #Molceular wt of Carbon\n",
- "#Calculations\n",
- "P = P0*1e5 + dGfCD*1e3/((1./rhoG-1./rhoD)*M*1e-3)\n",
- "\n",
- "#Results \n",
- "print 'Pressure at which graphite and dimond will be in equilibrium is %4.2e bar'%(P/1e5)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure at which graphite and dimond will be in equilibrium is 1.51e+04 bar\n"
- ]
- }
- ],
- "prompt_number": 98
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.14, Page Number 143"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "beta = 2.04e-4 #Thermal exapansion coefficient, /K\n",
- "kapa = 45.9e-6 #Isothermal compressibility, /bar\n",
- "T = 298.15 #Std. Temperature, K\n",
- "R = 8.206e-2 #Ideal gas constant, atm.L/(mol.K) \n",
- "T1 = 320.0 #Temperature, K\n",
- "Pi = 1.0 #Initial Pressure, bar\n",
- "V = 1.00 #Volume, m3\n",
- "a = 1.35 #van der Waals constant a for nitrogen, atm.L2/mol2\n",
- "\n",
- "#Calculations\n",
- "dUbydV = Pf = (beta*T1-kapa*P0)/kapa\n",
- "dVT = V*kapa*(Pf-Pi)\n",
- "dVbyV = dVT*100/V\n",
- "Vm = Pi/(R*T1)\n",
- "dUbydVm = a/(Vm**2)\n",
- "\n",
- "#Results \n",
- "print 'dUbydV = %4.2e bar'%(dUbydV)\n",
- "print 'dVbyV = %4.3f percent'%(dVbyV)\n",
- "print 'dUbydVm = %4.0e atm'%(dUbydVm)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dUbydV = 1.42e+03 bar\n",
- "dVbyV = 6.519 percent\n",
- "dUbydVm = 9e+02 atm\n"
- ]
- }
- ],
- "prompt_number": 113
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.15, Page Number 144"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp, log\n",
- "#Variable Declaration\n",
- "m = 1000.0 #mass of mercury, g\n",
- "Pi, Ti = 1.00, 300.0 #Intial pressure and temperature, bar, K\n",
- "Pf, Tf = 300., 600.0 #Final pressure and temperature, bar, K\n",
- "rho = 13534. #Density of mercury, kg/m3\n",
- "beta = 18.1e-4 #Thermal exapansion coefficient for Hg, /K \n",
- "kapa = 3.91e-6 #Isothermal compressibility for Hg, /Pa\n",
- "Cpm = 27.98 #Molar Specific heat at constant pressure, J/(mol.K) \n",
- "M = 200.59 #Molecular wt of Hg, g/mol\n",
- "\n",
- "#Calculations\n",
- "Vi = m*1e-3/rho\n",
- "Vf = Vi*exp(-kapa*(Pf-Pi))\n",
- "Ut = m*Cpm*(Tf-Ti)/M \n",
- "Up = (beta*Ti/kapa-Pi)*1e5*(Vf-Vi) + (Vi-Vf+Vf*log(Vf/Vi))*1e5/kapa\n",
- "dU = Ut + Up\n",
- "Ht = m*Cpm*(Tf-Ti)/M\n",
- "Hp = ((1 + beta*(Tf-Ti))*Vi*exp(-kapa*Pi)/kapa)*(exp(-kapa*Pi)-exp(-kapa*Pf))\n",
- "dH = Ht + Hp\n",
- "#Results\n",
- "print 'Internal energy change is %6.2e J/mol in which \\ncontribution of temeprature dependent term %6.4f percent'%(dU,Ut*100/dH)\n",
- "print 'Enthalpy change is %4.3e J/mol in which \\ncontribution of temeprature dependent term %4.1f percent'%(dH,Ht*100/dH)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Internal energy change is 4.06e+04 J/mol in which \n",
- "contribution of temeprature dependent term 99.9999 percent\n",
- "Enthalpy change is 4.185e+04 J/mol in which \n",
- "contribution of temeprature dependent term 100.0 percent\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.16, Page Number 145"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "T = 300.0 #Temperature of Hg, K \n",
- "beta = 18.1e-4 #Thermal exapansion coefficient for Hg, /K \n",
- "kapa = 3.91e-6 #Isothermal compressibility for Hg, /Pa\n",
- "M = 0.20059 #Molecular wt of Hg, kg/mol \n",
- "rho = 13534 #Density of mercury, kg/m3\n",
- "Cpm = 27.98 #Experimental Molar specif heat at const pressure for mercury, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "Vm = M/rho\n",
- "DCpmCv = T*Vm*beta**2/kapa\n",
- "Cvm = Cpm - DCpmCv\n",
- "#Results\n",
- "print 'Difference in molar specific heats \\nat constant volume and constant pressure %4.2e J/(mol.K)'%DCpmCv\n",
- "print 'Molar Specific heat of Hg at const. volume is %4.2f J/(mol.K)'%Cvm"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference in molar specific heats \n",
- "at constant volume and constant pressure 3.73e-03 J/(mol.K)\n",
- "Molar Specific heat of Hg at const. volume is 27.98 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 149
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.17, Page Number 147"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "T = 298.15 #Std. Temperature, K \n",
- "P = 1.0 #Initial Pressure, bar\n",
- "Hm0, Sm0 = 0.0,154.8 #Std. molar enthalpy and entropy of Ar(g), kJ, mol, K units\n",
- "Sm0H2, Sm0O2 = 130.7,205.2 #Std. molar entropy of O2 and H2 (g), kJ/(mol.K)\n",
- "dGfH2O = -237.1 #Gibbs energy of formation for H2O(l), kJ/mol \n",
- "nH2, nO2 = 1, 1./2 #Stoichiomentric coefficients for H2 and O2 in water formation reaction \n",
- "\n",
- "#Calculations\n",
- "Gm0 = Hm0 - T*Sm0\n",
- "dGmH2O = dGfH2O*1000 - T*(nH2*Sm0H2 + nO2*Sm0O2)\n",
- "#Results\n",
- "print 'Molar Gibbs energy of Ar %4.3f kJ/mol'%(Gm0/1e3)\n",
- "print 'Molar Gibbs energy of Water %4.3f kJ/mol'%(dGmH2O/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molar Gibbs energy of Ar -46.154 kJ/mol\n",
- "Molar Gibbs energy of Water -306.658 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06_2.ipynb deleted file mode 100755 index f5b2d065..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter06_2.ipynb +++ /dev/null @@ -1,713 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:267b6c699ea5b02708cd7837dc08f6cae55ca12f860a79f0fcaa34ccb60be757"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 06: Chemical Equilibrium"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.1, Page Number 117"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dHcCH4 = -891.0 #Std. heat of combustion for CH4, kJ/mol\n",
- "dHcC8H18 = -5471.0 #Std. heat of combustion for C8H18, kJ/mol\n",
- "\n",
- "T = 298.15\n",
- "SmCO2, SmCH4, SmH2O, SmO2, SmC8H18 = 213.8,186.3,70.0,205.2, 316.1\n",
- "dnCH4 = -2.\n",
- "dnC8H18 = 4.5\n",
- "R = 8.314\n",
- "#Calculations\n",
- "dACH4 = dHcCH4*1e3 - dnCH4*R*T - T*(SmCO2 + 2*SmH2O - SmCH4 - 2*SmO2)\n",
- "dAC8H18 = dHcC8H18*1e3 - dnC8H18*R*T - T*(8*SmCO2 + 9*SmH2O - SmC8H18 - 25.*SmO2/2) \n",
- "#Results \n",
- "print 'Maximum Available work through combustion of CH4 %4.1f kJ/mol'%(dACH4/1000)\n",
- "print 'Maximum Available work through combustion of C8H18 %4.1f kJ/mol'%(dAC8H18/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Available work through combustion of CH4 -813.6 kJ/mol\n",
- "Maximum Available work through combustion of C8H18 -5320.9 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.2, Page Number 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dHcCH4 = -891.0 #Std. heat of combustion for CH4, kJ/mol\n",
- "dHcC8H18 = -5471.0 #Std. heat of combustion for C8H18, kJ/mol\n",
- "\n",
- "T = 298.15\n",
- "SmCO2, SmCH4, SmH2O, SmO2, SmC8H18 = 213.8,186.3,70.0,205.2, 316.1\n",
- "dnCH4 = -2.\n",
- "dnC8H18 = 4.5\n",
- "R = 8.314\n",
- "#Calculations\n",
- "dGCH4 = dHcCH4*1e3 - T*(SmCO2 + 2*SmH2O - SmCH4 - 2*SmO2)\n",
- "dGC8H18 = dHcC8H18*1e3 - T*(8*SmCO2 + 9*SmH2O - SmC8H18 - 25.*SmO2/2) \n",
- "#Results \n",
- "print 'Maximum nonexapnasion work through combustion of CH4 %4.1f kJ/mol'%(dGCH4/1000)\n",
- "print 'Maximum nonexapnasion work through combustion of C8H18 %4.1f kJ/mol'%(dGC8H18/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum nonexapnasion work through combustion of CH4 -818.6 kJ/mol\n",
- "Maximum nonexapnasion work through combustion of C8H18 -5309.8 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.4, Page Number 123"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGf298 = 370.7 #Std. free energy of formation for Fe (g), kJ/mol\n",
- "dHf298 = 416.3 #Std. Enthalpy of formation for Fe (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "T = 400. #Temperature in K\n",
- "R = 8.314\n",
- "\n",
- "#Calculations\n",
- "\n",
- "dGf = T*(dGf298*1e3/T0 + dHf298*1e3*(1./T - 1./T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. free energy of formation for Fe(g) at 400 K is %4.1f kJ/mol'%(dGf/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. free energy of formation for Fe(g) at 400 K is 355.1 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.5, Page Number 127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "nHe = 1.0 #Number of moles of He\n",
- "nNe = 3.0 #Number of moles of Ne\n",
- "nAr = 2.0 #Number of moles of Ar\n",
- "nXe = 2.5 #Number of moles of Xe\n",
- "T = 298.15 #Temperature in K\n",
- "P = 1.0 #Pressure, bar\n",
- "R = 8.314\n",
- "\n",
- "#Calculations\n",
- "n = nHe + nNe + nAr + nXe\n",
- "dGmix = n*R*T*((nHe/n)*log(nHe/n) + (nNe/n)*log(nNe/n) +(nAr/n)*log(nAr/n) + (nXe/n)*log(nXe/n))\n",
- "dSmix = n*R*((nHe/n)*log(nHe/n) + (nNe/n)*log(nNe/n) +(nAr/n)*log(nAr/n) + (nXe/n)*log(nXe/n))\n",
- "\n",
- "#Results \n",
- "print 'Std. free energy Change on mixing is %3.1e J'%(dGmix)\n",
- "print 'Std. entropy Change on mixing is %4.1f J'%(dSmix)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. free energy Change on mixing is -2.8e+04 J\n",
- "Std. entropy Change on mixing is -93.3 J\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.6, Page Number 128"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGfFe = 0.0 #Std. Gibbs energy of formation for Fe (S), kJ/mol\n",
- "dGfH2O = -237.1 #Std. Gibbs energy of formation for Water (g), kJ/mol\n",
- "dGfFe2O3 = -1015.4 #Std. Gibbs energy of formation for Fe2O3 (s), kJ/mol\n",
- "dGfH2 = 0.0 #Std. Gibbs energy of formation for Hydrogen (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nFe, nH2, nFe2O3, nH2O = 3,-4,-1,4\n",
- "\n",
- "#Calculations\n",
- "dGR = nFe*dGfFe + nH2O*dGfH2O + nFe2O3*dGfFe2O3 + nH2*dGfH2 \n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %4.2f kJ/mol'%(dGR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 67.00 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.7, Page Number 128"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "dGR = 67.0 #Std. Gibbs energy of formation for reaction, kJ, from previous problem\n",
- "dHfFe = 0.0 #Enthalpy of formation for Fe (S), kJ/mol\n",
- "dHfH2O = -285.8 #Enthalpy of formation for Water (g), kJ/mol\n",
- "dHfFe2O3 = -1118.4 #Enthalpy of formation for Fe2O3 (s), kJ/mol\n",
- "dHfH2 = 0.0 #Enthalpy of formation for Hydrogen (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "T = 525. #Temperature in K\n",
- "R = 8.314\n",
- "nFe, nH2, nFe2O3, nH2O = 3,-4,-1,4\n",
- "\n",
- "#Calculations\n",
- "dHR = nFe*dHfFe + nH2O*dHfH2O + nFe2O3*dHfFe2O3 + nH2*dHfH2 \n",
- "dGR2 = T*(dGR*1e3/T0 + dHR*1e3*(1./T - 1./T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. Enthalpy change for reactionat %4.1f is %4.2f kJ/mol'%(T, dHR)\n",
- "print 'Std. Gibbs energy change for reactionat %4.1f is %4.0f kJ/mol'%(T, dGR2/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Enthalpy change for reactionat 525.0 is -24.80 kJ/mol\n",
- "Std. Gibbs energy change for reactionat 525.0 is 137 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.8, Page Number 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "dGfNO2 = 51.3 #Std. Gibbs energy of formation for NO2 (g), kJ/mol\n",
- "dGfN2O4 = 99.8 #Std. Gibbs energy of formation for N2O4 (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "pNO2 = 0.350 #Partial pressure of NO2, bar\n",
- "pN2O4 = 0.650 #Partial pressure of N2O4, bar\n",
- "R = 8.314\n",
- "nNO2, nN2O4 = -2, 1 #Stoichiomentric coeff of NO2 and N2O4 respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nN2O4*dGfN2O4*1e3 + nNO2*dGfNO2*1e3 + R*T0*log(pN2O4/(pNO2)**2)\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 1.337 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.9, Page Number 131"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCO2 = -394.4 #Std. Gibbs energy of formation for CO2 (g), kJ/mol\n",
- "dGfH2 = 0.0 #Std. Gibbs energy of formation for H2 (g), kJ/mol\n",
- "dGfCO = 237.1 #Std. Gibbs energy of formation for CO (g), kJ/mol\n",
- "dGfH2O = 137.2 #Std. Gibbs energy of formation for H24 (l), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCO2, nH2, nCO, nH2O = 1,1,1,1 #Stoichiomentric coeff of CO2,H2,CO,H2O respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nCO2*dGfCO2 + nH2*dGfH2 + nCO*dGfCO + nH2O*dGfH2O\n",
- "Kp = exp(-dGR*1e3/(R*T0))\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR/1e3)\n",
- "print 'Equilibrium constant for reaction is %5.3f '%(Kp)\n",
- "if Kp > 1: print 'Kp >> 1. hence, mixture will consists of product CO2 and H2'"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is -0.020 kJ/mol\n",
- "Equilibrium constant for reaction is 3323.254 \n",
- "Kp >> 1. hence, mixture will consists of product CO2 and H2\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.11, Page Number 133"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCl2 = 0.0 #Std. Gibbs energy of formation for CO2 (g), kJ/mol\n",
- "dGfCl = 105.7 #Std. Gibbs energy of formation for H2 (g), kJ/mol\n",
- "dHfCl2 = 0.0 #Std. Gibbs energy of formation for CO (g), kJ/mol\n",
- "dHfCl = 121.3 #Std. Gibbs energy of formation for H24 (l), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCl2, nCl= -1,2 #Stoichiomentric coeff of Cl2,Cl respectively in reaction\n",
- "PbyP0 = 0.01\n",
- "#Calculations\n",
- "dGR = nCl*dGfCl + nCl2*dGfCl2 \n",
- "dHR = nCl*dHfCl + nCl2*dHfCl2 \n",
- "func = lambda T: exp(-dGR*1e3/(R*T0) - dHR*1e3*(1./T - 1./T0)/R)\n",
- "Kp8 = func(800)\n",
- "Kp15 = func(1500)\n",
- "Kp20 = func(2000)\n",
- "DDiss = lambda K: sqrt(K/(K+4*PbyP0))\n",
- "alp8 = DDiss(Kp8)\n",
- "alp15 = DDiss(Kp15)\n",
- "alp20 = DDiss(Kp20)\n",
- "\n",
- "#Results \n",
- "print 'Part A'\n",
- "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR)\n",
- "print 'Std. Enthalpy change for reaction is %5.3f kJ/mol'%(dHR)\n",
- "print 'Equilibrium constants at 800, 1500, and 2000 K are %4.3e, %4.3e, and %4.3e'%(Kp8,Kp15,Kp20)\n",
- "\n",
- "print 'Part B'\n",
- "print 'Degree of dissociation at 800, 1500, and 2000 K are %4.3e, %4.3e, and %4.3e'%(alp8,alp15,alp20)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Part A\n",
- "Std. Gibbs energy change for reaction is 211.400 kJ/mol\n",
- "Std. Enthalpy change for reaction is 242.600 kJ/mol\n",
- "Equilibrium constants at 800, 1500, and 2000 K are 4.223e-11, 1.042e-03, and 1.349e-01\n",
- "Part B\n",
- "Degree of dissociation at 800, 1500, and 2000 K are 3.249e-05, 1.593e-01, and 8.782e-01\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.12, Page Number 134"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCaCO3 = -1128.8 #Std. Gibbs energy of formation for CaCO3 (s), kJ/mol\n",
- "dGfCaO = -603.3 #Std. Gibbs energy of formation for CaO (s), kJ/mol\n",
- "dGfCO2 = -394.4 #Std. Gibbs energy of formation for O2 (g), kJ/mol\n",
- "dHfCaCO3 = -1206.9 #Std. Enthalpy Change of formation for CaCO3 (s), kJ/mol\n",
- "dHfCaO = -634.9 #Std. Enthalpy Change of formation for CaO (s), kJ/mol\n",
- "dHfCO2 = -393.5 #Std. Enthalpy Change of formation for O2 (g), kJ/mol\n",
- "T0 = 298.15 #Temperature in K\n",
- "R = 8.314\n",
- "nCaCO3, nCaO, nO2 = -1,1,1 #Stoichiomentric coeff of CaCO3, CaO, O2 respectively in reaction\n",
- "\n",
- "#Calculations\n",
- "dGR = nCaO*dGfCaO + nO2*dGfCO2 + nCaCO3*dGfCaCO3\n",
- "dHR = nCaO*dHfCaO + nO2*dHfCO2 + nCaCO3*dHfCaCO3\n",
- "\n",
- "func = lambda T: exp(-dGR*1e3/(R*T0) - dHR*1e3*(1./T - 1./T0)/R)\n",
- "\n",
- "Kp10 = func(1000)\n",
- "Kp11 = func(1100)\n",
- "Kp12 = func(1200)\n",
- "\n",
- "#Results \n",
- "print 'Std. Gibbs energy change for reaction is %4.1f kJ/mol'%(dGR)\n",
- "print 'Std. Enthalpy change for reaction is %4.1f kJ/mol'%(dHR)\n",
- "print 'Equilibrium constants at 1000, 1100, and 1200 K are %4.4f, %4.3fe, and %4.3f'%(Kp10,Kp11,Kp12)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. Gibbs energy change for reaction is 131.1 kJ/mol\n",
- "Std. Enthalpy change for reaction is 178.5 kJ/mol\n",
- "Equilibrium constants at 1000, 1100, and 1200 K are 0.0956, 0.673e, and 3.423\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.13, Page Number 135"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "dGfCG = 0.0 #Std. Gibbs energy of formation for CaCO3 (s), kJ/mol\n",
- "dGfCD = 2.90 #Std. Gibbs energy of formation for CaO (s), kJ/mol\n",
- "rhoG = 2.25e3 #Density of Graphite, kg/m3\n",
- "rhoD = 3.52e3 #Density of dimond, kg/m3\n",
- "T0 = 298.15 #Std. Temperature, K\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K) \n",
- "P0 = 1.0 #Pressure, bar\n",
- "M = 12.01 #Molceular wt of Carbon\n",
- "#Calculations\n",
- "P = P0*1e5 + dGfCD*1e3/((1./rhoG-1./rhoD)*M*1e-3)\n",
- "\n",
- "#Results \n",
- "print 'Pressure at which graphite and dimond will be in equilibrium is %4.2e bar'%(P/1e5)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure at which graphite and dimond will be in equilibrium is 1.51e+04 bar\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.14, Page Number 143"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "beta = 2.04e-4 #Thermal exapansion coefficient, /K\n",
- "kapa = 45.9e-6 #Isothermal compressibility, /bar\n",
- "T = 298.15 #Std. Temperature, K\n",
- "R = 8.206e-2 #Ideal gas constant, atm.L/(mol.K) \n",
- "T1 = 320.0 #Temperature, K\n",
- "Pi = 1.0 #Initial Pressure, bar\n",
- "V = 1.00 #Volume, m3\n",
- "a = 1.35 #van der Waals constant a for nitrogen, atm.L2/mol2\n",
- "\n",
- "#Calculations\n",
- "dUbydV = Pf = (beta*T1-kapa*P0)/kapa\n",
- "dVT = V*kapa*(Pf-Pi)\n",
- "dVbyV = dVT*100/V\n",
- "Vm = Pi/(R*T1)\n",
- "dUbydVm = a/(Vm**2)\n",
- "\n",
- "#Results \n",
- "print 'dUbydV = %4.2e bar'%(dUbydV)\n",
- "print 'dVbyV = %4.3f percent'%(dVbyV)\n",
- "print 'dUbydVm = %4.0e atm'%(dUbydVm)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dUbydV = 1.42e+03 bar\n",
- "dVbyV = 6.519 percent\n",
- "dUbydVm = 9e+02 atm\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.15, Page Number 144"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp, log\n",
- "\n",
- "#Variable Declaration\n",
- "m = 1000.0 #mass of mercury, g\n",
- "Pi, Ti = 1.00, 300.0 #Intial pressure and temperature, bar, K\n",
- "Pf, Tf = 300., 600.0 #Final pressure and temperature, bar, K\n",
- "rho = 13534. #Density of mercury, kg/m3\n",
- "beta = 18.1e-4 #Thermal exapansion coefficient for Hg, /K \n",
- "kapa = 3.91e-6 #Isothermal compressibility for Hg, /Pa\n",
- "Cpm = 27.98 #Molar Specific heat at constant pressure, J/(mol.K) \n",
- "M = 200.59 #Molecular wt of Hg, g/mol\n",
- "\n",
- "#Calculations\n",
- "Vi = m*1e-3/rho\n",
- "Vf = Vi*exp(-kapa*(Pf-Pi))\n",
- "Ut = m*Cpm*(Tf-Ti)/M \n",
- "Up = (beta*Ti/kapa-Pi)*1e5*(Vf-Vi) + (Vi-Vf+Vf*log(Vf/Vi))*1e5/kapa\n",
- "dU = Ut + Up\n",
- "Ht = m*Cpm*(Tf-Ti)/M\n",
- "Hp = ((1 + beta*(Tf-Ti))*Vi*exp(-kapa*Pi)/kapa)*(exp(-kapa*Pi)-exp(-kapa*Pf))\n",
- "dH = Ht + Hp\n",
- "#Results\n",
- "print 'Internal energy change is %6.2e J/mol in which \\ncontribution of temeprature dependent term %6.4f percent'%(dU,Ut*100/dH)\n",
- "print 'Enthalpy change is %4.3e J/mol in which \\ncontribution of temeprature dependent term %4.1f percent'%(dH,Ht*100/dH)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Internal energy change is 4.06e+04 J/mol in which \n",
- "contribution of temeprature dependent term 99.9999 percent\n",
- "Enthalpy change is 4.185e+04 J/mol in which \n",
- "contribution of temeprature dependent term 100.0 percent\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.16, Page Number 145"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "T = 300.0 #Temperature of Hg, K \n",
- "beta = 18.1e-4 #Thermal exapansion coefficient for Hg, /K \n",
- "kapa = 3.91e-6 #Isothermal compressibility for Hg, /Pa\n",
- "M = 0.20059 #Molecular wt of Hg, kg/mol \n",
- "rho = 13534 #Density of mercury, kg/m3\n",
- "Cpm = 27.98 #Experimental Molar specif heat at const pressure for mercury, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "Vm = M/rho\n",
- "DCpmCv = T*Vm*beta**2/kapa\n",
- "Cvm = Cpm - DCpmCv\n",
- "#Results\n",
- "print 'Difference in molar specific heats \\nat constant volume and constant pressure %4.2e J/(mol.K)'%DCpmCv\n",
- "print 'Molar Specific heat of Hg at const. volume is %4.2f J/(mol.K)'%Cvm"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference in molar specific heats \n",
- "at constant volume and constant pressure 3.73e-03 J/(mol.K)\n",
- "Molar Specific heat of Hg at const. volume is 27.98 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 6.17, Page Number 147"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "T = 298.15 #Std. Temperature, K \n",
- "P = 1.0 #Initial Pressure, bar\n",
- "Hm0, Sm0 = 0.0,154.8 #Std. molar enthalpy and entropy of Ar(g), kJ, mol, K units\n",
- "Sm0H2, Sm0O2 = 130.7,205.2 #Std. molar entropy of O2 and H2 (g), kJ/(mol.K)\n",
- "dGfH2O = -237.1 #Gibbs energy of formation for H2O(l), kJ/mol \n",
- "nH2, nO2 = 1, 1./2 #Stoichiomentric coefficients for H2 and O2 in water formation reaction \n",
- "\n",
- "#Calculations\n",
- "Gm0 = Hm0 - T*Sm0\n",
- "dGmH2O = dGfH2O*1000 - T*(nH2*Sm0H2 + nO2*Sm0O2)\n",
- "#Results\n",
- "print 'Molar Gibbs energy of Ar %4.3f kJ/mol'%(Gm0/1e3)\n",
- "print 'Molar Gibbs energy of Water %4.3f kJ/mol'%(dGmH2O/1e3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molar Gibbs energy of Ar -46.154 kJ/mol\n",
- "Molar Gibbs energy of Water -306.658 kJ/mol\n"
- ]
- }
- ],
- "prompt_number": 15
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07.ipynb deleted file mode 100755 index d8ce946b..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07.ipynb +++ /dev/null @@ -1,72 +0,0 @@ -{
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- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 07: Properties of Real Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 7.3, Page Number 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "m = 1.0 #Mass of Methane, kg\n",
- "T = 230 #Temeprature of Methane, K\n",
- "P = 68.0 #Pressure, bar \n",
- "Tc = 190.56 #Critical Temeprature of Methane\n",
- "Pc = 45.99 #Critical Pressure of Methane\n",
- "R = 0.08314 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "M = 16.04 #Molecular wt of Methane\n",
- "\n",
- "#Calcualtions\n",
- "Tr = T/Tc\n",
- "Pr = P/Pc\n",
- "z = 0.63 #Methane compressibility factor\n",
- "n = m*1e3/M\n",
- "V = z*n*R*T/P\n",
- "Vig = n*R*T/P\n",
- "DV = (V - Vig)/V\n",
- "\n",
- "#Results\n",
- "print '(V-Videal) %4.2f L'%(V-Vig)\n",
- "print 'Percentage error %5.2f'%(DV*100)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(V-Videal) -6.49 L\n",
- "Percentage error -58.73\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07_1.ipynb deleted file mode 100755 index d8ce946b..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07_1.ipynb +++ /dev/null @@ -1,72 +0,0 @@ -{
- "metadata": {
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- "nbformat": 3,
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- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 07: Properties of Real Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 7.3, Page Number 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "m = 1.0 #Mass of Methane, kg\n",
- "T = 230 #Temeprature of Methane, K\n",
- "P = 68.0 #Pressure, bar \n",
- "Tc = 190.56 #Critical Temeprature of Methane\n",
- "Pc = 45.99 #Critical Pressure of Methane\n",
- "R = 0.08314 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "M = 16.04 #Molecular wt of Methane\n",
- "\n",
- "#Calcualtions\n",
- "Tr = T/Tc\n",
- "Pr = P/Pc\n",
- "z = 0.63 #Methane compressibility factor\n",
- "n = m*1e3/M\n",
- "V = z*n*R*T/P\n",
- "Vig = n*R*T/P\n",
- "DV = (V - Vig)/V\n",
- "\n",
- "#Results\n",
- "print '(V-Videal) %4.2f L'%(V-Vig)\n",
- "print 'Percentage error %5.2f'%(DV*100)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(V-Videal) -6.49 L\n",
- "Percentage error -58.73\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07_2.ipynb deleted file mode 100755 index d8ce946b..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter07_2.ipynb +++ /dev/null @@ -1,72 +0,0 @@ -{
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- "signature": "sha256:b92f3dd9a653fad8ff6778b3bdb8771b55861aa679332e3055f634cfd8668cd9"
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- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 07: Properties of Real Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 7.3, Page Number 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "m = 1.0 #Mass of Methane, kg\n",
- "T = 230 #Temeprature of Methane, K\n",
- "P = 68.0 #Pressure, bar \n",
- "Tc = 190.56 #Critical Temeprature of Methane\n",
- "Pc = 45.99 #Critical Pressure of Methane\n",
- "R = 0.08314 #Ideal Gas Constant, L.bar/(mol.K)\n",
- "M = 16.04 #Molecular wt of Methane\n",
- "\n",
- "#Calcualtions\n",
- "Tr = T/Tc\n",
- "Pr = P/Pc\n",
- "z = 0.63 #Methane compressibility factor\n",
- "n = m*1e3/M\n",
- "V = z*n*R*T/P\n",
- "Vig = n*R*T/P\n",
- "DV = (V - Vig)/V\n",
- "\n",
- "#Results\n",
- "print '(V-Videal) %4.2f L'%(V-Vig)\n",
- "print 'Percentage error %5.2f'%(DV*100)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(V-Videal) -6.49 L\n",
- "Percentage error -58.73\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08.ipynb deleted file mode 100755 index 3725247b..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08.ipynb +++ /dev/null @@ -1,157 +0,0 @@ -{
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- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 08: Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.2, Page Number 186 Check"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Varialble Declaration\n",
- "nbp = 353.24 #normal boiling point of Benzene, K\n",
- "pv20 = 1.19e4 #Vapor pressure of benzene at 20\u00b0C, Pa\n",
- "DHf = 9.95 #Latent heat of fusion, kJ/mol\n",
- "pv443 = 137 #Vapor pressure of benzene at -44.3\u00b0C, Pa\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "Pf = 101325 #Pa\n",
- "T20 = 293.15\n",
- "Po = 1\n",
- "Pl = 10000\n",
- "Ts = -44.3\n",
- "#Calculations\n",
- "Ts = Ts + 273.15\n",
- "DHf = DHf*1e3\n",
- "DHv = -R*log(Pf/pv20)/(1/nbp-1/T20)\n",
- "DSv = DHv/nbp\n",
- "Ttp = -DHf/(R*(log(Pl/Po)-log(pv443/Po)-(DHv+DHf)/(R*Ts) + DHv/(R*T20)))\n",
- "\n",
- "#Results\n",
- "print 'Latent heat of vaporization of benzene at 20\u00b0C %5.2f J/mol'%DHv\n",
- "print 'Entropy Change of vaporization of benzene at 20\u00b0C %3.1f J/mol'%DSv\n",
- "print 'Triple point of benzene %4.1f K'%Ttp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Latent heat of vaporization of benzene at 20\u00b0C 30686.36 J/mol\n",
- "Entropy Change of vaporization of benzene at 20\u00b0C 86.9 J/mol\n",
- "Triple point of benzene 267.3 K\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.3, Page Number 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import cos, pi\n",
- "\n",
- "#Varialble Declaration\n",
- "gama = 71.99e-3 #Surface tension of water, N/m\n",
- "r = 1.2e-4 #Radius of hemisphere, m\n",
- "theta = 0.0 #Contact angle, rad\n",
- "\n",
- "#Calculations\n",
- "DP = 2*gama*cos(theta)/r\n",
- "F = DP*pi*r**2\n",
- "\n",
- "#Results\n",
- "print 'Force exerted by one leg %5.3e N'%F"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force exerted by one leg 5.428e-05 N\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.4, Page Number 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import cos\n",
- "\n",
- "#Varialble Declaration\n",
- "gama = 71.99e-3 #Surface tension of water, N/m\n",
- "r = 2e-5 #Radius of xylem, m\n",
- "theta = 0.0 #Contact angle, rad\n",
- "rho = 997.0 #Density of water, kg/m3\n",
- "g = 9.81 #gravitational acceleration, m/s2\n",
- "H = 100 #Height at top of redwood tree, m\n",
- "\n",
- "#Calculations\n",
- "h = 2*gama/(rho*g*r*cos(theta))\n",
- "\n",
- "#Results\n",
- "print 'Height to which water can rise by capillary action is %3.2f m'%h\n",
- "print 'This is very less than %4.1f n, hence water can not reach top of tree'%H"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Height to which water can rise by capillary action is 0.74 m\n",
- "This is very less than 100.0 n, hence water can not reach top of tree\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08_1.ipynb deleted file mode 100755 index 3725247b..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08_1.ipynb +++ /dev/null @@ -1,157 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f7a31615b3beb83b8b17b3507af2b7505643c9385577352b3394c24d378e5d0b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 08: Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.2, Page Number 186 Check"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Varialble Declaration\n",
- "nbp = 353.24 #normal boiling point of Benzene, K\n",
- "pv20 = 1.19e4 #Vapor pressure of benzene at 20\u00b0C, Pa\n",
- "DHf = 9.95 #Latent heat of fusion, kJ/mol\n",
- "pv443 = 137 #Vapor pressure of benzene at -44.3\u00b0C, Pa\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "Pf = 101325 #Pa\n",
- "T20 = 293.15\n",
- "Po = 1\n",
- "Pl = 10000\n",
- "Ts = -44.3\n",
- "#Calculations\n",
- "Ts = Ts + 273.15\n",
- "DHf = DHf*1e3\n",
- "DHv = -R*log(Pf/pv20)/(1/nbp-1/T20)\n",
- "DSv = DHv/nbp\n",
- "Ttp = -DHf/(R*(log(Pl/Po)-log(pv443/Po)-(DHv+DHf)/(R*Ts) + DHv/(R*T20)))\n",
- "\n",
- "#Results\n",
- "print 'Latent heat of vaporization of benzene at 20\u00b0C %5.2f J/mol'%DHv\n",
- "print 'Entropy Change of vaporization of benzene at 20\u00b0C %3.1f J/mol'%DSv\n",
- "print 'Triple point of benzene %4.1f K'%Ttp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Latent heat of vaporization of benzene at 20\u00b0C 30686.36 J/mol\n",
- "Entropy Change of vaporization of benzene at 20\u00b0C 86.9 J/mol\n",
- "Triple point of benzene 267.3 K\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.3, Page Number 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import cos, pi\n",
- "\n",
- "#Varialble Declaration\n",
- "gama = 71.99e-3 #Surface tension of water, N/m\n",
- "r = 1.2e-4 #Radius of hemisphere, m\n",
- "theta = 0.0 #Contact angle, rad\n",
- "\n",
- "#Calculations\n",
- "DP = 2*gama*cos(theta)/r\n",
- "F = DP*pi*r**2\n",
- "\n",
- "#Results\n",
- "print 'Force exerted by one leg %5.3e N'%F"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force exerted by one leg 5.428e-05 N\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.4, Page Number 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import cos\n",
- "\n",
- "#Varialble Declaration\n",
- "gama = 71.99e-3 #Surface tension of water, N/m\n",
- "r = 2e-5 #Radius of xylem, m\n",
- "theta = 0.0 #Contact angle, rad\n",
- "rho = 997.0 #Density of water, kg/m3\n",
- "g = 9.81 #gravitational acceleration, m/s2\n",
- "H = 100 #Height at top of redwood tree, m\n",
- "\n",
- "#Calculations\n",
- "h = 2*gama/(rho*g*r*cos(theta))\n",
- "\n",
- "#Results\n",
- "print 'Height to which water can rise by capillary action is %3.2f m'%h\n",
- "print 'This is very less than %4.1f n, hence water can not reach top of tree'%H"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Height to which water can rise by capillary action is 0.74 m\n",
- "This is very less than 100.0 n, hence water can not reach top of tree\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08_2.ipynb deleted file mode 100755 index 920a2998..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter08_2.ipynb +++ /dev/null @@ -1,167 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7cd73e591684a9b750091e874ca08a72b326eee20de62e33004e879fa5300f72"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 08: Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.2, Page Number 186 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, exp\n",
- "\n",
- "#Varialble Declaration\n",
- "Tn = 353.24 #normal boiling point of Benzene, K\n",
- "pi = 1.19e4 #Vapor pressure of benzene at 20\u00b0C, Pa\n",
- "DHf = 9.95 #Latent heat of fusion, kJ/mol\n",
- "pv443 = 137. #Vapor pressure of benzene at -44.3\u00b0C, Pa\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "Pf = 101325 #Std. atmospheric pressure, Pa\n",
- "T20 = 293.15 #Temperature in K\n",
- "P0 = 1.\n",
- "Pl = 10000.\n",
- "Ts = -44.3 #Temperature of solid benzene, \u00b0C\n",
- "\n",
- "#Calculations\n",
- "Ts = Ts + 273.15\n",
- "#Part a\n",
- "\n",
- "DHv = -(R*log(Pf/pi))/(1./Tn-1./T20)\n",
- "#Part b\n",
- "\n",
- "DSv = DHv/Tn\n",
- "DHf = DHf*1e3\n",
- "#Part c\n",
- "\n",
- "Ttp = -DHf/(R*(log(Pl/P0)-log(pv443/P0)-(DHv+DHf)/(R*Ts)+DHv/(R*T20)))\n",
- "Ptp = exp(-DHv/R*(1./Ttp-1./Tn))*101325\n",
- "\n",
- "#Results\n",
- "print 'Latent heat of vaporization of benzene at 20\u00b0C %4.1f kJ/mol'%(DHv/1000)\n",
- "print 'Entropy Change of vaporization of benzene at 20\u00b0C %3.1f J/mol'%DSv\n",
- "print 'Triple point temperature = %4.1f K for benzene'%Ttp\n",
- "print 'Triple point pressure = %4.2e Pa for benzene'%Ptp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Latent heat of vaporization of benzene at 20\u00b0C 30.7 kJ/mol\n",
- "Entropy Change of vaporization of benzene at 20\u00b0C 86.9 J/mol\n",
- "Triple point temperature = 267.3 K for benzene\n",
- "Triple point pressure = 3.53e+03 Pa for benzene\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.3, Page Number 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import cos, pi\n",
- "\n",
- "#Varialble Declaration\n",
- "gama = 71.99e-3 #Surface tension of water, N/m\n",
- "r = 1.2e-4 #Radius of hemisphere, m\n",
- "theta = 0.0 #Contact angle, rad\n",
- "\n",
- "#Calculations\n",
- "DP = 2*gama*cos(theta)/r\n",
- "F = DP*pi*r**2\n",
- "\n",
- "#Results\n",
- "print 'Force exerted by one leg %5.3e N'%F"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force exerted by one leg 5.428e-05 N\n"
- ]
- }
- ],
- "prompt_number": 40
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 8.4, Page Number 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import cos\n",
- "\n",
- "#Varialble Declaration\n",
- "gama = 71.99e-3 #Surface tension of water, N/m\n",
- "r = 2e-5 #Radius of xylem, m\n",
- "theta = 0.0 #Contact angle, rad\n",
- "rho = 997.0 #Density of water, kg/m3\n",
- "g = 9.81 #gravitational acceleration, m/s2\n",
- "H = 100 #Height at top of redwood tree, m\n",
- "\n",
- "#Calculations\n",
- "h = 2*gama/(rho*g*r*cos(theta))\n",
- "\n",
- "#Results\n",
- "print 'Height to which water can rise by capillary action is %3.2f m'%h\n",
- "print 'This is very less than %4.1f n, hence water can not reach top of tree'%H"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Height to which water can rise by capillary action is 0.74 m\n",
- "This is very less than 100.0 n, hence water can not reach top of tree\n"
- ]
- }
- ],
- "prompt_number": 41
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09.ipynb deleted file mode 100755 index 3141af69..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09.ipynb +++ /dev/null @@ -1,545 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c2f0eb9f813e96ecfb0ebcea9c89faad6a2b8ee530cc39d385b0a4ab04407ef5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 09: Ideal and Real Solutions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.2, Page Number 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "P = 1.0 #Pressure, bar\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "xb = nb/n\n",
- "xt = 1. - xb\n",
- "dGmix = n*R*T*(xb*log(xb)+xt*log(xt))\n",
- "dSmix = n*R*(xb*log(xb)+xt*log(xt))\n",
- "\n",
- "#Results\n",
- "print 'Gibbs energy change of mixing is %4.3e J'%dGmix\n",
- "print 'Gibbs energy change of mixing is < 0, hence the mixing is spontaneous'\n",
- "print 'Entropy change of mixing is %4.2f J/K'%dSmix"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Gibbs energy change of mixing is -1.371e+04 J\n",
- "Gibbs energy change of mixing is < 0, hence the mixing is spontaneous\n",
- "Entropy change of mixing is -45.99 J/K\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.3, Page Number 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "P0b = 96.4 #Vapor pressure of Benzene, torr\n",
- "P0t = 28.9 #Vapor pressure of Toluene, torr\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "xb = nb/n\n",
- "xt = 1. - xb\n",
- "P = xb*P0b + xt*P0t\n",
- "y = (P0b*P - P0t*P0b)/(P*(P0b-P0t))\n",
- "yt = 1.-yb\n",
- "\n",
- "#Results\n",
- "print 'Total pressure of the vapor is %4.1f torr'%P\n",
- "print 'Benzene fraction in vapor is %4.3f '%yb\n",
- "print 'Toulene fraction in vapor is %4.3f '%yt"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total pressure of the vapor is 69.8 torr\n",
- "Benzene fraction in vapor is 0.837 \n",
- "Toulene fraction in vapor is 0.163 \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.4, Page Number 206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "P0b = 96.4 #Vapor pressure of Benzene, torr\n",
- "P0t = 28.9 #Vapor pressure of Toluene, torr\n",
- "nv = 1.5 #moles vaporized, mol\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "nl = n - nv\n",
- "zb = nb/n\n",
- "\n",
- "x,y, P = symbols('x y P')\n",
- "e1 = nv*(y-zb)-nl*(zb-x)\n",
- "print 'Mass Balance:', e1\n",
- "e2 = P - (x*P0b + (1-x)*P0t)\n",
- "print 'Pressure and x:',e2\n",
- "e3 = y - (P0b*P - P0t*P0b)/(P*(P0b-P0t))\n",
- "print 'Pressure and y:', e3\n",
- "equations = [e1,e2,e3]\n",
- "sol = solve(equations)\n",
- "\n",
- "#Results\n",
- "for i in sol:\n",
- " if ((i[x] > 0.0 and i[x] <1.0) and (i[P] > 0.0) and (i[y]>zb and i[y]<1.0)):\n",
- " print 'Pressure is %4.2f torr' %i[P]\n",
- " print 'Mole fraction of benzene in liquid phase %4.3f' %i[x]\n",
- " print 'Mole fraction of benzene in vapor phase %4.3f' %i[y]\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mass Balance: 6.75*x + 1.5*y - 5.0\n",
- "Pressure and x: P - 67.5*x - 28.9\n",
- "Pressure and y: y - 0.0148148148148148*(96.4*P - 2785.96)/P\n",
- "Pressure is 66.75 torr"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Mole fraction of benzene in liquid phase 0.561\n",
- "Mole fraction of benzene in vapor phase 0.810\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.6, Page Number 212"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 4.50 #Mass of substance dissolved, g\n",
- "ms = 125.0 #Mass of slovent (CCl4), g\n",
- "TbE = 0.65 #Boiling point elevation, \u00b0C\n",
- "Kf, Kb = 30.0, 4.95 #Constants for freezing point elevation \n",
- " # and boiling point depression for CCl4, K kg/mol\n",
- "Msolvent = 153.8 #Molecualr wt of solvent, g/mol\n",
- "#Calculations\n",
- "DTf = -Kf*TbE/Kb\n",
- "Msolute = Kb*m/(ms*1e-3*TbE)\n",
- "nsolute = m/Msolute\n",
- "nsolvent = ms/Msolvent \n",
- "x = 1.0 - nsolute/(nsolute + nsolvent)\n",
- "\n",
- "#Results\n",
- "print 'Freezing point depression %5.2f K'%DTf\n",
- "print 'Molecualr wt of solute %4.1f g/mol'%Msolute\n",
- "print 'Vapor pressure of solvent is reduced by a factor of %4.3f'%x"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Freezing point depression -3.94 K\n",
- "Molecualr wt of solute 274.2 g/mol\n",
- "Vapor pressure of solvent is reduced by a factor of 0.980\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.7, Page Number 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "csolute = 0.500 #Concentration of solute, g/L\n",
- "R = 8.206e-2 #Gas constant L.atm/(mol.K)\n",
- "T = 298.15 #Temperature of the solution, K\n",
- "\n",
- "#Calculations\n",
- "pii = csolute*R*T\n",
- "\n",
- "#Results\n",
- "print 'Osmotic pressure %4.2f atm'%pii\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Osmotic pressure 12.23 atm\n"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.8, Page Number 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "xCS2 = 0.3502 #Mol fraction of CS2, g/L\n",
- "pCS2 = 358.3 #Partial pressure of CS2, torr\n",
- "p0CS2 = 512.3 #Total pressure, torr\n",
- "\n",
- "#Calculations\n",
- "alpha = pCS2/p0CS2\n",
- "gama = alpha/xCS2\n",
- "\n",
- "#Results\n",
- "print 'Activity of CS2 %5.4f atm'%alpha\n",
- "print 'Activity coefficinet of CS2 %5.4f atm'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity of CS2 0.6994 atm\n",
- "Activity coefficinet of CS2 1.9971 atm\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.9, Page Number 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "xCS2 = 0.3502 #Mol fraction of CS2, g/L\n",
- "pCS2 = 358.3 #Partial pressure of CS2, torr\n",
- "kHCS2 = 2010. #Total pressure, torr\n",
- "\n",
- "#Calculations\n",
- "alpha = pCS2/kHCS2\n",
- "gama = alpha/xCS2\n",
- "\n",
- "#Results\n",
- "print 'Activity of CS2 %5.4f atm'%alpha\n",
- "print 'Activity coefficinet of CS2 %5.4f atm'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity of CS2 0.1783 atm\n",
- "Activity coefficinet of CS2 0.5090 atm\n"
- ]
- }
- ],
- "prompt_number": 73
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.10, Page Number 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "rho = 789.9 #Density of acetone, g/L\n",
- "n = 1.0 #moles of acetone, mol\n",
- "M = 58.08 #Molecular wt of acetone, g/mol\n",
- "kHacetone = 1950 #Henrys law constant, torr\n",
- "#Calculations\n",
- "H = n*M*kHacetone/rho\n",
- "\n",
- "#Results\n",
- "print 'Henrys constant = %5.2f torr'%H"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Henrys constant = 143.38 torr\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.11, Page Number 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 0.5 #Mass of water, kg\n",
- "ms = 24.0 #Mass of solute, g\n",
- "Ms = 241.0 #Molecular wt of solute, g/mol\n",
- "Tfd = 0.359 #Freezinf point depression, \u00b0C or K\n",
- "kf = 1.86 #Constants for freezing point depression for water, K kg/mol\n",
- "\n",
- "#Calculations\n",
- "msolute = ms/(Ms*m)\n",
- "gama = Tfd/(kf*msolute)\n",
- "\n",
- "#Results\n",
- "print 'Activity coefficient = %4.3f'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity coefficient = 0.969\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.12, Page Number 223"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 70.0 #Mass of human body, kg\n",
- "V = 5.00 #Volume of blood, L\n",
- "HN2 = 9.04e4 #Henry law constant for N2 solubility in blood, bar\n",
- "T = 298.0 #Temperature, K\n",
- "rho = 1.00 #density of blood, kg/L\n",
- "Mw = 18.02 #Molecualr wt of water, g/mol\n",
- "X = 80 #Percent of N2 at sea level\n",
- "p1, p2 = 1.0, 50.0 #Pressures, bar\n",
- "R = 8.314e-2 #Ideal Gas constant, L.bar/(mol.K)\n",
- "#Calculations\n",
- "nN21 = (V*rho*1e3/Mw)*(p1*X/100)/HN2\n",
- "nN22 = (V*rho*1e3/Mw)*(p2*X/100)/HN2\n",
- "V = (nN22-nN21)*R*T/p1\n",
- "#Results\n",
- "print 'Number of moles of nitrogen in blood at 1 and 50 bar are %3.2e,%3.3f mol'%(nN21,nN22)\n",
- "print 'Volume of nitrogen released from blood at reduced pressure %4.3f L'%V"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of moles of nitrogen in blood at 1 and 50 bar are 2.46e-03,0.123 mol\n",
- "Volume of nitrogen released from blood at reduced pressure 2.981 L\n"
- ]
- }
- ],
- "prompt_number": 90
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.14, Page Number 226"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg, divide\n",
- "from pylab import plot,show\n",
- "\n",
- "#Variable Declaration\n",
- "cCB = array([1e-6,2e-6,3e-6,5e-6,10e-6])\n",
- "nu = array([0.006,0.012,0.018,0.028,0.052])\n",
- "y = nu/cCB\n",
- "print y\n",
- "xlim(0.0, 0.06)\n",
- "ylim(5000,6300)\n",
- "#Calculations\n",
- "A = array([ nu, ones(size(nu))])\n",
- "print A\n",
- "# linearly generated sequence\n",
- "\n",
- "w = linalg.lstsq(A.T,y)[0] # obtaining the parameters\n",
- "print 'slope %8.1f'%w[0]\n",
- "print 'Intercept %8.1f' %w[1]\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = w[0]*nu+w[1] # regression line\n",
- " \n",
- "#Results\n",
- "plot(nu,line,'r-',nu,y,'o')\n",
- "show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "[ 6000. 6000. 6000. 5600. 5200.]\n",
- "[[ 0.006 0.012 0.018 0.028 0.052]\n",
- " [ 1. 1. 1. 1. 1. ]]"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "slope -19188.2\n",
- "Intercept 6205.2\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
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MlievdHc3M6+f8EREJBvMPfXPcTMbB2wELiRcR1gXlYBecPdOZjYKwN1viLaf\nDowD3o626Ry1nw30d/eLqxxfSUVEpA7cvWo5v1Z2eWZgZs2BAnf/l5m1AAYC44GngHOBG6PvT0a7\nPAU8Yma3EcpAHYGK6Oxhg5n1AiqAc4CyTHdGRETqpqYyUSvgieiGoELgj+4+w8zmAY+Z2VBgNXAG\ngLtXmtljQCWwFRjm/z71GAY8AOwOTHP36Rnui4iI1FGtykQiIpKfsjYC2cwGRQPRVpjZz75mm7Jo\n/UIz61KbfeOWZv/uM7P3zWxx9iJOXV37ZmbtzOwFM1tqZkvMrCS7kacmjf4VmdkcM1tgZpVmdn12\nI09NOr+b0bqCaNDp09mJuHbS/L/3H4Nqc0mafdvLzCab2bLo97P3Lt/M3ev9CyggjDk4ANgNWAB0\nrrLNyYTyEUAv4PVU9437K53+Ra/7Al2AxXH3JcM/u28BR0XLewB/zcOfXfPoeyHwOnBM3H3KZP+i\ntiuAPwJPxd2fevj5vQXsE3c/6qlvDwIXJP1+7rmr98vWmUFPYKW7r3b3LcCfCAPUku0cyObuc4C9\nzOxbKe4bt3T6h7vPBv6ZxXhro659a+Xu69x9QdS+EVhGGHOSS+rcv+j159E2TQn/eT/OStSpS6t/\nZtaW8IFzL/85+DQXpNW/SC72C9Lom5ntCfR19/uidVvd/dNdvVm2kkEb4J2k1zsGo6WyzX4p7Bu3\ndPqX6+rat7bJG5jZAYSznzkZjzA9afUvKqEsIAy+fMHdK+sx1rpI93fzdmAksL2+AkxTuv2rblBt\nrkjnd7M9sN7M7jezN83snuju0K+VrWSQ6lXqXM3QNalr/xrC1fu0+2ZmewCTgRHRGUIuSat/7r7N\n3Y8i/AfsZ2aJDMaWCXXtn5nZqcAH7j6/mvW5It3PlmPcvQtwEnCJmfXNTFgZkc7vZiHQFbjL3bsC\nnxHNIfd1spUM1gLtkl6346vTU1S3Tdtom1T2jVtd+7e2nuPKhLT6Zma7AX8BHnb3J8k9GfnZRafg\nU4Hu9RBjOtLp33eA083sLWASMMDMHqrHWOsirZ+fu78bfV8PPEEozeSKdPq2Bljj7nOj9smE5PD1\nsnQhpBBYRbgQ0pSaL4T05t8XIWvcN+6vdPqXtP4AcvMCcjo/OyNMSnh73P2op/59E9grWt4deAk4\nLu4+Zfp3M2rvDzwdd38y/PNrTphAE6AF8AowMO4+ZepnF/0+HhwtlwI37vL9stixkwh3k6wERkdt\nFwEXJW2QtLXXAAAAj0lEQVRzZ7R+IdB1V/vm2lea/ZsEvAtsJtT/zo+7P5noG3AModa8AJgffQ2K\nuz8Z7N/hwJtR/xYBI+PuS6Z/N5PW9ycH7yZK8+d3YPSzWwAsycXPljQ/V44E5kbtj1PD3UQadCYi\nInrspYiIKBmIiAhKBiIigpKBiIigZCAiIigZiIgISgYiIoKSgYiIAP8PBAcNAieTgSEAAAAASUVO\nRK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x59f38d0>"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": []
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09_1.ipynb deleted file mode 100755 index fab68570..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09_1.ipynb +++ /dev/null @@ -1,537 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:046710f7ffcadc799ac5e6b952be69b28c410bd2f124592b0fc1c1a580da7ec0"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 09: Ideal and Real Solutions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.2, Page Number 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "P = 1.0 #Pressure, bar\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "xb = nb/n\n",
- "xt = 1. - xb\n",
- "dGmix = n*R*T*(xb*log(xb)+xt*log(xt))\n",
- "dSmix = n*R*(xb*log(xb)+xt*log(xt))\n",
- "\n",
- "#Results\n",
- "print 'Gibbs energy change of mixing is %4.3e J'%dGmix\n",
- "print 'Gibbs energy change of mixing is < 0, hence the mixing is spontaneous'\n",
- "print 'Entropy change of mixing is %4.2f J/K'%dSmix"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Gibbs energy change of mixing is -1.371e+04 J\n",
- "Gibbs energy change of mixing is < 0, hence the mixing is spontaneous\n",
- "Entropy change of mixing is -45.99 J/K\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.3, Page Number 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "P0b = 96.4 #Vapor pressure of Benzene, torr\n",
- "P0t = 28.9 #Vapor pressure of Toluene, torr\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "xb = nb/n\n",
- "xt = 1. - xb\n",
- "P = xb*P0b + xt*P0t\n",
- "y = (P0b*P - P0t*P0b)/(P*(P0b-P0t))\n",
- "yt = 1.-yb\n",
- "\n",
- "#Results\n",
- "print 'Total pressure of the vapor is %4.1f torr'%P\n",
- "print 'Benzene fraction in vapor is %4.3f '%yb\n",
- "print 'Toulene fraction in vapor is %4.3f '%yt"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total pressure of the vapor is 69.8 torr\n",
- "Benzene fraction in vapor is 0.837 \n",
- "Toulene fraction in vapor is 0.163 \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.4, Page Number 206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "P0b = 96.4 #Vapor pressure of Benzene, torr\n",
- "P0t = 28.9 #Vapor pressure of Toluene, torr\n",
- "nv = 1.5 #moles vaporized, mol\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "nl = n - nv\n",
- "zb = nb/n\n",
- "\n",
- "x,y, P = symbols('x y P')\n",
- "e1 = nv*(y-zb)-nl*(zb-x)\n",
- "print 'Mass Balance:', e1\n",
- "e2 = P - (x*P0b + (1-x)*P0t)\n",
- "print 'Pressure and x:',e2\n",
- "e3 = y - (P0b*P - P0t*P0b)/(P*(P0b-P0t))\n",
- "print 'Pressure and y:', e3\n",
- "equations = [e1,e2,e3]\n",
- "sol = solve(equations)\n",
- "\n",
- "#Results\n",
- "for i in sol:\n",
- " if ((i[x] > 0.0 and i[x] <1.0) and (i[P] > 0.0) and (i[y]>zb and i[y]<1.0)):\n",
- " print 'Pressure is %4.2f torr' %i[P]\n",
- " print 'Mole fraction of benzene in liquid phase %4.3f' %i[x]\n",
- " print 'Mole fraction of benzene in vapor phase %4.3f' %i[y]\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mass Balance: 6.75*x + 1.5*y - 5.0\n",
- "Pressure and x: P - 67.5*x - 28.9\n",
- "Pressure and y: y - 0.0148148148148148*(96.4*P - 2785.96)/P\n",
- "Pressure is 66.75 torr"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Mole fraction of benzene in liquid phase 0.561\n",
- "Mole fraction of benzene in vapor phase 0.810\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.6, Page Number 212"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 4.50 #Mass of substance dissolved, g\n",
- "ms = 125.0 #Mass of slovent (CCl4), g\n",
- "TbE = 0.65 #Boiling point elevation, \u00b0C\n",
- "Kf, Kb = 30.0, 4.95 #Constants for freezing point elevation \n",
- " # and boiling point depression for CCl4, K kg/mol\n",
- "Msolvent = 153.8 #Molecualr wt of solvent, g/mol\n",
- "#Calculations\n",
- "DTf = -Kf*TbE/Kb\n",
- "Msolute = Kb*m/(ms*1e-3*TbE)\n",
- "nsolute = m/Msolute\n",
- "nsolvent = ms/Msolvent \n",
- "x = 1.0 - nsolute/(nsolute + nsolvent)\n",
- "\n",
- "#Results\n",
- "print 'Freezing point depression %5.2f K'%DTf\n",
- "print 'Molecualr wt of solute %4.1f g/mol'%Msolute\n",
- "print 'Vapor pressure of solvent is reduced by a factor of %4.3f'%x"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Freezing point depression -3.94 K\n",
- "Molecualr wt of solute 274.2 g/mol\n",
- "Vapor pressure of solvent is reduced by a factor of 0.980\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.7, Page Number 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "csolute = 0.500 #Concentration of solute, g/L\n",
- "R = 8.206e-2 #Gas constant L.atm/(mol.K)\n",
- "T = 298.15 #Temperature of the solution, K\n",
- "\n",
- "#Calculations\n",
- "pii = csolute*R*T\n",
- "\n",
- "#Results\n",
- "print 'Osmotic pressure %4.2f atm'%pii\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Osmotic pressure 12.23 atm\n"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.8, Page Number 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "xCS2 = 0.3502 #Mol fraction of CS2, g/L\n",
- "pCS2 = 358.3 #Partial pressure of CS2, torr\n",
- "p0CS2 = 512.3 #Total pressure, torr\n",
- "\n",
- "#Calculations\n",
- "alpha = pCS2/p0CS2\n",
- "gama = alpha/xCS2\n",
- "\n",
- "#Results\n",
- "print 'Activity of CS2 %5.4f atm'%alpha\n",
- "print 'Activity coefficinet of CS2 %5.4f atm'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity of CS2 0.6994 atm\n",
- "Activity coefficinet of CS2 1.9971 atm\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.9, Page Number 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "xCS2 = 0.3502 #Mol fraction of CS2, g/L\n",
- "pCS2 = 358.3 #Partial pressure of CS2, torr\n",
- "kHCS2 = 2010. #Total pressure, torr\n",
- "\n",
- "#Calculations\n",
- "alpha = pCS2/kHCS2\n",
- "gama = alpha/xCS2\n",
- "\n",
- "#Results\n",
- "print 'Activity of CS2 %5.4f atm'%alpha\n",
- "print 'Activity coefficinet of CS2 %5.4f atm'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity of CS2 0.1783 atm\n",
- "Activity coefficinet of CS2 0.5090 atm\n"
- ]
- }
- ],
- "prompt_number": 73
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.10, Page Number 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "rho = 789.9 #Density of acetone, g/L\n",
- "n = 1.0 #moles of acetone, mol\n",
- "M = 58.08 #Molecular wt of acetone, g/mol\n",
- "kHacetone = 1950 #Henrys law constant, torr\n",
- "#Calculations\n",
- "H = n*M*kHacetone/rho\n",
- "\n",
- "#Results\n",
- "print 'Henrys constant = %5.2f torr'%H"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Henrys constant = 143.38 torr\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.11, Page Number 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 0.5 #Mass of water, kg\n",
- "ms = 24.0 #Mass of solute, g\n",
- "Ms = 241.0 #Molecular wt of solute, g/mol\n",
- "Tfd = 0.359 #Freezinf point depression, \u00b0C or K\n",
- "kf = 1.86 #Constants for freezing point depression for water, K kg/mol\n",
- "\n",
- "#Calculations\n",
- "msolute = ms/(Ms*m)\n",
- "gama = Tfd/(kf*msolute)\n",
- "\n",
- "#Results\n",
- "print 'Activity coefficient = %4.3f'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity coefficient = 0.969\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.12, Page Number 223"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 70.0 #Mass of human body, kg\n",
- "V = 5.00 #Volume of blood, L\n",
- "HN2 = 9.04e4 #Henry law constant for N2 solubility in blood, bar\n",
- "T = 298.0 #Temperature, K\n",
- "rho = 1.00 #density of blood, kg/L\n",
- "Mw = 18.02 #Molecualr wt of water, g/mol\n",
- "X = 80 #Percent of N2 at sea level\n",
- "p1, p2 = 1.0, 50.0 #Pressures, bar\n",
- "R = 8.314e-2 #Ideal Gas constant, L.bar/(mol.K)\n",
- "#Calculations\n",
- "nN21 = (V*rho*1e3/Mw)*(p1*X/100)/HN2\n",
- "nN22 = (V*rho*1e3/Mw)*(p2*X/100)/HN2\n",
- "V = (nN22-nN21)*R*T/p1\n",
- "#Results\n",
- "print 'Number of moles of nitrogen in blood at 1 and 50 bar are %3.2e,%3.3f mol'%(nN21,nN22)\n",
- "print 'Volume of nitrogen released from blood at reduced pressure %4.3f L'%V"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of moles of nitrogen in blood at 1 and 50 bar are 2.46e-03,0.123 mol\n",
- "Volume of nitrogen released from blood at reduced pressure 2.981 L\n"
- ]
- }
- ],
- "prompt_number": 90
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.14, Page Number 226"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg, divide\n",
- "from pylab import plot,show\n",
- "\n",
- "#Variable Declaration\n",
- "cCB = array([1e-6,2e-6,3e-6,5e-6,10e-6])\n",
- "nu = array([0.006,0.012,0.018,0.028,0.052])\n",
- "y = nu/cCB\n",
- "print y\n",
- "xlim(0.0, 0.06)\n",
- "ylim(5000,6300)\n",
- "#Calculations\n",
- "A = array([ nu, ones(size(nu))])\n",
- "print A\n",
- "# linearly generated sequence\n",
- "\n",
- "w = linalg.lstsq(A.T,y)[0] # obtaining the parameters\n",
- "print 'slope %8.1f'%w[0]\n",
- "print 'Intercept %8.1f' %w[1]\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = w[0]*nu+w[1] # regression line\n",
- " \n",
- "#Results\n",
- "plot(nu,line,'r-',nu,y,'o')\n",
- "show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "[ 6000. 6000. 6000. 5600. 5200.]\n",
- "[[ 0.006 0.012 0.018 0.028 0.052]\n",
- " [ 1. 1. 1. 1. 1. ]]"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "slope -19188.2\n",
- "Intercept 6205.2\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Utm3cUYnUmq4ZiKTDDM44I5SOOnSAI4+EG26AzZvjjkwk65QMRFq0gF/9Ktx19OqrYVbU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\nRK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x59f38d0>"
- ]
- }
- ],
- "prompt_number": 16
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09_2.ipynb deleted file mode 100755 index 263d750b..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter09_2.ipynb +++ /dev/null @@ -1,540 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ada203a80b82700a1d56c68f95bd6837a954d4e31eab856fecd94550e07ac6e1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 09: Ideal and Real Solutions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.2, Page Number 202"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "P = 1.0 #Pressure, bar\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "xb = nb/n\n",
- "xt = 1. - xb\n",
- "dGmix = n*R*T*(xb*log(xb)+xt*log(xt))\n",
- "dSmix = -n*R*(xb*log(xb)+xt*log(xt))\n",
- "\n",
- "#Results\n",
- "print 'Gibbs energy change of mixing is %4.3e J'%dGmix\n",
- "print 'Gibbs energy change of mixing is < 0, hence the mixing is spontaneous'\n",
- "print 'Entropy change of mixing is %4.2f J/K'%dSmix"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Gibbs energy change of mixing is -1.371e+04 J\n",
- "Gibbs energy change of mixing is < 0, hence the mixing is spontaneous\n",
- "Entropy change of mixing is 45.99 J/K\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.3, Page Number 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "P0b = 96.4 #Vapor pressure of Benzene, torr\n",
- "P0t = 28.9 #Vapor pressure of Toluene, torr\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "xb = nb/n\n",
- "xt = 1. - xb\n",
- "P = xb*P0b + xt*P0t\n",
- "y = (P0b*P - P0t*P0b)/(P*(P0b-P0t))\n",
- "yt = 1.-yb\n",
- "\n",
- "#Results\n",
- "print 'Total pressure of the vapor is %4.1f torr'%P\n",
- "print 'Benzene fraction in vapor is %4.3f '%yb\n",
- "print 'Toulene fraction in vapor is %4.3f '%yt"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total pressure of the vapor is 69.8 torr\n",
- "Benzene fraction in vapor is 0.837 \n",
- "Toulene fraction in vapor is 0.163 \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.4, Page Number 206"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "#Variable Declaration\n",
- "nb = 5.00 #Number of moles of Benzene, mol\n",
- "nt = 3.25 #Number of moles of Toluene, mol\n",
- "T = 298.15 #Temperature, K\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "P0b = 96.4 #Vapor pressure of Benzene, torr\n",
- "P0t = 28.9 #Vapor pressure of Toluene, torr\n",
- "nv = 1.5 #moles vaporized, mol\n",
- "\n",
- "#Calculations\n",
- "n = nb + nt\n",
- "nl = n - nv\n",
- "zb = nb/n\n",
- "\n",
- "x,y, P = symbols('x y P')\n",
- "e1 = nv*(y-zb)-nl*(zb-x)\n",
- "print 'Mass Balance:', e1\n",
- "e2 = P - (x*P0b + (1-x)*P0t)\n",
- "print 'Pressure and x:',e2\n",
- "e3 = y - (P0b*P - P0t*P0b)/(P*(P0b-P0t))\n",
- "print 'Pressure and y:', e3\n",
- "equations = [e1,e2,e3]\n",
- "sol = solve(equations)\n",
- "\n",
- "#Results\n",
- "for i in sol:\n",
- " if ((i[x] > 0.0 and i[x] <1.0) and (i[P] > 0.0) and (i[y]>zb and i[y]<1.0)):\n",
- " print 'Pressure is %4.1f torr' %i[P]\n",
- " print 'Mole fraction of benzene in liquid phase %4.3f' %i[x]\n",
- " print 'Mole fraction of benzene in vapor phase %4.3f' %i[y]\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mass Balance: 6.75*x + 1.5*y - 5.0\n",
- "Pressure and x: P - 67.5*x - 28.9\n",
- "Pressure and y: y - 0.0148148148148148*(96.4*P - 2785.96)/P\n",
- "Pressure is 66.8 torr"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Mole fraction of benzene in liquid phase 0.561\n",
- "Mole fraction of benzene in vapor phase 0.810\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.6, Page Number 212"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 4.50 #Mass of substance dissolved, g\n",
- "ms = 125.0 #Mass of slovent (CCl4), g\n",
- "TbE = 0.65 #Boiling point elevation, \u00b0C\n",
- "Kf, Kb = 30.0, 4.95 #Constants for freezing point elevation \n",
- " # and boiling point depression for CCl4, K kg/mol\n",
- "Msolvent = 153.8 #Molecualr wt of solvent, g/mol\n",
- "#Calculations\n",
- "DTf = -Kf*TbE/Kb\n",
- "Msolute = Kb*m/(ms*1e-3*TbE)\n",
- "nsolute = m/Msolute\n",
- "nsolvent = ms/Msolvent \n",
- "x = 1.0 - nsolute/(nsolute + nsolvent)\n",
- "\n",
- "#Results\n",
- "print 'Freezing point depression %5.2f K'%DTf\n",
- "print 'Molecualr wt of solute %4.1f g/mol'%Msolute\n",
- "print 'Vapor pressure of solvent is reduced by a factor of %4.3f'%x"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Freezing point depression -3.94 K\n",
- "Molecualr wt of solute 274.2 g/mol\n",
- "Vapor pressure of solvent is reduced by a factor of 0.980\n"
- ]
- }
- ],
- "prompt_number": 69
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.7, Page Number 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "csolute = 0.500 #Concentration of solute, g/L\n",
- "R = 8.206e-2 #Gas constant L.atm/(mol.K)\n",
- "T = 298.15 #Temperature of the solution, K\n",
- "\n",
- "#Calculations\n",
- "pii = csolute*R*T\n",
- "\n",
- "#Results\n",
- "print 'Osmotic pressure %4.2f atm'%pii\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Osmotic pressure 12.23 atm\n"
- ]
- }
- ],
- "prompt_number": 70
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.8, Page Number 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "xCS2 = 0.3502 #Mol fraction of CS2, g/L\n",
- "pCS2 = 358.3 #Partial pressure of CS2, torr\n",
- "p0CS2 = 512.3 #Total pressure, torr\n",
- "\n",
- "#Calculations\n",
- "alpha = pCS2/p0CS2\n",
- "gama = alpha/xCS2\n",
- "\n",
- "#Results\n",
- "print 'Activity of CS2 %5.4f atm'%alpha\n",
- "print 'Activity coefficinet of CS2 %5.4f atm'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity of CS2 0.6994 atm\n",
- "Activity coefficinet of CS2 1.9971 atm\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.9, Page Number 220"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "xCS2 = 0.3502 #Mol fraction of CS2, g/L\n",
- "pCS2 = 358.3 #Partial pressure of CS2, torr\n",
- "kHCS2 = 2010. #Total pressure, torr\n",
- "\n",
- "#Calculations\n",
- "alpha = pCS2/kHCS2\n",
- "gama = alpha/xCS2\n",
- "\n",
- "#Results\n",
- "print 'Activity of CS2 %5.4f atm'%alpha\n",
- "print 'Activity coefficinet of CS2 %5.4f atm'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity of CS2 0.1783 atm\n",
- "Activity coefficinet of CS2 0.5090 atm\n"
- ]
- }
- ],
- "prompt_number": 73
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.10, Page Number 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "rho = 789.9 #Density of acetone, g/L\n",
- "n = 1.0 #moles of acetone, mol\n",
- "M = 58.08 #Molecular wt of acetone, g/mol\n",
- "kHacetone = 1950 #Henrys law constant, torr\n",
- "#Calculations\n",
- "H = n*M*kHacetone/rho\n",
- "\n",
- "#Results\n",
- "print 'Henrys constant = %5.2f torr'%H"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Henrys constant = 143.38 torr\n"
- ]
- }
- ],
- "prompt_number": 76
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.11, Page Number 221"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 0.5 #Mass of water, kg\n",
- "ms = 24.0 #Mass of solute, g\n",
- "Ms = 241.0 #Molecular wt of solute, g/mol\n",
- "Tfd = 0.359 #Freezinf point depression, \u00b0C or K\n",
- "kf = 1.86 #Constants for freezing point depression for water, K kg/mol\n",
- "\n",
- "#Calculations\n",
- "msolute = ms/(Ms*m)\n",
- "gama = Tfd/(kf*msolute)\n",
- "\n",
- "#Results\n",
- "print 'Activity coefficient = %4.3f'%gama"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activity coefficient = 0.969\n"
- ]
- }
- ],
- "prompt_number": 81
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.12, Page Number 223"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "m = 70.0 #Mass of human body, kg\n",
- "V = 5.00 #Volume of blood, L\n",
- "HN2 = 9.04e4 #Henry law constant for N2 solubility in blood, bar\n",
- "T = 298.0 #Temperature, K\n",
- "rho = 1.00 #density of blood, kg/L\n",
- "Mw = 18.02 #Molecualr wt of water, g/mol\n",
- "X = 80 #Percent of N2 at sea level\n",
- "p1, p2 = 1.0, 50.0 #Pressures, bar\n",
- "R = 8.314e-2 #Ideal Gas constant, L.bar/(mol.K)\n",
- "#Calculations\n",
- "nN21 = (V*rho*1e3/Mw)*(p1*X/100)/HN2\n",
- "nN22 = (V*rho*1e3/Mw)*(p2*X/100)/HN2\n",
- "V = (nN22-nN21)*R*T/p1\n",
- "#Results\n",
- "print 'Number of moles of nitrogen in blood at 1 and 50 bar are %3.2e,%3.3f mol'%(nN21,nN22)\n",
- "print 'Volume of nitrogen released from blood at reduced pressure %4.3f L'%V"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of moles of nitrogen in blood at 1 and 50 bar are 2.46e-03,0.123 mol\n",
- "Volume of nitrogen released from blood at reduced pressure 2.981 L\n"
- ]
- }
- ],
- "prompt_number": 90
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 9.14, Page Number 226"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg, divide\n",
- "from matplotlib import pyplot\n",
- "%matplotlib inline\n",
- "#Variable Declaration\n",
- "cCB = array([1e-6,2e-6,3e-6,5e-6,10e-6])\n",
- "nu = array([0.006,0.012,0.018,0.028,0.052])\n",
- "y = nu/cCB\n",
- "print y\n",
- "xlim(0.0, 0.06)\n",
- "ylim(5000,6300)\n",
- "#Calculations\n",
- "A = array([ nu, ones(size(nu))])\n",
- "print A\n",
- "# linearly generated sequence\n",
- "\n",
- "w = linalg.lstsq(A.T,y)[0] # obtaining the parameters\n",
- "print 'slope %8.1f'%w[0]\n",
- "print 'Intercept %8.1f' %w[1]\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = w[0]*nu+w[1] # regression line\n",
- " \n",
- "#Results\n",
- "plot(nu,line,'r-',nu,y,'o')\n",
- "xlabel('$ \\overline{\\upsilon} $')\n",
- "ylabel('$ \\overline{\\upsilon}/C_{CB}, M^{-1} $')\n",
- "#ylabel('$ \\dfrac{\\overline{\\upsilon}}{C_{CB}} $')\n",
- "show()"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "[ 6000. 6000. 6000. 5600. 5200.]\n",
- "[[ 0.006 0.012 0.018 0.028 0.052]\n",
- " [ 1. 1. 1. 1. 1. ]]"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "slope -19188.2\n",
- "Intercept 6205.2\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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095XAcqC/mbUDdnL3Wcl+jwOnpx+2iMg2dO0KEydCRQWVF13BSXsOoaz/NZx0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\n9ASq8UBFKwAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0xa4dd2d0>"
- ]
- }
- ],
- "prompt_number": 39
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10.ipynb deleted file mode 100755 index de8342a7..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10.ipynb +++ /dev/null @@ -1,59 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1d42669bd8107d861c6e3eefe6f0700977fa1e9c23ddda22a03d5e75fa945558"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: Electrolyte Solutions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 10.2, Page Number 242"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "M = 0.050 #Molarity of NaCl solution and also forNa2SO4 solution, mol/kg\n",
- "\n",
- "\n",
- "#Calculations\n",
- "Ia = M*(1+1)/2\n",
- "Ib = M*(2+4)/2\n",
- "\n",
- "#Results\n",
- "print 'Ionic streangth for NaCl solution is %4.3f and for Na2SO4 solution is %4.3f, mol/kg'%(Ia,Ib)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ionic streangth for NaCl solution is 0.050 and for Na2SO4 solution is 0.150, mol/kg\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10_1.ipynb deleted file mode 100755 index 746159a0..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10_1.ipynb +++ /dev/null @@ -1,62 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:83e7e02c224bb370929cff9832918ae9c98d9e287b7f595b2c76bce75548416a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: Electrolyte Solutions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 10.2, Page Number 242"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "M = 0.050 #Molarity for NaCl and Na2SO4 solution, mol/kg\n",
- "npa, zpa = 1, 1\n",
- "nma, zma = 1, 1\n",
- "npb, zpb = 2, 1\n",
- "nmb, zmb = 1, 2\n",
- "\n",
- "#Calculations\n",
- "Ia = M*(npa*zpa**2 + nma*zma**2)/2\n",
- "Ib = M*(npb*zpb**2 + nmb*zmb**2)/2\n",
- "\n",
- "#Results\n",
- "print 'Ionic streangth for NaCl solution is %4.3f and for Na2SO4 solution is %4.3f, mol/kg'%(Ia,Ib)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ionic streangth for NaCl solution is 0.050 and for Na2SO4 solution is 0.150, mol/kg\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10_2.ipynb deleted file mode 100755 index 746159a0..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter10_2.ipynb +++ /dev/null @@ -1,62 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:83e7e02c224bb370929cff9832918ae9c98d9e287b7f595b2c76bce75548416a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: Electrolyte Solutions"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 10.2, Page Number 242"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "M = 0.050 #Molarity for NaCl and Na2SO4 solution, mol/kg\n",
- "npa, zpa = 1, 1\n",
- "nma, zma = 1, 1\n",
- "npb, zpb = 2, 1\n",
- "nmb, zmb = 1, 2\n",
- "\n",
- "#Calculations\n",
- "Ia = M*(npa*zpa**2 + nma*zma**2)/2\n",
- "Ib = M*(npb*zpb**2 + nmb*zmb**2)/2\n",
- "\n",
- "#Results\n",
- "print 'Ionic streangth for NaCl solution is %4.3f and for Na2SO4 solution is %4.3f, mol/kg'%(Ia,Ib)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ionic streangth for NaCl solution is 0.050 and for Na2SO4 solution is 0.150, mol/kg\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11.ipynb deleted file mode 100755 index b8b395f7..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11.ipynb +++ /dev/null @@ -1,308 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:d710461b968e986411ee49bb4f3e24c283e59536cefcb736abd7424eef7f3627"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Electrochemical Cells, Batteries, and Fuel Cells"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.1, Page Number 256"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "aH = 0.770 #Activity of \n",
- "fH2 = 1.13 #Fugacity of Hydrogen gas\n",
- "E0 = 0.0 #Std. electrode potential, V\n",
- "n = 1.0 #Number of electrons transfered\n",
- "\n",
- "#Calculations\n",
- "E = E0 - (0.05916/n)*log(aH/sqrt(fH2),10)\n",
- "\n",
- "#Results\n",
- "print 'The potential of H+/H2 half cell %5.4f V'%E"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "THe potential of H+/H2 half cell 0.00829 V\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.2, Page Number 256"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "E0r1 = -0.877 #Std Electrod potential for Rx2 : Al3+ + 3e- ------> Al (s) \n",
- "E0r2 = -1.660 #Std Electrod potential for Rx2 : Al3+ + 3e- ------> Al (s)\n",
- "E0r3 = +0.071 #Std Electrod potential for Rx3 : AgBr (s) + e- ------> Ag(s) +Br- (aq.)\n",
- "\n",
- "#Calculations\n",
- "#3Fe(OH)2 (s)+ 2Al (s) <---------> 3Fe (s) + 6(OH-) + 2Al3+\n",
- "E0a = 3*E0r1 + (-2)*E0r2\n",
- "#Fe (s) + 2OH- + 2AgBr (s) -------> Fe(OH)2 (s) + 2Ag(s) + 2Br- (aq.)\n",
- "E0b = -E0r1 + (2)*E0r3\n",
- "\n",
- "#Results\n",
- "print '%5.3f %5.3f'%(E0a,E0b)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.689 1.019\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.3, Page Number 257"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "E01 = 0.771 #Rx1 : Fe3+ + e- -----> Fe2+\n",
- "E02 = -0.447 #Rx2 : Fe2+ + 2e- -----> Fe\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n1,n2,n3 = 1.,2.,3.\n",
- "\n",
- "#Calculations\n",
- "dG01 = -n1*F*E01\n",
- "dG02 = -n2*F*E02\n",
- " #For overall reaction\n",
- "dG0 = dG01 + dG02\n",
- "E0Fe3byFe = -dG0/(n3*F)\n",
- "\n",
- "#Results\n",
- "print 'E0 for overall reaction is %5.3f V'%(E0Fe3byFe)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "E0 for overall reaction is -0.041 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.4, Page Number 258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E01 = +1.36 #Std. electrode potential for Cl2/Cl\n",
- "dE0bydT = -1.20e-3 #V/K\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n = 2.\n",
- "S0H = 0.0 #Std. entropy J/(K.mol) for H+ ,Cl-,H2, Cl2 \n",
- "S0Cl = 56.5\n",
- "S0H2 = 130.7\n",
- "S0Cl2 = 223.1\n",
- "nH, nCl, nH2, nCl2 = 2, 2, -1,-1\n",
- "#Calculations\n",
- "dS01 = n*F*dE0bydT\n",
- "dS02 =nH*S0H + nCl*S0Cl + nH2*S0H2 + nCl2*S0Cl2\n",
- "\n",
- "#Results\n",
- "print 'Std. entropy change of reaction from dE0bydT is %4.2e and\\nStd entropy values is %4.2e V'%(dS01,dS02)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. entropy change of reaction from dE0bydT is -2.32e+02 and\n",
- "Std entropy values is -2.41e+02 V\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.5, Page Number 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "#Variable Declaration\n",
- "E0 = +1.10 #Std. electrode potential for Danniel cell, V\n",
- " #Zn(s) + Cu++ -----> Zn2+ + Cu\n",
- "T = 298.15 #V/K\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n = 2.\n",
- "R = 8.314 #Gas constant, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "K = exp(n*F*E0/(R*T))\n",
- "\n",
- "#Results\n",
- "print 'Equilibrium constant for reaction is %4.2e'%(K)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equilibrium constant for reaction is 1.55e+37\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.6, Page Number 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E = +0.29 #Cell emf, V\n",
- "\n",
- "#Calculations\n",
- "Ksp = 10**(-n*E/0.05916)\n",
- "\n",
- "#Results\n",
- "print 'Equilibrium constant for reaction is %4.2e'%(Ksp)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equilibrium constant for reaction is 1.57e-10\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.8, Page Number 262"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E = +1.51 #EMF for reduction of permangnet, V\n",
- "E01 = -0.7618 #Zn2+ + 2e- --------> Zn (s)\n",
- "E02 = +0.7996 #Ag+ + e- --------> Ag (s)\n",
- "E03 = +1.6920 #Au+ + e- --------> Au (s) \n",
- "\n",
- "#Calculations\n",
- "EZn = E - E01\n",
- "EAg = E - E02\n",
- "EAu = E - E03\n",
- "animals = {\"parrot\": 2, \"fish\": 6}\n",
- "Er = {\"Zn\":EZn,\"Ag\":EAg,\"Au\":EAu}\n",
- "#Results\n",
- "print 'Cell potentials for Zn, Ag, Au are %4.2f V, %4.2f V, and %4.2f V'%(EZn, EAg,EAu)\n",
- "for i in Er:\n",
- " if Er[i] >0.0:\n",
- " print '%s has positive cell potential of %4.3f V and Can be oxidized bypermangnate ion' %(i,Er[i])\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cell potentials for Zn, Ag, Au are 2.27 V, 0.71 V, and -0.18 V\n",
- "Zn has positive cell potential of 2.272 V and Can be oxidized bypermangnate ion\n",
- "Ag has positive cell potential of 0.710 V and Can be oxidized bypermangnate ion\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11_1.ipynb deleted file mode 100755 index 6248a97e..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11_1.ipynb +++ /dev/null @@ -1,307 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a9874fbd8c9c1e7bd5f52f872f15a435d2027cd53f7e4dc989a569aebbff426a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Electrochemical Cells, Batteries, and Fuel Cells"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.1, Page Number 256"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "aH = 0.770 #Activity of \n",
- "fH2 = 1.13 #Fugacity of Hydrogen gas\n",
- "E0 = 0.0 #Std. electrode potential, V\n",
- "n = 1.0 #Number of electrons transfered\n",
- "\n",
- "#Calculations\n",
- "E = E0 - (0.05916/n)*log(aH/sqrt(fH2),10)\n",
- "\n",
- "#Results\n",
- "print 'The potential of H+/H2 half cell %5.4f V'%E"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "THe potential of H+/H2 half cell 0.00829 V\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.2, Page Number 256"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E0r1 = -0.877 #Std Electrod potential for Rx2 : Al3+ + 3e- ------> Al (s) \n",
- "E0r2 = -1.660 #Std Electrod potential for Rx2 : Al3+ + 3e- ------> Al (s)\n",
- "E0r3 = +0.071 #Std Electrod potential for Rx3 : AgBr (s) + e- ------> Ag(s) +Br- (aq.)\n",
- "\n",
- "#Calculations\n",
- "#3Fe(OH)2 (s)+ 2Al (s) <---------> 3Fe (s) + 6(OH-) + 2Al3+\n",
- "E0a = 3*E0r1 + (-2)*E0r2\n",
- "#Fe (s) + 2OH- + 2AgBr (s) -------> Fe(OH)2 (s) + 2Ag(s) + 2Br- (aq.)\n",
- "E0b = -E0r1 + (2)*E0r3\n",
- "\n",
- "#Results\n",
- "print '%5.3f %5.3f'%(E0a,E0b)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.689 1.019\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.3, Page Number 257"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "E01 = 0.771 #Rx1 : Fe3+ + e- -----> Fe2+\n",
- "E02 = -0.447 #Rx2 : Fe2+ + 2e- -----> Fe\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n1,n2,n3 = 1.,2.,3.\n",
- "\n",
- "#Calculations\n",
- "dG01 = -n1*F*E01\n",
- "dG02 = -n2*F*E02\n",
- " #For overall reaction\n",
- "dG0 = dG01 + dG02\n",
- "E0Fe3byFe = -dG0/(n3*F)\n",
- "\n",
- "#Results\n",
- "print 'E0 for overall reaction is %5.3f V'%(E0Fe3byFe)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "E0 for overall reaction is -0.041 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.4, Page Number 258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E01 = +1.36 #Std. electrode potential for Cl2/Cl\n",
- "dE0bydT = -1.20e-3 #V/K\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n = 2.\n",
- "S0H = 0.0 #Std. entropy J/(K.mol) for H+ ,Cl-,H2, Cl2 \n",
- "S0Cl = 56.5\n",
- "S0H2 = 130.7\n",
- "S0Cl2 = 223.1\n",
- "nH, nCl, nH2, nCl2 = 2, 2, -1,-1\n",
- "#Calculations\n",
- "dS01 = n*F*dE0bydT\n",
- "dS02 =nH*S0H + nCl*S0Cl + nH2*S0H2 + nCl2*S0Cl2\n",
- "\n",
- "#Results\n",
- "print 'Std. entropy change of reaction from dE0bydT is %4.2e and\\nStd entropy values is %4.2e V'%(dS01,dS02)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. entropy change of reaction from dE0bydT is -2.32e+02 and\n",
- "Std entropy values is -2.41e+02 V\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.5, Page Number 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "#Variable Declaration\n",
- "E0 = +1.10 #Std. electrode potential for Danniel cell, V\n",
- " #Zn(s) + Cu++ -----> Zn2+ + Cu\n",
- "T = 298.15 #V/K\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n = 2.\n",
- "R = 8.314 #Gas constant, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "K = exp(n*F*E0/(R*T))\n",
- "\n",
- "#Results\n",
- "print 'Equilibrium constant for reaction is %4.2e'%(K)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equilibrium constant for reaction is 1.55e+37\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.6, Page Number 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E = +0.29 #Cell emf, V\n",
- "n = 2.\n",
- "\n",
- "#Calculations\n",
- "Ksp = 10**(-n*E/0.05916)\n",
- "\n",
- "#Results\n",
- "print 'Equilibrium constant for reaction is %4.2e'%(Ksp)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equilibrium constant for reaction is 1.57e-10\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.8, Page Number 262"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E = +1.51 #EMF for reduction of permangnet, V\n",
- "E01 = -0.7618 #Zn2+ + 2e- --------> Zn (s)\n",
- "E02 = +0.7996 #Ag+ + e- --------> Ag (s)\n",
- "E03 = +1.6920 #Au+ + e- --------> Au (s) \n",
- "\n",
- "#Calculations\n",
- "EZn = E - E01\n",
- "EAg = E - E02\n",
- "EAu = E - E03\n",
- "animals = {\"parrot\": 2, \"fish\": 6}\n",
- "Er = {\"Zn\":EZn,\"Ag\":EAg,\"Au\":EAu}\n",
- "#Results\n",
- "print 'Cell potentials for Zn, Ag, Au are %4.2f V, %4.2f V, and %4.2f V'%(EZn, EAg,EAu)\n",
- "for i in Er:\n",
- " if Er[i] >0.0:\n",
- " print '%s has positive cell potential of %4.3f V and Can be oxidized bypermangnate ion' %(i,Er[i])\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cell potentials for Zn, Ag, Au are 2.27 V, 0.71 V, and -0.18 V\n",
- "Zn has positive cell potential of 2.272 V and Can be oxidized bypermangnate ion\n",
- "Ag has positive cell potential of 0.710 V and Can be oxidized bypermangnate ion\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11_2.ipynb deleted file mode 100755 index 723caafc..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter11_2.ipynb +++ /dev/null @@ -1,310 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:d2becfaa562cf3368a669798b21524329e478ea72b3f08ab186dd0dd13556e25"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Electrochemical Cells, Batteries, and Fuel Cells"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.1, Page Number 256"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "aH = 0.770 #Activity of \n",
- "fH2 = 1.13 #Fugacity of Hydrogen gas\n",
- "E0 = 0.0 #Std. electrode potential, V\n",
- "n = 1.0 #Number of electrons transfered\n",
- "\n",
- "#Calculations\n",
- "E = E0 - (0.05916/n)*log(aH/sqrt(fH2),10)\n",
- "\n",
- "#Results\n",
- "print 'The potential of H+/H2 half cell %5.4f V'%E"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "THe potential of H+/H2 half cell 0.00829 V\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.2, Page Number 256"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "E0r1 = -0.877 #Std Electrod potential for Rx2 : Al3+ + 3e- ------> Al (s) \n",
- "E0r2 = -1.660 #Std Electrod potential for Rx2 : Al3+ + 3e- ------> Al (s)\n",
- "E0r3 = +0.071 #Std Electrod potential for Rx3 : AgBr (s) + e- ------> Ag(s) +Br- (aq.)\n",
- "\n",
- "#Calculations\n",
- "#3Fe(OH)2 (s)+ 2Al (s) <---------> 3Fe (s) + 6(OH-) + 2Al3+\n",
- "E0a = 3*E0r1 + (-2)*E0r2\n",
- "#Fe (s) + 2OH- + 2AgBr (s) -------> Fe(OH)2 (s) + 2Ag(s) + 2Br- (aq.)\n",
- "E0b = -E0r1 + (2)*E0r3\n",
- "\n",
- "#Results\n",
- "print '%5.3f %5.3f'%(E0a,E0b)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.689 1.019\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.3, Page Number 257"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "E01 = 0.771 #Rx1 : Fe3+ + e- -----> Fe2+\n",
- "E02 = -0.447 #Rx2 : Fe2+ + 2e- -----> Fe\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n1,n2,n3 = 1.,2.,3.\n",
- "\n",
- "#Calculations\n",
- "dG01 = -n1*F*E01\n",
- "dG02 = -n2*F*E02\n",
- " #For overall reaction\n",
- "dG0 = dG01 + dG02\n",
- "E0Fe3byFe = -dG0/(n3*F)\n",
- "\n",
- "#Results\n",
- "print 'E0 for overall reaction is %5.3f V'%(E0Fe3byFe)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "E0 for overall reaction is -0.041 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.4, Page Number 258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Variable Declaration\n",
- "E01 = +1.36 #Std. electrode potential for Cl2/Cl\n",
- "dE0bydT = -1.20e-3 #V/K\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n = 2.\n",
- "S0H = 0.0 #Std. entropy J/(K.mol) for H+ ,Cl-,H2, Cl2 \n",
- "S0Cl = 56.5\n",
- "S0H2 = 130.7\n",
- "S0Cl2 = 223.1\n",
- "nH, nCl, nH2, nCl2 = 2, 2, -1,-1\n",
- "#Calculations\n",
- "dS01 = n*F*dE0bydT\n",
- "dS02 =nH*S0H + nCl*S0Cl + nH2*S0H2 + nCl2*S0Cl2\n",
- "\n",
- "#Results\n",
- "print 'Std. entropy change of reaction from dE0bydT is %4.2e and\\nStd entropy values is %4.2e V'%(dS01,dS02)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Std. entropy change of reaction from dE0bydT is -2.32e+02 and\n",
- "Std entropy values is -2.41e+02 V\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.5, Page Number 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "E0 = +1.10 #Std. electrode potential for Danniel cell, V\n",
- " #Zn(s) + Cu++ -----> Zn2+ + Cu\n",
- "T = 298.15 #V/K\n",
- "F = 96485 #Faraday constant, C/mol\n",
- "n = 2.\n",
- "R = 8.314 #Gas constant, J/(mol.K)\n",
- "\n",
- "#Calculations\n",
- "K = exp(n*F*E0/(R*T))\n",
- "\n",
- "#Results\n",
- "print 'Equilibrium constant for reaction is %4.2e'%(K)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equilibrium constant for reaction is 1.55e+37\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.6, Page Number 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E = +0.29 #Cell emf, V\n",
- "n = 2.\n",
- "\n",
- "#Calculations\n",
- "Ksp = 10**(-n*E/0.05916)\n",
- "\n",
- "#Results\n",
- "print 'Equilibrium constant for reaction is %4.2e'%(Ksp)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equilibrium constant for reaction is 1.57e-10\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 11.8, Page Number 262"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "E = +1.51 #EMF for reduction of permangnet, V\n",
- "E01 = -0.7618 #Zn2+ + 2e- --------> Zn (s)\n",
- "E02 = +0.7996 #Ag+ + e- --------> Ag (s)\n",
- "E03 = +1.6920 #Au+ + e- --------> Au (s) \n",
- "\n",
- "#Calculations\n",
- "EZn = E - E01\n",
- "EAg = E - E02\n",
- "EAu = E - E03\n",
- "animals = {\"parrot\": 2, \"fish\": 6}\n",
- "Er = {\"Zn\":EZn,\"Ag\":EAg,\"Au\":EAu}\n",
- "#Results\n",
- "print 'Cell potentials for Zn, Ag, Au are %4.2f V, %4.2f V, and %4.2f V'%(EZn, EAg,EAu)\n",
- "for i in Er:\n",
- " if Er[i] >0.0:\n",
- " print '%s has positive cell potential of %4.3f V and Can be oxidized bypermangnate ion' %(i,Er[i])\n",
- " "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cell potentials for Zn, Ag, Au are 2.27 V, 0.71 V, and -0.18 V\n",
- "Zn has positive cell potential of 2.272 V and Can be oxidized bypermangnate ion\n",
- "Ag has positive cell potential of 0.710 V and Can be oxidized bypermangnate ion\n"
- ]
- }
- ],
- "prompt_number": 25
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12.ipynb deleted file mode 100755 index c12bf202..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12.ipynb +++ /dev/null @@ -1,462 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:41b66ed6dbdebdc3dd934ef963f179576dd8d1497050c4ff934c3909d90dbee3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12: Probability"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.1, Page Number 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": [],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.2, Page Number 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "n = 52 #Total cards\n",
- "nheart = 13 #Number of cards with hearts\n",
- "\n",
- "#Calculations\n",
- "Pe = Fraction(nheart,n)\n",
- "\n",
- "#Results\n",
- "print 'Probability of one (heart)card picked from a std. stack of %d cards is'%n,Pe"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of one (heart)card picked from a std. stack of 52 cards is 1/4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.3, Page Number 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": [],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.4, Page Number 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "n1 = 2 #Two spin states for 1st electron in orbit 1\n",
- "n2 = 2 #Two spin states for 2nd electron in orbit 2\n",
- "\n",
- "#Calculation\n",
- "M = n1*n1\n",
- "\n",
- "#Results\n",
- "print 'Possible spin states for excited state are %2d'%M"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Possible spin states for excited state are 4\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.5, Page Number 286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "n = 12 #Total Number of players \n",
- "j = 5 #Number player those can play match\n",
- "\n",
- "#Calculation\n",
- "P = factorial(n)/factorial(n-j)\n",
- "\n",
- "#Results\n",
- "print 'Maximum Possible permutations for 5 player to play are %8d'%P"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Possible permutations for 5 player to play are 95040\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.6, Page Number 287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "n = 52 #Number of cards in std . pack\n",
- "j = 5 #Number of cards in subset\n",
- "\n",
- "#Calculation\n",
- "C = factorial(n)/(factorial(j)*factorial(n-j))\n",
- "\n",
- "#Results\n",
- "print 'Maximum Possible 5-card combinations are %8d'%C"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Possible 5-card combinations are 2598960\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.7, Page Number 288"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "x = 6 #Number of electrons\n",
- "n = 2 #Number of states\n",
- "\n",
- "#Calculation\n",
- "P = factorial(x)/(factorial(n)*factorial(x-n))\n",
- "\n",
- "#Results\n",
- "print 'Total number of quantum states are %3d'%P"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total number of quantum states are 15\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.8, Page Number 289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "n = 50 #Number of separate experiments\n",
- "j1 = 25 #Number of sucessful expt with heads up\n",
- "j2 = 10 #Number of sucessful expt with heads up\n",
- "\n",
- "#Calculation\n",
- "C25 = factorial(n)/(factorial(j1)*factorial(n-j1))\n",
- "PE25 = Fraction(1,2)**j1\n",
- "PEC25 = (1-Fraction(1,2))**(n-j1)\n",
- "P25 = C25*PE25*PEC25\n",
- "\n",
- "C10 = factorial(n)/(factorial(j2)*factorial(n-j2))\n",
- "PE10 = Fraction(1,2)**j2\n",
- "PEC10 = (1-Fraction(1,2))**(n-j2)\n",
- "P10 = C10*PE10*PEC10\n",
- "\n",
- "#Results\n",
- "print 'Probability of getting 25 head out of 50 tossing is %4.3f'%(float(P25))\n",
- "print 'Probability of getting 10 head out of 50 tossing is %4.3e'%(float(P10))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of getting 25 head out of 50 tossing is 0.112\n",
- "Probability of getting 10 head out of 50 tossing is 9.124e-06\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.9, Page Number 290"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial, log\n",
- "#Variable Declaration\n",
- "N = [10,50,100] #Valures for N\n",
- "\n",
- "#Calculations\n",
- "print ' N ln(N!) ln(N!)sterling Error'\n",
- "for i in N:\n",
- " lnN = log(factorial(i))\n",
- " lnNs = i*log(i)-i\n",
- " err = abs(lnN-lnNs)\n",
- " print '%3d %5.2f %5.2f %4.2f'%(i,lnN,lnNs, err)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " N ln(N!) ln(N!)sterling Error\n",
- " 10 15.10 13.03 2.08\n",
- " 50 148.48 145.60 2.88\n",
- "100 363.74 360.52 3.22\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.10, Page Number 293"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "fi = 1 #Probability of receiving any card\n",
- "n = 52 #Number od Cards\n",
- "\n",
- "#Calculations\n",
- "sum = 0\n",
- "for i in range(52):\n",
- " sum = sum + fi\n",
- "\n",
- "Pxi = Fraction(fi,sum)\n",
- "\n",
- "#Results\n",
- "print 'Probability of receiving any card', Pxi"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of receiving any card 1/52\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.11, Page Number 295"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "from scipy import integrate\n",
- "#Variable Declaration\n",
- "\n",
- "#Calculations\n",
- "fun = lambda x: exp(-0.05*x)\n",
- "Pt = 0\n",
- "for i in range(0,101):\n",
- " Pt = Pt + fun(i)\n",
- " \n",
- "Ptot = integrate.quad(fun, 0.0, 100.)\n",
- "\n",
- "#Results\n",
- "print 'Sum of Px considering it as discrete function %4.1f'%Pt\n",
- "print 'Sum of Px considering it as contineous function %4.1f'%Ptot[0]"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(19.865241060018292, 2.205484801456136e-13)\n",
- "Sum of Px considering it as discrete function 20.4\n",
- "Sum of Px considering it as contineous function 19.9\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.12, Page Number 296"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "r = Symbol('r') #Radius of inner circle\n",
- "C = [5,2,0]\n",
- "#Calculations\n",
- "A1 = pi*r**2\n",
- "A2 = pi*(2*r)**2 - A1\n",
- "A3 = pi*(3*r)**2 - (A1 + A2)\n",
- "At = A1 + A2 + A3\n",
- "f1 = A1/At\n",
- "f2 = A2/At\n",
- "f3 = A3/At\n",
- "sf = f1 + f2 + f3\n",
- "\n",
- "ns = (f1*C[0]+f2*C[1]+f3*C[2])/sf\n",
- "\n",
- "#Results\n",
- "print 'A1, A2, A3: ', A1, A2, A3\n",
- "print 'f1, f2, f3: ', f1,f2,f3\n",
- "print 'Average payout $', round(float(ns),2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "A1, A2, A3: pi*r**2 3*pi*r**2 5*pi*r**2\n",
- "f1, f2, f3: 1/9 1/3 5/9\n",
- "Average payout $ 1.22\n"
- ]
- }
- ],
- "prompt_number": 60
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12_1.ipynb deleted file mode 100755 index 9fb89e58..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12_1.ipynb +++ /dev/null @@ -1,479 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c19a90db6a87eae57c23bc4670f56b33da607fbe1caf5a0bc049c2dde83bee35"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12: Probability"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.1, Page Number 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varible declaration\n",
- "n = range(1,51,1)\n",
- "Prob = 0\n",
- "for x in n:\n",
- " Prob = 1./len(n) + Prob\n",
- "\n",
- "#Results\n",
- "print 'Probability of picking up any one ball is %3.1f'%Prob"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of picking up any one ball is 1.0\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.2, Page Number 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "n = 52 #Total cards\n",
- "nheart = 13 #Number of cards with hearts\n",
- "\n",
- "#Calculations\n",
- "Pe = Fraction(nheart,n)\n",
- "\n",
- "#Results\n",
- "print 'Probability of one (heart)card picked from a std. stack of %d cards is'%n,Pe"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of one (heart)card picked from a std. stack of 52 cards is 1/4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.3, Page Number 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": [],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.4, Page Number 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "n1 = 2 #Two spin states for 1st electron in orbit 1\n",
- "n2 = 2 #Two spin states for 2nd electron in orbit 2\n",
- "\n",
- "#Calculation\n",
- "M = n1*n1\n",
- "\n",
- "#Results\n",
- "print 'Possible spin states for excited state are %2d'%M"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Possible spin states for excited state are 4\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.5, Page Number 286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "n = 12 #Total Number of players \n",
- "j = 5 #Number player those can play match\n",
- "\n",
- "#Calculation\n",
- "P = factorial(n)/factorial(n-j)\n",
- "\n",
- "#Results\n",
- "print 'Maximum Possible permutations for 5 player to play are %8d'%P"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Possible permutations for 5 player to play are 95040\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.6, Page Number 287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "n = 52 #Number of cards in std . pack\n",
- "j = 5 #Number of cards in subset\n",
- "\n",
- "#Calculation\n",
- "C = factorial(n)/(factorial(j)*factorial(n-j))\n",
- "\n",
- "#Results\n",
- "print 'Maximum Possible 5-card combinations are %8d'%C"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Possible 5-card combinations are 2598960\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.7, Page Number 288"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "x = 6 #Number of electrons\n",
- "n = 2 #Number of states\n",
- "\n",
- "#Calculation\n",
- "P = factorial(x)/(factorial(n)*factorial(x-n))\n",
- "\n",
- "#Results\n",
- "print 'Total number of quantum states are %3d'%P"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total number of quantum states are 15\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.8, Page Number 289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "n = 50 #Number of separate experiments\n",
- "j1 = 25 #Number of sucessful expt with heads up\n",
- "j2 = 10 #Number of sucessful expt with heads up\n",
- "\n",
- "#Calculation\n",
- "C25 = factorial(n)/(factorial(j1)*factorial(n-j1))\n",
- "PE25 = Fraction(1,2)**j1\n",
- "PEC25 = (1-Fraction(1,2))**(n-j1)\n",
- "P25 = C25*PE25*PEC25\n",
- "\n",
- "C10 = factorial(n)/(factorial(j2)*factorial(n-j2))\n",
- "PE10 = Fraction(1,2)**j2\n",
- "PEC10 = (1-Fraction(1,2))**(n-j2)\n",
- "P10 = C10*PE10*PEC10\n",
- "\n",
- "#Results\n",
- "print 'Probability of getting 25 head out of 50 tossing is %4.3f'%(float(P25))\n",
- "print 'Probability of getting 10 head out of 50 tossing is %4.3e'%(float(P10))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of getting 25 head out of 50 tossing is 0.112\n",
- "Probability of getting 10 head out of 50 tossing is 9.124e-06\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.9, Page Number 290"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial, log\n",
- "#Variable Declaration\n",
- "N = [10,50,100] #Valures for N\n",
- "\n",
- "#Calculations\n",
- "print ' N ln(N!) ln(N!)sterling Error'\n",
- "for i in N:\n",
- " lnN = log(factorial(i))\n",
- " lnNs = i*log(i)-i\n",
- " err = abs(lnN-lnNs)\n",
- " print '%3d %5.2f %5.2f %4.2f'%(i,lnN,lnNs, err)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " N ln(N!) ln(N!)sterling Error\n",
- " 10 15.10 13.03 2.08\n",
- " 50 148.48 145.60 2.88\n",
- "100 363.74 360.52 3.22\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.10, Page Number 293"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "fi = 1 #Probability of receiving any card\n",
- "n = 52 #Number od Cards\n",
- "\n",
- "#Calculations\n",
- "sum = 0\n",
- "for i in range(52):\n",
- " sum = sum + fi\n",
- "\n",
- "Pxi = Fraction(fi,sum)\n",
- "\n",
- "#Results\n",
- "print 'Probability of receiving any card', Pxi"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of receiving any card 1/52\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.11, Page Number 295"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "from scipy import integrate\n",
- "#Variable Declaration\n",
- "\n",
- "#Calculations\n",
- "fun = lambda x: exp(-0.05*x)\n",
- "Pt = 0\n",
- "for i in range(0,101):\n",
- " Pt = Pt + fun(i)\n",
- " \n",
- "Ptot = integrate.quad(fun, 0.0, 100.)\n",
- "\n",
- "#Results\n",
- "print 'Sum of Px considering it as discrete function %4.1f'%Pt\n",
- "print 'Sum of Px considering it as contineous function %4.1f'%Ptot[0]"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(19.865241060018292, 2.205484801456136e-13)\n",
- "Sum of Px considering it as discrete function 20.4\n",
- "Sum of Px considering it as contineous function 19.9\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.12, Page Number 296"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "r = Symbol('r') #Radius of inner circle\n",
- "C = [5,2,0]\n",
- "#Calculations\n",
- "A1 = pi*r**2\n",
- "A2 = pi*(2*r)**2 - A1\n",
- "A3 = pi*(3*r)**2 - (A1 + A2)\n",
- "At = A1 + A2 + A3\n",
- "f1 = A1/At\n",
- "f2 = A2/At\n",
- "f3 = A3/At\n",
- "sf = f1 + f2 + f3\n",
- "\n",
- "ns = (f1*C[0]+f2*C[1]+f3*C[2])/sf\n",
- "\n",
- "#Results\n",
- "print 'A1, A2, A3: ', A1, A2, A3\n",
- "print 'f1, f2, f3: ', f1,f2,f3\n",
- "print 'Average payout $', round(float(ns),2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "A1, A2, A3: pi*r**2 3*pi*r**2 5*pi*r**2\n",
- "f1, f2, f3: 1/9 1/3 5/9\n",
- "Average payout $ 1.22\n"
- ]
- }
- ],
- "prompt_number": 60
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12_2.ipynb deleted file mode 100755 index 3706fbca..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter12_2.ipynb +++ /dev/null @@ -1,479 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:829a63088d9b470e9c4646990e2a5951ca622e4ff802f1bdf0a653073bcbae67"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12: Probability"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.1, Page Number 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Varible declaration\n",
- "n = range(1,51,1)\n",
- "Prob = 0\n",
- "for x in n:\n",
- " Prob = 1./len(n) + Prob\n",
- "\n",
- "#Results\n",
- "print 'Probability of picking up any one ball is %3.1f'%Prob"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of picking up any one ball is 1.0\n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.2, Page Number 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "n = 52 #Total cards\n",
- "nheart = 13 #Number of cards with hearts\n",
- "\n",
- "#Calculations\n",
- "Pe = Fraction(nheart,n)\n",
- "\n",
- "#Results\n",
- "print 'Probability of one (heart)card picked from a std. stack of %d cards is'%n,Pe"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of one (heart)card picked from a std. stack of 52 cards is 1/4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.3, Page Number 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [],
- "language": "python",
- "metadata": {},
- "outputs": [],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.4, Page Number 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "n1 = 2 #Two spin states for 1st electron in orbit 1\n",
- "n2 = 2 #Two spin states for 2nd electron in orbit 2\n",
- "\n",
- "#Calculation\n",
- "M = n1*n1\n",
- "\n",
- "#Results\n",
- "print 'Possible spin states for excited state are %2d'%M"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Possible spin states for excited state are 4\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.5, Page Number 286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "n = 12 #Total Number of players \n",
- "j = 5 #Number player those can play match\n",
- "\n",
- "#Calculation\n",
- "P = factorial(n)/factorial(n-j)\n",
- "\n",
- "#Results\n",
- "print 'Maximum Possible permutations for 5 player to play are %8d'%P"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Possible permutations for 5 player to play are 95040\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.6, Page Number 287"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "n = 52 #Number of cards in std . pack\n",
- "j = 5 #Number of cards in subset\n",
- "\n",
- "#Calculation\n",
- "C = factorial(n)/(factorial(j)*factorial(n-j))\n",
- "\n",
- "#Results\n",
- "print 'Maximum Possible 5-card combinations are %8d'%C"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum Possible 5-card combinations are 2598960\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.7, Page Number 288"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "x = 6 #Number of electrons\n",
- "n = 2 #Number of states\n",
- "\n",
- "#Calculation\n",
- "P = factorial(x)/(factorial(n)*factorial(x-n))\n",
- "\n",
- "#Results\n",
- "print 'Total number of quantum states are %3d'%P"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total number of quantum states are 15\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.8, Page Number 289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "n = 50 #Number of separate experiments\n",
- "j1 = 25 #Number of sucessful expt with heads up\n",
- "j2 = 10 #Number of sucessful expt with heads up\n",
- "\n",
- "#Calculation\n",
- "C25 = factorial(n)/(factorial(j1)*factorial(n-j1))\n",
- "PE25 = Fraction(1,2)**j1\n",
- "PEC25 = (1-Fraction(1,2))**(n-j1)\n",
- "P25 = C25*PE25*PEC25\n",
- "\n",
- "C10 = factorial(n)/(factorial(j2)*factorial(n-j2))\n",
- "PE10 = Fraction(1,2)**j2\n",
- "PEC10 = (1-Fraction(1,2))**(n-j2)\n",
- "P10 = C10*PE10*PEC10\n",
- "\n",
- "#Results\n",
- "print 'Probability of getting 25 head out of 50 tossing is %4.3f'%(float(P25))\n",
- "print 'Probability of getting 10 head out of 50 tossing is %4.3e'%(float(P10))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of getting 25 head out of 50 tossing is 0.112\n",
- "Probability of getting 10 head out of 50 tossing is 9.124e-06\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.9, Page Number 290"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial, log\n",
- "#Variable Declaration\n",
- "N = [10,50,100] #Valures for N\n",
- "\n",
- "#Calculations\n",
- "print ' N ln(N!) ln(N!)sterling Error'\n",
- "for i in N:\n",
- " lnN = log(factorial(i))\n",
- " lnNs = i*log(i)-i\n",
- " err = abs(lnN-lnNs)\n",
- " print '%3d %5.2f %5.2f %4.2f'%(i,lnN,lnNs, err)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " N ln(N!) ln(N!)sterling Error\n",
- " 10 15.10 13.03 2.08\n",
- " 50 148.48 145.60 2.88\n",
- "100 363.74 360.52 3.22\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.10, Page Number 293"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from fractions import Fraction\n",
- "\n",
- "#Variable Declaration\n",
- "fi = 1 #Probability of receiving any card\n",
- "n = 52 #Number od Cards\n",
- "\n",
- "#Calculations\n",
- "sum = 0\n",
- "for i in range(52):\n",
- " sum = sum + fi\n",
- "\n",
- "Pxi = Fraction(fi,sum)\n",
- "\n",
- "#Results\n",
- "print 'Probability of receiving any card', Pxi"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of receiving any card 1/52\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.11, Page Number 295"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "from scipy import integrate\n",
- "#Variable Declaration\n",
- "\n",
- "#Calculations\n",
- "fun = lambda x: exp(-0.05*x)\n",
- "Pt = 0\n",
- "for i in range(0,101):\n",
- " Pt = Pt + fun(i)\n",
- " \n",
- "Ptot = integrate.quad(fun, 0.0, 100.)\n",
- "\n",
- "#Results\n",
- "print 'Sum of Px considering it as discrete function %4.1f'%Pt\n",
- "print 'Sum of Px considering it as contineous function %4.1f'%Ptot[0]"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(19.865241060018292, 2.205484801456136e-13)\n",
- "Sum of Px considering it as discrete function 20.4\n",
- "Sum of Px considering it as contineous function 19.9\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 12.12, Page Number 296"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import Symbol\n",
- "\n",
- "#Variable Declaration\n",
- "r = Symbol('r') #Radius of inner circle\n",
- "C = [5,2,0]\n",
- "#Calculations\n",
- "A1 = pi*r**2\n",
- "A2 = pi*(2*r)**2 - A1\n",
- "A3 = pi*(3*r)**2 - (A1 + A2)\n",
- "At = A1 + A2 + A3\n",
- "f1 = A1/At\n",
- "f2 = A2/At\n",
- "f3 = A3/At\n",
- "sf = f1 + f2 + f3\n",
- "\n",
- "ns = (f1*C[0]+f2*C[1]+f3*C[2])/sf\n",
- "\n",
- "#Results\n",
- "print 'A1, A2, A3: ', A1,', ', A2,', ', A3\n",
- "print 'f1, f2, f3: ', f1,f2,f3\n",
- "print 'Average payout $', round(float(ns),2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "A1, A2, A3: 3.14159265358979*r**2 , 9.42477796076938*r**2 , 15.707963267949*r**2\n",
- "f1, f2, f3: 0.111111111111111 0.333333333333333 0.555555555555556\n",
- "Average payout $ 1.22\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13.ipynb deleted file mode 100755 index da21e674..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13.ipynb +++ /dev/null @@ -1,223 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a0be5f6d37f6b1696997f58d0cc492cc18c5483e1f769555c228f3249920c51f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Boltzmann Distribution"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.1, Page Number 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "aH = 40\n",
- "N = 100\n",
- "\n",
- "#Calculations\n",
- "aT = 100 - aH\n",
- "We = factorial(N)/(factorial(aT)*factorial(aH))\n",
- "Wexpected = factorial(N)/(factorial(N/2)*factorial(N/2))\n",
- "\n",
- "#Results\n",
- "print 'The observed weight %5.2e compared to %5.2e'%(We,Wexpected)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The observed weight 1.37e+28 compared to 1.01e+29\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.2, Page Number 310"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.3, Page Number 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "p0 = 0.633 #Probabilities of Energy level 1,2,3 \n",
- "p1 = 0.233\n",
- "p2 = 0.086\n",
- "\n",
- "#Calculation\n",
- "p4 = 1. -(p0+p1+p2)\n",
- "\n",
- "#Results\n",
- "print 'Probability of finding an oscillator at energy level of n>3 is %4.3f i.e.%4.1f percent'%(p4,p4*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of finding an oscillator at energy level of n>3 is 0.048 i.e. 4.8 percent\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.4, Page Number 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "p0 = 0.394 #Probabilities of Energy level 1,2,3 \n",
- "p1by2 = 0.239\n",
- "p2 = 0.145\n",
- "\n",
- "#Calculation\n",
- "p4 = 1. -(p0+p1by2+p2)\n",
- "\n",
- "#Results\n",
- "print 'Probability of finding an oscillator at energy level of n>3 is %4.3f'%(p4)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of finding an oscillator at energy level of n>3 is 0.222\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.5, Page Number 321"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "I2 = 208 #Vibrational frequency, cm-1 \n",
- "T = 298 #Molecular Temperature, K\n",
- "c = 3.00e10 #speed of light, cm/s\n",
- "h = 6.626e-34 #Planks constant, J/K\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "#Calculation\n",
- "q = 1./(1.-exp(-h*c*I2/(k*T)))\n",
- "p2 = exp(-2*h*c*I2/(k*T))/q\n",
- "\n",
- "#Results\n",
- "print 'Partition function is %4.3f'%(q)\n",
- "print 'Probability of occupying the second vibrational state n=2 is %4.3f'%(p2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Partition function is 1.577\n",
- "Probability of occupying the second vibrational state n=2 is 0.085\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.6, Page Number 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "B = 1.45 #Magnetic field streangth, Teslas \n",
- "T = 298 #Molecular Temperature, K\n",
- "c = 3.00e10 #speed of light, cm/s\n",
- "h = 6.626e-34 #Planks constant, J/K\n",
- "k = 1.38e-23 #Boltzman constant, J/K \n",
- "gnbn = 2.82e-26 #J/T\n",
- "#Calculation\n",
- "ahpbyahm = exp(-gnbn*B/(k*T))\n",
- "\n",
- "#Results\n",
- "print 'Occupation Number is %7.6f'%(ahpbyahm)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Occupation Number is 0.999990\n"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13_1.ipynb deleted file mode 100755 index 8aaf79b0..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13_1.ipynb +++ /dev/null @@ -1,280 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:78db264f9210996c6e59dc097dbc6696221fac03055a9d5ba5033673d3e33c9f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Boltzmann Distribution"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.1, Page Number 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "aH = 40 #Number of heads\n",
- "N = 100 #Total events\n",
- "\n",
- "#Calculations\n",
- "aT = 100 - aH\n",
- "We = factorial(N)/(factorial(aT)*factorial(aH))\n",
- "Wexpected = factorial(N)/(factorial(N/2)*factorial(N/2))\n",
- "\n",
- "#Results\n",
- "print 'The observed weight %5.2e compared to %5.2e'%(We,Wexpected)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The observed weight 1.37e+28 compared to 1.01e+29\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.2, Page Number 310"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Varialbe declaration\n",
- "n = 10000 #Total number of particles\n",
- "\n",
- "\n",
- "#Calcualtions\n",
- "def ster(i):\n",
- " return i*log(i)-i\n",
- "\n",
- "n1, n2, n3, W = symbols('n1 n2 n3 W',positive=True)\n",
- "\n",
- "n2 = 5000 - 2*n3\n",
- "n1 = 10000 - n2 -n3\n",
- "logW = ster(n) - ster(n1) - ster(n2) - ster(n3) \n",
- "fun = diff(logW, n3)\n",
- "dfun = diff(fun, n3)\n",
- "x0 = 10.0\n",
- "err = 1.0\n",
- "while err>0.001:\n",
- " f = fun.subs(n3,x0)\n",
- " df = dfun.subs(n3,x0)\n",
- " xnew = x0 - f/df\n",
- " err = abs(x0-xnew)/x0\n",
- " x0 = xnew\n",
- "\n",
- "x0 = int(x0)\n",
- "N2 = n2.subs(n3,x0)\n",
- "N3 = x0\n",
- "n1 = n1.subs(n3,x0)\n",
- "N1 = n1.subs(n2,N2)\n",
- "lnW = logW.subs(n3,N3)\n",
- "\n",
- "#Results\n",
- "print 'At maximum value of ln(W)'\n",
- "print 'Values of N1 : %4d, N2: %4d and N3: %4d '%(N1, N2,N3)\n",
- "print 'Maximum value of ln(W)= %6d'%lnW"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At maximum value of ln(W)\n",
- "Values of N1 : 6162, N2: 2676 and N3: 1162 \n",
- "Maximum value of ln(W)= 9012\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.3, Page Number 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "p0 = 0.633 #Probabilities of Energy level 1,2,3 \n",
- "p1 = 0.233\n",
- "p2 = 0.086\n",
- "\n",
- "#Calculation\n",
- "p4 = 1. -(p0+p1+p2)\n",
- "\n",
- "#Results\n",
- "print 'Probability of finding an oscillator at energy level of n>3 is %4.3f i.e.%4.1f percent'%(p4,p4*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of finding an oscillator at energy level of n>3 is 0.048 i.e. 4.8 percent\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.4, Page Number 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "p0 = 0.394 #Probabilities of Energy level 1,2,3 \n",
- "p1by2 = 0.239\n",
- "p2 = 0.145\n",
- "\n",
- "#Calculation\n",
- "p4 = 1. -(p0+p1by2+p2)\n",
- "\n",
- "#Results\n",
- "print 'Probability of finding an oscillator at energy level of n>3 is %4.3f'%(p4)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of finding an oscillator at energy level of n>3 is 0.222\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.5, Page Number 321"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "I2 = 208 #Vibrational frequency, cm-1 \n",
- "T = 298 #Molecular Temperature, K\n",
- "c = 3.00e10 #speed of light, cm/s\n",
- "h = 6.626e-34 #Planks constant, J/K\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "#Calculation\n",
- "q = 1./(1.-exp(-h*c*I2/(k*T)))\n",
- "p2 = exp(-2*h*c*I2/(k*T))/q\n",
- "\n",
- "#Results\n",
- "print 'Partition function is %4.3f'%(q)\n",
- "print 'Probability of occupying the second vibrational state n=2 is %4.3f'%(p2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Partition function is 1.577\n",
- "Probability of occupying the second vibrational state n=2 is 0.085\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.6, Page Number 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "B = 1.45 #Magnetic field streangth, Teslas \n",
- "T = 298 #Molecular Temperature, K\n",
- "c = 3.00e10 #speed of light, cm/s\n",
- "h = 6.626e-34 #Planks constant, J/K\n",
- "k = 1.38e-23 #Boltzman constant, J/K \n",
- "gnbn = 2.82e-26 #J/T\n",
- "#Calculation\n",
- "ahpbyahm = exp(-gnbn*B/(k*T))\n",
- "\n",
- "#Results\n",
- "print 'Occupation Number is %7.6f'%(ahpbyahm)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Occupation Number is 0.999990\n"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13_2.ipynb deleted file mode 100755 index 8aaf79b0..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter13_2.ipynb +++ /dev/null @@ -1,280 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:78db264f9210996c6e59dc097dbc6696221fac03055a9d5ba5033673d3e33c9f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13: Boltzmann Distribution"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.1, Page Number 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import factorial\n",
- "\n",
- "#Variable Declaration\n",
- "\n",
- "aH = 40 #Number of heads\n",
- "N = 100 #Total events\n",
- "\n",
- "#Calculations\n",
- "aT = 100 - aH\n",
- "We = factorial(N)/(factorial(aT)*factorial(aH))\n",
- "Wexpected = factorial(N)/(factorial(N/2)*factorial(N/2))\n",
- "\n",
- "#Results\n",
- "print 'The observed weight %5.2e compared to %5.2e'%(We,Wexpected)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The observed weight 1.37e+28 compared to 1.01e+29\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.2, Page Number 310"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Varialbe declaration\n",
- "n = 10000 #Total number of particles\n",
- "\n",
- "\n",
- "#Calcualtions\n",
- "def ster(i):\n",
- " return i*log(i)-i\n",
- "\n",
- "n1, n2, n3, W = symbols('n1 n2 n3 W',positive=True)\n",
- "\n",
- "n2 = 5000 - 2*n3\n",
- "n1 = 10000 - n2 -n3\n",
- "logW = ster(n) - ster(n1) - ster(n2) - ster(n3) \n",
- "fun = diff(logW, n3)\n",
- "dfun = diff(fun, n3)\n",
- "x0 = 10.0\n",
- "err = 1.0\n",
- "while err>0.001:\n",
- " f = fun.subs(n3,x0)\n",
- " df = dfun.subs(n3,x0)\n",
- " xnew = x0 - f/df\n",
- " err = abs(x0-xnew)/x0\n",
- " x0 = xnew\n",
- "\n",
- "x0 = int(x0)\n",
- "N2 = n2.subs(n3,x0)\n",
- "N3 = x0\n",
- "n1 = n1.subs(n3,x0)\n",
- "N1 = n1.subs(n2,N2)\n",
- "lnW = logW.subs(n3,N3)\n",
- "\n",
- "#Results\n",
- "print 'At maximum value of ln(W)'\n",
- "print 'Values of N1 : %4d, N2: %4d and N3: %4d '%(N1, N2,N3)\n",
- "print 'Maximum value of ln(W)= %6d'%lnW"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At maximum value of ln(W)\n",
- "Values of N1 : 6162, N2: 2676 and N3: 1162 \n",
- "Maximum value of ln(W)= 9012\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.3, Page Number 314"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "p0 = 0.633 #Probabilities of Energy level 1,2,3 \n",
- "p1 = 0.233\n",
- "p2 = 0.086\n",
- "\n",
- "#Calculation\n",
- "p4 = 1. -(p0+p1+p2)\n",
- "\n",
- "#Results\n",
- "print 'Probability of finding an oscillator at energy level of n>3 is %4.3f i.e.%4.1f percent'%(p4,p4*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of finding an oscillator at energy level of n>3 is 0.048 i.e. 4.8 percent\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.4, Page Number 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "p0 = 0.394 #Probabilities of Energy level 1,2,3 \n",
- "p1by2 = 0.239\n",
- "p2 = 0.145\n",
- "\n",
- "#Calculation\n",
- "p4 = 1. -(p0+p1by2+p2)\n",
- "\n",
- "#Results\n",
- "print 'Probability of finding an oscillator at energy level of n>3 is %4.3f'%(p4)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Probability of finding an oscillator at energy level of n>3 is 0.222\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.5, Page Number 321"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "I2 = 208 #Vibrational frequency, cm-1 \n",
- "T = 298 #Molecular Temperature, K\n",
- "c = 3.00e10 #speed of light, cm/s\n",
- "h = 6.626e-34 #Planks constant, J/K\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "#Calculation\n",
- "q = 1./(1.-exp(-h*c*I2/(k*T)))\n",
- "p2 = exp(-2*h*c*I2/(k*T))/q\n",
- "\n",
- "#Results\n",
- "print 'Partition function is %4.3f'%(q)\n",
- "print 'Probability of occupying the second vibrational state n=2 is %4.3f'%(p2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Partition function is 1.577\n",
- "Probability of occupying the second vibrational state n=2 is 0.085\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 13.6, Page Number 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "B = 1.45 #Magnetic field streangth, Teslas \n",
- "T = 298 #Molecular Temperature, K\n",
- "c = 3.00e10 #speed of light, cm/s\n",
- "h = 6.626e-34 #Planks constant, J/K\n",
- "k = 1.38e-23 #Boltzman constant, J/K \n",
- "gnbn = 2.82e-26 #J/T\n",
- "#Calculation\n",
- "ahpbyahm = exp(-gnbn*B/(k*T))\n",
- "\n",
- "#Results\n",
- "print 'Occupation Number is %7.6f'%(ahpbyahm)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Occupation Number is 0.999990\n"
- ]
- }
- ],
- "prompt_number": 18
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14.ipynb deleted file mode 100755 index daf732e9..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14.ipynb +++ /dev/null @@ -1,513 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1671a89ef38b6b7afbc72d0d6b374367e1ac4b1abd5ab2d480c98bd2b6a92ca8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: Ensemble and Molecular Partition Function"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.1, Page Number 332"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "l = 0.01 #Box length, m \n",
- "n1,n2 = 2,1 #Energy levels states\n",
- "m = 5.31e-26 #mass of oxygen molecule, kg\n",
- "\n",
- "#Calculations \n",
- "dE = (n1+n2)*h**2/(8*m*l**2)\n",
- "dEcm = dE/(h*c*1e2)\n",
- "#Results\n",
- "print 'Difference in energy levels is %3.2e J or %3.2e 1/cm'%(dE,dEcm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference in energy levels is 3.10e-38 J or 1.56e-15\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.2, Page Number 333"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, sqrt\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "v = 1.0 #Volume, L\n",
- "T = 298.0 #Temeprature of Ar, K\n",
- "m = 6.63e-26 #Mass of Argon molecule, kg \n",
- "\n",
- "#Calculations \n",
- "GAMA = h/sqrt(2*pi*m*k*T)\n",
- "v = v*1e-3\n",
- "qT3D = v/GAMA**3\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave length is %3.2e m and\\nTranslational partition function is %3.2e'%(GAMA,qT3D)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave length is 1.60e-11 m and\n",
- "Translational partition function is 2.44e+29\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.4, Page Number 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, sqrt\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "J = 4 #Rotational energy level\n",
- "B = 8.46 #Spectrum, 1/cm\n",
- "\n",
- "#Calculations \n",
- "T = (2*J+1)**2*h*c*100*B/(2*k)\n",
- "#Results\n",
- "print 'Spectrum will be observed at %4.0f K'%(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Spectrum will be observed at 494 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.5, Page Number 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "B = 60.589 #Spectrum for H2, 1/cm\n",
- "T = 1000 #Temperture of Hydrogen, K\n",
- "#Calculations \n",
- "qR = k*T/(2*h*c*100*B)\n",
- "qRs = 0.0\n",
- "#for J in range(101):\n",
- "# print J\n",
- "# if (J%2 == 0):\n",
- "# qRs = qRs + (2*J+1)*exp(-h*c*100*B*J*(J+1)/(k*T)\n",
- "# else:\n",
- "# qRs = qRs + 3*(2*J+1)*exp(-h*c*100*B*J*(J+1)/(k*T))\n",
- "#print qRs/4\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function of H2 at %4.0f is %4.3f'%(T,qR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function of H2 at 1000 is 5.729\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.6, Page Number 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "B = 0.0374 #Spectrum for H2, 1/cm\n",
- "T = 100.0 #Temperture of Hydrogen, K\n",
- "sigma = 2.\n",
- "#Calculations\n",
- "ThetaR = h*c*100*B/k\n",
- "qR = T/(sigma*ThetaR)\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function of H2 at %4.0f K is %4.3f'%(T,qR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function of H2 at 100 K is 928.121\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.7, Page Number 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, sqrt\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "Ba = 1.48 #Spectrum for OCS, 1/cm\n",
- "Bb = [2.84,0.191,0.179] #Spectrum for ONCI, 1/cm\n",
- "Bc = [9.40,1.29,1.13] #Spectrum for CH2O, 1/cm\n",
- "T = 298.0 #Temperture of Hydrogen, K\n",
- "sigmab = 1\n",
- "sigmac = 2\n",
- "\n",
- "#Calculations\n",
- "qRa = k*T/(h*c*100*Ba)\n",
- "qRb = (sqrt(pi)/sigmab)*(k*T/(h*c*100))**(3./2)*sqrt(1/Bb[0])*sqrt(1/Bb[1])*sqrt(1/Bb[2])\n",
- "qRc = (sqrt(pi)/sigmac)*(k*T/(h*c*100))**(3./2)*sqrt(1/Bc[0])*sqrt(1/Bc[1])*sqrt(1/Bc[2])\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function for OCS, ONCI, CH2O at %4.0f K are %4.0f, %4.0f, and %4.0f respectively'%(T,qRa,qRb,qRc)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function for OCS, ONCI, CH2O at 298 K are 140, 16926, and 712 respectively\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.8, Page Number 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "Ba = 1.48 #Frequency for OCS, 1/cm\n",
- "Bb = [2.84,0.191,0.179] #Frequency for ONCI, 1/cm\n",
- "Bc = [9.40,1.29,1.13] #Frequency for CH2O, 1/cm\n",
- "T298 = 298.0 #Temperture of Hydrogen, K\n",
- "T1000 = 1000 #Temperture of Hydrogen, K\n",
- "nubar = 208\n",
- "\n",
- "#Calculations\n",
- "qv298 = 1./(1.-exp(-h*c*100*nubar/(k*T298)))\n",
- "qv1000 = 1./(1.-exp(-h*c*100*nubar/(k*T1000)))\n",
- "\n",
- "#Results\n",
- "print 'Vibrational partition function for I2 at %4d and %4d are %4.2f K and %4.2f respectively'%(T298, T1000,qv298, qv1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Vibrational partition function for I2 at 298 and 1000 are 1.58 K and 3.86 respectively\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.9, Page Number 346"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "T = 298 #Temeprature, K\n",
- "nubar = [450, 945, 1100] #Vibrational mode frequencies for OClO, 1/cm\n",
- "\n",
- "#Calculations\n",
- "Qv = 1.\n",
- "for i in nubar:\n",
- " qv = 1./(1.-exp(-h*c*100*i/(k*T)))\n",
- " print 'At %4.0f 1/cm the q = %4.3f'%(i,qv)\n",
- " Qv = Qv*qv\n",
- "#Results\n",
- "print 'Total Vibrational partition function for OClO at %4.1f K is %4.3f'%(T, Qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At 450 1/cm the q = 1.128\n",
- "At 945 1/cm the q = 1.010\n",
- "At 1100 1/cm the q = 1.005\n",
- "Total Vibrational partition function for OClO at 298.0 K is 1.146 respectively\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.10, Page Number 348"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "T = 298 #Temeprature, K\n",
- "nubar = 917 #Vibrational mode frequencies for F2, 1/cm\n",
- "\n",
- "#Calculations\n",
- "ThetaV = h*c*100*nubar/k\n",
- "Th = 10*ThetaV\n",
- "qv = 1/(1.-exp(-ThetaV/Th))\n",
- "\n",
- "#Results\n",
- "print 'Vibrational partition function for F2 at %4.1f K is %4.3f'%(T, qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Vibrational partition function for F2 at 298.0 K is 10.508\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.11, Page Number 348"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "T = 1000 #Temeprature, K\n",
- "nubar = [1388, 667.4,667.4,2349] #Vibrational mode frequencies for CO2, 1/cm\n",
- "\n",
- "#Calculations\n",
- "Qv = 1.\n",
- "for i in nubar:\n",
- " qv = 1./(1.-exp(-h*c*100*i/(k*T)))\n",
- " print 'At %4.0f 1/cm the q = %4.3f'%(i,qv)\n",
- " Qv = Qv*qv\n",
- "#Results\n",
- "print 'Total Vibrational partition function for OClO at %4.1f K is %4.3f'%(T, Qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At 1388 1/cm the q = 1.157\n",
- "At 667 1/cm the q = 1.619\n",
- "At 667 1/cm the q = 1.619\n",
- "At 2349 1/cm the q = 1.035\n",
- "Total Vibrational partition function for OClO at 1000.0 K is 3.139\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.12, Page Number 352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "T = 298. #Temeprature, K\n",
- "n = [0,1,2,3,4,5,6,7,8] #Energy levels\n",
- "E0 = [0,137.38,323.46,552.96,2112.28,2153.21,2220.11,2311.36,2424.78] #Energies, 1/cm\n",
- "g0 = [4,6,8,10,2,4,6,8,10]\n",
- "\n",
- "#Calculations\n",
- "qE = 0.0\n",
- "for i in range(9):\n",
- " a =g0[i]*exp(-h*c*100*E0[i]/(k*T))\n",
- " qE = qE + a\n",
- "\n",
- "#Results\n",
- "print 'Electronic partition function for F2 at %4.1f K is %4.2f'%(T, qE)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electronic partition function for F2 at 298.0 K is 9.45\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14_1.ipynb deleted file mode 100755 index 74ee404f..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14_1.ipynb +++ /dev/null @@ -1,505 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:bb7917df7f0f53519b214ba6078e165a23f61bf03cb0026cea4d88d66f714bcc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: Ensemble and Molecular Partition Function"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.1, Page Number 332"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "l = 0.01 #Box length, m \n",
- "n1,n2 = 2,1 #Energy levels states\n",
- "m = 5.31e-26 #mass of oxygen molecule, kg\n",
- "\n",
- "#Calculations \n",
- "dE = (n1+n2)*h**2/(8*m*l**2)\n",
- "dEcm = dE/(h*c*1e2)\n",
- "#Results\n",
- "print 'Difference in energy levels is %3.2e J or %3.2e 1/cm'%(dE,dEcm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference in energy levels is 3.10e-38 J or 1.56e-15\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.2, Page Number 333"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, sqrt\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "v = 1.0 #Volume, L\n",
- "T = 298.0 #Temeprature of Ar, K\n",
- "m = 6.63e-26 #Mass of Argon molecule, kg \n",
- "\n",
- "#Calculations \n",
- "GAMA = h/sqrt(2*pi*m*k*T)\n",
- "v = v*1e-3\n",
- "qT3D = v/GAMA**3\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave length is %3.2e m and\\nTranslational partition function is %3.2e'%(GAMA,qT3D)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave length is 1.60e-11 m and\n",
- "Translational partition function is 2.44e+29\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.4, Page Number 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "J = 4 #Rotational energy level\n",
- "B = 8.46 #Spectrum, 1/cm\n",
- "\n",
- "#Calculations \n",
- "T = (2*J+1)**2*h*c*100*B/(2*k)\n",
- "#Results\n",
- "print 'Spectrum will be observed at %4.0f K'%(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Spectrum will be observed at 494 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.5, Page Number 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "B = 60.589 #Spectrum for H2, 1/cm\n",
- "T = 1000 #Temperture of Hydrogen, K\n",
- "#Calculations \n",
- "qR = k*T/(2*h*c*100*B)\n",
- "qRs = 0.0\n",
- "#for J in range(101):\n",
- "# print J\n",
- "# if (J%2 == 0):\n",
- "# qRs = qRs + (2*J+1)*exp(-h*c*100*B*J*(J+1)/(k*T)\n",
- "# else:\n",
- "# qRs = qRs + 3*(2*J+1)*exp(-h*c*100*B*J*(J+1)/(k*T))\n",
- "#print qRs/4\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function of H2 at %4.0f is %4.3f'%(T,qR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function of H2 at 1000 is 5.729\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.6, Page Number 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "B = 0.0374 #Spectrum for H2, 1/cm\n",
- "T = 100.0 #Temperture of Hydrogen, K\n",
- "sigma = 2.\n",
- "\n",
- "#Calculations\n",
- "ThetaR = h*c*100*B/k\n",
- "qR = T/(sigma*ThetaR)\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function of H2 at %4.0f K is %4.3f'%(T,qR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function of H2 at 100 K is 928.121\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.7, Page Number 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, sqrt\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "Ba = 1.48 #Spectrum for OCS, 1/cm\n",
- "Bb = [2.84,0.191,0.179] #Spectrum for ONCI, 1/cm\n",
- "Bc = [9.40,1.29,1.13] #Spectrum for CH2O, 1/cm\n",
- "T = 298.0 #Temperture of Hydrogen, K\n",
- "sigmab = 1\n",
- "sigmac = 2\n",
- "\n",
- "#Calculations\n",
- "qRa = k*T/(h*c*100*Ba)\n",
- "qRb = (sqrt(pi)/sigmab)*(k*T/(h*c*100))**(3./2)*sqrt(1/Bb[0])*sqrt(1/Bb[1])*sqrt(1/Bb[2])\n",
- "qRc = (sqrt(pi)/sigmac)*(k*T/(h*c*100))**(3./2)*sqrt(1/Bc[0])*sqrt(1/Bc[1])*sqrt(1/Bc[2])\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function for OCS, ONCI, CH2O at %4.0f K are %4.0f, %4.0f, and %4.0f respectively'%(T,qRa,qRb,qRc)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function for OCS, ONCI, CH2O at 298 K are 140, 16926, and 712 respectively\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.8, Page Number 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "Ba = 1.48 #Frequency for OCS, 1/cm\n",
- "Bb = [2.84,0.191,0.179] #Frequency for ONCI, 1/cm\n",
- "Bc = [9.40,1.29,1.13] #Frequency for CH2O, 1/cm\n",
- "T298 = 298.0 #Temperture of Hydrogen, K\n",
- "T1000 = 1000 #Temperture of Hydrogen, K\n",
- "nubar = 208\n",
- "\n",
- "#Calculations\n",
- "qv298 = 1./(1.-exp(-h*c*100*nubar/(k*T298)))\n",
- "qv1000 = 1./(1.-exp(-h*c*100*nubar/(k*T1000)))\n",
- "\n",
- "#Results\n",
- "print 'Vibrational partition function for I2 at %4d and %4d are %4.2f K and %4.2f respectively'%(T298, T1000,qv298, qv1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Vibrational partition function for I2 at 298 and 1000 are 1.58 K and 3.86 respectively\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.9, Page Number 346"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "T = 298 #Temeprature, K\n",
- "nubar = [450, 945, 1100] #Vibrational mode frequencies for OClO, 1/cm\n",
- "\n",
- "#Calculations\n",
- "Qv = 1.\n",
- "for i in nubar:\n",
- " qv = 1./(1.-exp(-h*c*100*i/(k*T)))\n",
- " print 'At %4.0f 1/cm the q = %4.3f'%(i,qv)\n",
- " Qv = Qv*qv\n",
- "#Results\n",
- "print 'Total Vibrational partition function for OClO at %4.1f K is %4.3f'%(T, Qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At 450 1/cm the q = 1.128\n",
- "At 945 1/cm the q = 1.010\n",
- "At 1100 1/cm the q = 1.005\n",
- "Total Vibrational partition function for OClO at 298.0 K is 1.146 respectively\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.10, Page Number 348"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "T = 298 #Temeprature, K\n",
- "nubar = 917 #Vibrational mode frequencies for F2, 1/cm\n",
- "\n",
- "#Calculations\n",
- "ThetaV = h*c*100*nubar/k\n",
- "Th = 10*ThetaV\n",
- "qv = 1/(1.-exp(-ThetaV/Th))\n",
- "\n",
- "#Results\n",
- "print 'Vibrational partition function for F2 at %4.1f K is %4.3f'%(T, qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Vibrational partition function for F2 at 298.0 K is 10.508\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.11, Page Number 348"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "T = 1000 #Temeprature, K\n",
- "nubar = [1388, 667.4,667.4,2349] #Vibrational mode frequencies for CO2, 1/cm\n",
- "\n",
- "#Calculations\n",
- "Qv = 1.\n",
- "for i in nubar:\n",
- " qv = 1./(1.-exp(-h*c*100*i/(k*T)))\n",
- " print 'At %4.0f 1/cm the q = %4.3f'%(i,qv)\n",
- " Qv = Qv*qv\n",
- "#Results\n",
- "print 'Total Vibrational partition function for OClO at %4.1f K is %4.3f'%(T, Qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At 1388 1/cm the q = 1.157\n",
- "At 667 1/cm the q = 1.619\n",
- "At 667 1/cm the q = 1.619\n",
- "At 2349 1/cm the q = 1.035\n",
- "Total Vibrational partition function for OClO at 1000.0 K is 3.139\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.12, Page Number 352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "T = 298. #Temeprature, K\n",
- "n = [0,1,2,3,4,5,6,7,8] #Energy levels\n",
- "E0 = [0,137.38,323.46,552.96,2112.28,2153.21,2220.11,2311.36,2424.78] #Energies, 1/cm\n",
- "g0 = [4,6,8,10,2,4,6,8,10]\n",
- "\n",
- "#Calculations\n",
- "qE = 0.0\n",
- "for i in range(9):\n",
- " a =g0[i]*exp(-h*c*100*E0[i]/(k*T))\n",
- " qE = qE + a\n",
- "\n",
- "#Results\n",
- "print 'Electronic partition function for F2 at %4.1f K is %4.2f'%(T, qE)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electronic partition function for F2 at 298.0 K is 9.45\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14_2.ipynb deleted file mode 100755 index 74ee404f..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter14_2.ipynb +++ /dev/null @@ -1,505 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:bb7917df7f0f53519b214ba6078e165a23f61bf03cb0026cea4d88d66f714bcc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: Ensemble and Molecular Partition Function"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.1, Page Number 332"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "l = 0.01 #Box length, m \n",
- "n1,n2 = 2,1 #Energy levels states\n",
- "m = 5.31e-26 #mass of oxygen molecule, kg\n",
- "\n",
- "#Calculations \n",
- "dE = (n1+n2)*h**2/(8*m*l**2)\n",
- "dEcm = dE/(h*c*1e2)\n",
- "#Results\n",
- "print 'Difference in energy levels is %3.2e J or %3.2e 1/cm'%(dE,dEcm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference in energy levels is 3.10e-38 J or 1.56e-15\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.2, Page Number 333"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, sqrt\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "v = 1.0 #Volume, L\n",
- "T = 298.0 #Temeprature of Ar, K\n",
- "m = 6.63e-26 #Mass of Argon molecule, kg \n",
- "\n",
- "#Calculations \n",
- "GAMA = h/sqrt(2*pi*m*k*T)\n",
- "v = v*1e-3\n",
- "qT3D = v/GAMA**3\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave length is %3.2e m and\\nTranslational partition function is %3.2e'%(GAMA,qT3D)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave length is 1.60e-11 m and\n",
- "Translational partition function is 2.44e+29\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.4, Page Number 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "J = 4 #Rotational energy level\n",
- "B = 8.46 #Spectrum, 1/cm\n",
- "\n",
- "#Calculations \n",
- "T = (2*J+1)**2*h*c*100*B/(2*k)\n",
- "#Results\n",
- "print 'Spectrum will be observed at %4.0f K'%(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Spectrum will be observed at 494 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.5, Page Number 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "B = 60.589 #Spectrum for H2, 1/cm\n",
- "T = 1000 #Temperture of Hydrogen, K\n",
- "#Calculations \n",
- "qR = k*T/(2*h*c*100*B)\n",
- "qRs = 0.0\n",
- "#for J in range(101):\n",
- "# print J\n",
- "# if (J%2 == 0):\n",
- "# qRs = qRs + (2*J+1)*exp(-h*c*100*B*J*(J+1)/(k*T)\n",
- "# else:\n",
- "# qRs = qRs + 3*(2*J+1)*exp(-h*c*100*B*J*(J+1)/(k*T))\n",
- "#print qRs/4\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function of H2 at %4.0f is %4.3f'%(T,qR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function of H2 at 1000 is 5.729\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.6, Page Number 341"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "B = 0.0374 #Spectrum for H2, 1/cm\n",
- "T = 100.0 #Temperture of Hydrogen, K\n",
- "sigma = 2.\n",
- "\n",
- "#Calculations\n",
- "ThetaR = h*c*100*B/k\n",
- "qR = T/(sigma*ThetaR)\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function of H2 at %4.0f K is %4.3f'%(T,qR)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function of H2 at 100 K is 928.121\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.7, Page Number 342"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, sqrt\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "Ba = 1.48 #Spectrum for OCS, 1/cm\n",
- "Bb = [2.84,0.191,0.179] #Spectrum for ONCI, 1/cm\n",
- "Bc = [9.40,1.29,1.13] #Spectrum for CH2O, 1/cm\n",
- "T = 298.0 #Temperture of Hydrogen, K\n",
- "sigmab = 1\n",
- "sigmac = 2\n",
- "\n",
- "#Calculations\n",
- "qRa = k*T/(h*c*100*Ba)\n",
- "qRb = (sqrt(pi)/sigmab)*(k*T/(h*c*100))**(3./2)*sqrt(1/Bb[0])*sqrt(1/Bb[1])*sqrt(1/Bb[2])\n",
- "qRc = (sqrt(pi)/sigmac)*(k*T/(h*c*100))**(3./2)*sqrt(1/Bc[0])*sqrt(1/Bc[1])*sqrt(1/Bc[2])\n",
- "\n",
- "#Results\n",
- "print 'Rotation partition function for OCS, ONCI, CH2O at %4.0f K are %4.0f, %4.0f, and %4.0f respectively'%(T,qRa,qRb,qRc)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rotation partition function for OCS, ONCI, CH2O at 298 K are 140, 16926, and 712 respectively\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.8, Page Number 344"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi, exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "Ba = 1.48 #Frequency for OCS, 1/cm\n",
- "Bb = [2.84,0.191,0.179] #Frequency for ONCI, 1/cm\n",
- "Bc = [9.40,1.29,1.13] #Frequency for CH2O, 1/cm\n",
- "T298 = 298.0 #Temperture of Hydrogen, K\n",
- "T1000 = 1000 #Temperture of Hydrogen, K\n",
- "nubar = 208\n",
- "\n",
- "#Calculations\n",
- "qv298 = 1./(1.-exp(-h*c*100*nubar/(k*T298)))\n",
- "qv1000 = 1./(1.-exp(-h*c*100*nubar/(k*T1000)))\n",
- "\n",
- "#Results\n",
- "print 'Vibrational partition function for I2 at %4d and %4d are %4.2f K and %4.2f respectively'%(T298, T1000,qv298, qv1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Vibrational partition function for I2 at 298 and 1000 are 1.58 K and 3.86 respectively\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.9, Page Number 346"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "\n",
- "T = 298 #Temeprature, K\n",
- "nubar = [450, 945, 1100] #Vibrational mode frequencies for OClO, 1/cm\n",
- "\n",
- "#Calculations\n",
- "Qv = 1.\n",
- "for i in nubar:\n",
- " qv = 1./(1.-exp(-h*c*100*i/(k*T)))\n",
- " print 'At %4.0f 1/cm the q = %4.3f'%(i,qv)\n",
- " Qv = Qv*qv\n",
- "#Results\n",
- "print 'Total Vibrational partition function for OClO at %4.1f K is %4.3f'%(T, Qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At 450 1/cm the q = 1.128\n",
- "At 945 1/cm the q = 1.010\n",
- "At 1100 1/cm the q = 1.005\n",
- "Total Vibrational partition function for OClO at 298.0 K is 1.146 respectively\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.10, Page Number 348"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "T = 298 #Temeprature, K\n",
- "nubar = 917 #Vibrational mode frequencies for F2, 1/cm\n",
- "\n",
- "#Calculations\n",
- "ThetaV = h*c*100*nubar/k\n",
- "Th = 10*ThetaV\n",
- "qv = 1/(1.-exp(-ThetaV/Th))\n",
- "\n",
- "#Results\n",
- "print 'Vibrational partition function for F2 at %4.1f K is %4.3f'%(T, qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Vibrational partition function for F2 at 298.0 K is 10.508\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.11, Page Number 348"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "T = 1000 #Temeprature, K\n",
- "nubar = [1388, 667.4,667.4,2349] #Vibrational mode frequencies for CO2, 1/cm\n",
- "\n",
- "#Calculations\n",
- "Qv = 1.\n",
- "for i in nubar:\n",
- " qv = 1./(1.-exp(-h*c*100*i/(k*T)))\n",
- " print 'At %4.0f 1/cm the q = %4.3f'%(i,qv)\n",
- " Qv = Qv*qv\n",
- "#Results\n",
- "print 'Total Vibrational partition function for OClO at %4.1f K is %4.3f'%(T, Qv)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At 1388 1/cm the q = 1.157\n",
- "At 667 1/cm the q = 1.619\n",
- "At 667 1/cm the q = 1.619\n",
- "At 2349 1/cm the q = 1.035\n",
- "Total Vibrational partition function for OClO at 1000.0 K is 3.139\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 14.12, Page Number 352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declarations\n",
- "h = 6.626e-34 #Planks constant, J.s\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "c = 3.0e8 #speed of light, m/s\n",
- "T = 298. #Temeprature, K\n",
- "n = [0,1,2,3,4,5,6,7,8] #Energy levels\n",
- "E0 = [0,137.38,323.46,552.96,2112.28,2153.21,2220.11,2311.36,2424.78] #Energies, 1/cm\n",
- "g0 = [4,6,8,10,2,4,6,8,10]\n",
- "\n",
- "#Calculations\n",
- "qE = 0.0\n",
- "for i in range(9):\n",
- " a =g0[i]*exp(-h*c*100*E0[i]/(k*T))\n",
- " qE = qE + a\n",
- "\n",
- "#Results\n",
- "print 'Electronic partition function for F2 at %4.1f K is %4.2f'%(T, qE)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electronic partition function for F2 at 298.0 K is 9.45\n"
- ]
- }
- ],
- "prompt_number": 26
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15.ipynb deleted file mode 100755 index e19500e1..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15.ipynb +++ /dev/null @@ -1,214 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f8110fb6016a142f9c7bb6eab2fb7171930a565929eb5cb2f65ff0d315425e0a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Statistical Thermodyanamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.2, Page Number 362"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "hnu = 1.00e-20\n",
- "NA = 6.023e23\n",
- "k = 1.38e-23\n",
- "U = 1.00e3\n",
- "n = 1\n",
- "#Calcualtions\n",
- "T = hnu/(k*log(n*NA*hnu/U-1.))\n",
- "\n",
- "#Results\n",
- "print 'For Internal energy to be %4.1f J temperature will be %4.1f K'%(U,T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For Internal energy to be 1000.0 J temperature will be 449.0 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.3, Page Number 367"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "g0 = 3.0\n",
- "c = 3.00e8\n",
- "h = 6.626e-34\n",
- "NA = 6.023e23\n",
- "k = 1.38e-23\n",
- "labda = 1263e-9\n",
- "T = 500.\n",
- "n = 1.0\n",
- "#Calcualtions\n",
- "beta = 1./(k*T)\n",
- "eps = h*c/labda\n",
- "qE = g0 + exp(-beta*eps)\n",
- "UE = n*NA*eps*exp(-beta*eps)/qE\n",
- "\n",
- "#Results\n",
- "print 'Energy of excited state is %4.2e J'%eps\n",
- "print 'Electronic partition function qE is %4.3e'%qE\n",
- "print 'Electronic partition function UE is %4.3e J'%UE"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy of excited state is 1.57e-19\n",
- "Electronic partition function qE is 3.000e+00\n",
- "Electronic partition function UE is 3.922e-06 J\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.5, Page Number 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, pi, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "Mne = 0.0201797\n",
- "Mkr = 0.0837980\n",
- "Vmne = 0.0224\n",
- "Vmkr = 0.0223\n",
- "h = 6.626e-34\n",
- "NA = 6.023e23\n",
- "k = 1.38e-23\n",
- "T = 298\n",
- "R = 8.314\n",
- "n = 1.0\n",
- "\n",
- "#Calcualtions\n",
- "mne = Mne/NA\n",
- "mkr = Mkr/NA\n",
- "Labdane = sqrt(h**2/(2*pi*mne*k*T))\n",
- "Labdakr = sqrt(h**2/(2*pi*mkr*k*T))\n",
- "Sne = 5.*R/2 + R*log(Vmne/Labdane**3)-R*log(NA)\n",
- "Skr = 5.*R/2 + R*log(Vmkr/Labdakr**3)-R*log(NA)\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave lengths for Ne is %4.2e m3'%Labdane\n",
- "print 'Std. Molar entropy for Ne is %4.2f J/(mol.K)'%Sne\n",
- "print 'Thermal wave lengths for Kr is %4.2e m3'%Labdakr\n",
- "print 'Std. Molar entropy for Kr is %4.2f J/(mol.K)'%Skr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave lengths for Ne is 2.25e-11 m3\n",
- "Std. Molar entropy for Ne is 145.46 J/(mol.K)\n",
- "Thermal wave lengths for Kr is 1.11e-11 m3\n",
- "Std. Molar entropy for Kr is 163.18 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.8, Page Number 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, pi\n",
- "\n",
- "#Variable Declaration\n",
- "M = 0.040\n",
- "h = 6.626e-34\n",
- "NA = 6.023e23\n",
- "k = 1.38e-23\n",
- "T = 298.15\n",
- "P = 1e5\n",
- "R = 8.314\n",
- "n = 1.0\n",
- "\n",
- "#Calcualtions\n",
- "m = M/NA\n",
- "Labda3 = (h**2/(2*pi*m*k*T))**(3./2)\n",
- "G0 = -n*R*T*log(k*T/(P*Labda3))\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave lengths for Ne is %4.2e m3'%Labda3\n",
- "print 'The Gibbs energy for 1 mol of Ar is %6.2f kJ'%(G0/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave lengths for Ne is 4.09e-33 m3\n",
- "The Gibbs energy for 1 mol of Ar is -39.97 kJ\n"
- ]
- }
- ],
- "prompt_number": 39
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15_1.ipynb deleted file mode 100755 index 103b8a14..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15_1.ipynb +++ /dev/null @@ -1,216 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:6dc49c200858406b9f3434d1ee04fae39e6605a618c992670722f8ceac7e7c3e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Statistical Thermodyanamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.2, Page Number 362"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "U = 1.00e3 #Total internal energy, J\n",
- "hnu = 1.00e-20 #Energy level separation, J\n",
- "NA = 6.022e23 #Avagadro's Number, 1/mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "n = 1 #Number of moles, mol\n",
- "\n",
- "#Calcualtions\n",
- "T = hnu/(k*log(n*NA*hnu/U-1.))\n",
- "\n",
- "#Results\n",
- "print 'For Internal energy to be %4.1f J temperature will be %4.1f K'%(U,T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For Internal energy to be 1000.0 J temperature will be 449.0 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.3, Page Number 367"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "g0 = 3.0 #Ground State partition function\n",
- "labda = 1263e-9 #Wave length in nm\n",
- "T = 500. #Temperature, K\n",
- "c = 3.00e8 #Speed of light, m/s\n",
- "NA = 6.022e23 #Avagadro's Number, 1/mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "n = 1.0 #Number of moles, mol\n",
- "h = 6.626e-34 #Planks's Constant, J.s\n",
- "\n",
- "#Calcualtions\n",
- "beta = 1./(k*T)\n",
- "eps = h*c/labda\n",
- "qE = g0 + exp(-beta*eps)\n",
- "UE = n*NA*eps*exp(-beta*eps)/qE\n",
- "\n",
- "#Results\n",
- "print 'Energy of excited state is %4.2e J'%eps\n",
- "print 'Electronic partition function qE is %4.3e'%qE\n",
- "print 'Electronic contribution to internal enrgy is %4.3e J'%UE"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy of excited state is 1.57e-19 J\n",
- "Electronic partition function qE is 3.000e+00\n",
- "Electronic contribution to internal enrgy is 3.921e-06 J\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.5, Page Number 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, pi, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "Mne = 0.0201797 #Molecular wt of ne, kg/mol \n",
- "Mkr = 0.0837980 #Molecular wt of kr, kg/mol\n",
- "Vmne = 0.0224 #Std. state molar volume of ne, m3\n",
- "Vmkr = 0.0223 #Std. state molar volume of kr, m3\n",
- "h = 6.626e-34 #Planks's Constant, J.s\n",
- "NA = 6.022e23 #Avagadro's Number, 1/mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "T = 298 #Std. state temeprature,K \n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "n = 1.0 #Number of mole, mol\n",
- "\n",
- "#Calcualtions\n",
- "mne = Mne/NA\n",
- "mkr = Mkr/NA\n",
- "Labdane = sqrt(h**2/(2*pi*mne*k*T))\n",
- "Labdakr = sqrt(h**2/(2*pi*mkr*k*T))\n",
- "Sne = 5.*R/2 + R*log(Vmne/Labdane**3)-R*log(NA)\n",
- "Skr = 5.*R/2 + R*log(Vmkr/Labdakr**3)-R*log(NA)\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave lengths for Ne is %4.2e m3'%Labdane\n",
- "print 'Std. Molar entropy for Ne is %4.2f J/(mol.K)'%Sne\n",
- "print 'Thermal wave lengths for Kr is %4.2e m3'%Labdakr\n",
- "print 'Std. Molar entropy for Kr is %4.2f J/(mol.K)'%Skr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave lengths for Ne is 2.25e-11 m3\n",
- "Std. Molar entropy for Ne is 145.46 J/(mol.K)\n",
- "Thermal wave lengths for Kr is 1.11e-11 m3\n",
- "Std. Molar entropy for Kr is 163.18 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.8, Page Number 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, pi\n",
- "\n",
- "#Variable Declaration\n",
- "M = 0.040 #Moleculat wt of Ar, kg/mol\n",
- "h = 6.626e-34 #Planks's Constant, J.s\n",
- "NA = 6.022e23 #Avagadro's Number, 1/mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "T = 298.15 #Std. state temeprature,K \n",
- "P = 1e5 #Std. state pressure, Pa\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "n = 1.0 #Number of mole, mol\n",
- "\n",
- "#Calcualtions\n",
- "m = M/NA\n",
- "Labda3 = (h**2/(2*pi*m*k*T))**(3./2)\n",
- "G0 = -n*R*T*log(k*T/(P*Labda3))\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave lengths for Ne is %4.2e m3'%Labda3\n",
- "print 'The Gibbs energy for 1 mol of Ar is %6.2f kJ'%(G0/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave lengths for Ne is 4.09e-33 m3\n",
- "The Gibbs energy for 1 mol of Ar is -39.97 kJ\n"
- ]
- }
- ],
- "prompt_number": 39
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15_2.ipynb deleted file mode 100755 index 103b8a14..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter15_2.ipynb +++ /dev/null @@ -1,216 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:6dc49c200858406b9f3434d1ee04fae39e6605a618c992670722f8ceac7e7c3e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Statistical Thermodyanamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.2, Page Number 362"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "U = 1.00e3 #Total internal energy, J\n",
- "hnu = 1.00e-20 #Energy level separation, J\n",
- "NA = 6.022e23 #Avagadro's Number, 1/mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "n = 1 #Number of moles, mol\n",
- "\n",
- "#Calcualtions\n",
- "T = hnu/(k*log(n*NA*hnu/U-1.))\n",
- "\n",
- "#Results\n",
- "print 'For Internal energy to be %4.1f J temperature will be %4.1f K'%(U,T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For Internal energy to be 1000.0 J temperature will be 449.0 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.3, Page Number 367"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "g0 = 3.0 #Ground State partition function\n",
- "labda = 1263e-9 #Wave length in nm\n",
- "T = 500. #Temperature, K\n",
- "c = 3.00e8 #Speed of light, m/s\n",
- "NA = 6.022e23 #Avagadro's Number, 1/mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "n = 1.0 #Number of moles, mol\n",
- "h = 6.626e-34 #Planks's Constant, J.s\n",
- "\n",
- "#Calcualtions\n",
- "beta = 1./(k*T)\n",
- "eps = h*c/labda\n",
- "qE = g0 + exp(-beta*eps)\n",
- "UE = n*NA*eps*exp(-beta*eps)/qE\n",
- "\n",
- "#Results\n",
- "print 'Energy of excited state is %4.2e J'%eps\n",
- "print 'Electronic partition function qE is %4.3e'%qE\n",
- "print 'Electronic contribution to internal enrgy is %4.3e J'%UE"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy of excited state is 1.57e-19 J\n",
- "Electronic partition function qE is 3.000e+00\n",
- "Electronic contribution to internal enrgy is 3.921e-06 J\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.5, Page Number 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, pi, sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "Mne = 0.0201797 #Molecular wt of ne, kg/mol \n",
- "Mkr = 0.0837980 #Molecular wt of kr, kg/mol\n",
- "Vmne = 0.0224 #Std. state molar volume of ne, m3\n",
- "Vmkr = 0.0223 #Std. state molar volume of kr, m3\n",
- "h = 6.626e-34 #Planks's Constant, J.s\n",
- "NA = 6.022e23 #Avagadro's Number, 1/mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "T = 298 #Std. state temeprature,K \n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "n = 1.0 #Number of mole, mol\n",
- "\n",
- "#Calcualtions\n",
- "mne = Mne/NA\n",
- "mkr = Mkr/NA\n",
- "Labdane = sqrt(h**2/(2*pi*mne*k*T))\n",
- "Labdakr = sqrt(h**2/(2*pi*mkr*k*T))\n",
- "Sne = 5.*R/2 + R*log(Vmne/Labdane**3)-R*log(NA)\n",
- "Skr = 5.*R/2 + R*log(Vmkr/Labdakr**3)-R*log(NA)\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave lengths for Ne is %4.2e m3'%Labdane\n",
- "print 'Std. Molar entropy for Ne is %4.2f J/(mol.K)'%Sne\n",
- "print 'Thermal wave lengths for Kr is %4.2e m3'%Labdakr\n",
- "print 'Std. Molar entropy for Kr is %4.2f J/(mol.K)'%Skr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave lengths for Ne is 2.25e-11 m3\n",
- "Std. Molar entropy for Ne is 145.46 J/(mol.K)\n",
- "Thermal wave lengths for Kr is 1.11e-11 m3\n",
- "Std. Molar entropy for Kr is 163.18 J/(mol.K)\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 15.8, Page Number 381"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, pi\n",
- "\n",
- "#Variable Declaration\n",
- "M = 0.040 #Moleculat wt of Ar, kg/mol\n",
- "h = 6.626e-34 #Planks's Constant, J.s\n",
- "NA = 6.022e23 #Avagadro's Number, 1/mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "T = 298.15 #Std. state temeprature,K \n",
- "P = 1e5 #Std. state pressure, Pa\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "n = 1.0 #Number of mole, mol\n",
- "\n",
- "#Calcualtions\n",
- "m = M/NA\n",
- "Labda3 = (h**2/(2*pi*m*k*T))**(3./2)\n",
- "G0 = -n*R*T*log(k*T/(P*Labda3))\n",
- "\n",
- "#Results\n",
- "print 'Thermal wave lengths for Ne is %4.2e m3'%Labda3\n",
- "print 'The Gibbs energy for 1 mol of Ar is %6.2f kJ'%(G0/1000)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal wave lengths for Ne is 4.09e-33 m3\n",
- "The Gibbs energy for 1 mol of Ar is -39.97 kJ\n"
- ]
- }
- ],
- "prompt_number": 39
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16.ipynb deleted file mode 100755 index 3aac10d8..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ae5c10cebc67e59fefae332eaff3aec21c49c3c4a8e87dfaab8a18fadc8340a5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Kinetic Theory of Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.2, Page Number 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperatureof Gas, K\n",
- "MNe = 0.020 #Molecular wt of Ne, kg/mol\n",
- "MKr = 0.083 #Molecular wt of Kr, kg/mol\n",
- "\n",
- "#Calculations\n",
- "vmpNe = sqrt(2*R*T/MNe)\n",
- "vmpKr = sqrt(2*R*T/MKr)\n",
- "\n",
- "#Results\n",
- "print 'Most probable speed of Ne and Krypton at 298 K are %4.0f, %4.0f m/s'%(vmpNe,vmpKr)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Most probable speed of Ne and Krypton at 298 K are 498, 244\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.2, Page Number 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperatureof Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "\n",
- "\n",
- "#Calculations\n",
- "vmp = sqrt(2*R*T/M)\n",
- "vave = sqrt(8*R*T/(M*pi))\n",
- "vrms = sqrt(3*R*T/M)\n",
- "\n",
- "#Results\n",
- "print 'Maximum, average, root mean square speed of Ar\\nat 298 K are %4.0f, %4.0f, %4.0f m/s'%(vmp,vave,vrms)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum, average, root mean square speed of Ar\n",
- "at 298 K are 352, 397, 431 m/s\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.4, Page Numbe 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "P = 101325 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "V = 1.0 #Volume of Container, L\n",
- "\n",
- "#Calculations\n",
- "Zc = P*NA/sqrt(2*pi*R*T*M)\n",
- "Nc = Zc*A\n",
- "#Results\n",
- "print 'Number of Collisions %4.2e per s'%(Nc)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of Collisions 2.45e+27 per s\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.5, Page Number 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "P0 = 1013.25 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "V = 1.0 #Volume of Container, L\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "t = 3600 #time of effusion, s\n",
- "A = 0.01 #Area, um2\n",
- "\n",
- "#Calculations\n",
- "A = A*1e-12\n",
- "V = V*1e-3\n",
- "expo = (A*t/V)*(k*T/(2*pi*M/NA))\n",
- "P = P0*exp(-expo)\n",
- "#Results\n",
- "print 'Pressure after 1 hr of effusion is %4.3e Pa'%(P/101325)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure after 1 hr of effusion is 1.00e-02 Pa\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.6, Page Number 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.044 #Molecular wt of CO2, kg/mol\n",
- "P = 101325 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "sigm = 5.2e-19 #m2\n",
- "\n",
- "#Calculations\n",
- "zCO2 = (P*NA/(R*T))*sigm*sqrt(2)*sqrt(8*R*T/(pi*M)) \n",
- "#Results\n",
- "print 'Single particle collisional frequency is %4.1e per s'%(zCO2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Single particle collisional frequency is 6.9e+09 per s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.7, Page Number 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "MAr = 0.04 #Molecular wt of Ar, kg/mol\n",
- "MKr = 0.084 #Molecular wt of Kr, kg/mol\n",
- "pAr = 360 #Partial Pressure Ar, torr\n",
- "pKr = 400 #Partial Pressure Kr, torr\n",
- "rAr = 0.17e-9 #Hard sphere radius of Ar, m\n",
- "rKr = 0.20e-9 #Hard sphere radius of Kr, m\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "\n",
- "#Calculations\n",
- "pAr = pAr*101325/760\n",
- "pKr = pKr*101325/760\n",
- "p1 = pAr*NA/(R*T)\n",
- "p2 = pKr*NA/(R*T)\n",
- "sigm = pi*(rAr+rKr)**2\n",
- "mu = MAr*MKr/((MAr+MKr)*NA)\n",
- "p3 = sqrt(8*k*T/(pi*mu)) \n",
- "zArKr = p1*p2*sigm*p3\n",
- "\n",
- "#Results\n",
- "print 'Collisional frequency is %4.2e m-3s-1'%(zArKr)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Collisional frequency is 3.14e+34 m-3s-1\n"
- ]
- }
- ],
- "prompt_number": 12
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16_1.ipynb deleted file mode 100755 index ca56b415..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16_1.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e4f3286f01de06b8e633d17405a3c84337f344a5c6fc5eb4c6bbd0b064d6eee8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Kinetic Theory of Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.2, Page Number 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperatureof Gas, K\n",
- "MNe = 0.020 #Molecular wt of Ne, kg/mol\n",
- "MKr = 0.083 #Molecular wt of Kr, kg/mol\n",
- "\n",
- "#Calculations\n",
- "vmpNe = sqrt(2*R*T/MNe)\n",
- "vmpKr = sqrt(2*R*T/MKr)\n",
- "\n",
- "#Results\n",
- "print 'Most probable speed of Ne and Krypton at 298 K are %4.0f, %4.0f m/s'%(vmpNe,vmpKr)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Most probable speed of Ne and Krypton at 298 K are 498, 244\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.2, Page Number 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperatureof Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "\n",
- "\n",
- "#Calculations\n",
- "vmp = sqrt(2*R*T/M)\n",
- "vave = sqrt(8*R*T/(M*pi))\n",
- "vrms = sqrt(3*R*T/M)\n",
- "\n",
- "#Results\n",
- "print 'Maximum, average, root mean square speed of Ar\\nat 298 K are %4.0f, %4.0f, %4.0f m/s'%(vmp,vave,vrms)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum, average, root mean square speed of Ar\n",
- "at 298 K are 352, 397, 431 m/s\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.4, Page Numbe 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "P = 101325 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "V = 1.0 #Volume of Container, L\n",
- "\n",
- "#Calculations\n",
- "Zc = P*NA/sqrt(2*pi*R*T*M)\n",
- "Nc = Zc*A\n",
- "#Results\n",
- "print 'Number of Collisions %4.2e per s'%(Nc)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of Collisions 2.45e+27 per s\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.5, Page Number 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "P0 = 1013.25 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "V = 1.0 #Volume of Container, L\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "t = 3600 #time of effusion, s\n",
- "A = 0.01 #Area, um2\n",
- "\n",
- "#Calculations\n",
- "A = A*1e-12\n",
- "V = V*1e-3\n",
- "expo = (A*t/V)*(k*T/(2*pi*M/NA))\n",
- "P = P0*exp(-expo)\n",
- "#Results\n",
- "print 'Pressure after 1 hr of effusion is %4.3e Pa'%(P/101325)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure after 1 hr of effusion is 1.00e-02 Pa\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.6, Page Number 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.044 #Molecular wt of CO2, kg/mol\n",
- "P = 101325 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "sigm = 5.2e-19 #m2\n",
- "\n",
- "#Calculations\n",
- "zCO2 = (P*NA/(R*T))*sigm*sqrt(2)*sqrt(8*R*T/(pi*M)) \n",
- "#Results\n",
- "print 'Single particle collisional frequency is %4.1e per s'%(zCO2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Single particle collisional frequency is 6.9e+09 per s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.7, Page Number 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "MAr = 0.04 #Molecular wt of Ar, kg/mol\n",
- "MKr = 0.084 #Molecular wt of Kr, kg/mol\n",
- "pAr = 360 #Partial Pressure Ar, torr\n",
- "pKr = 400 #Partial Pressure Kr, torr\n",
- "rAr = 0.17e-9 #Hard sphere radius of Ar, m\n",
- "rKr = 0.20e-9 #Hard sphere radius of Kr, m\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "\n",
- "#Calculations\n",
- "pAr = pAr*101325/760\n",
- "pKr = pKr*101325/760\n",
- "p1 = pAr*NA/(R*T)\n",
- "p2 = pKr*NA/(R*T)\n",
- "sigm = pi*(rAr+rKr)**2\n",
- "mu = MAr*MKr/((MAr+MKr)*NA)\n",
- "p3 = sqrt(8*k*T/(pi*mu)) \n",
- "zArKr = p1*p2*sigm*p3\n",
- "\n",
- "#Results\n",
- "print 'Collisional frequency is %4.2e m-3s-1'%(zArKr)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Collisional frequency is 3.14e+34 m-3s-1\n"
- ]
- }
- ],
- "prompt_number": 12
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16_2.ipynb deleted file mode 100755 index ca56b415..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter16_2.ipynb +++ /dev/null @@ -1,284 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e4f3286f01de06b8e633d17405a3c84337f344a5c6fc5eb4c6bbd0b064d6eee8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16: Kinetic Theory of Gases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.2, Page Number 400"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperatureof Gas, K\n",
- "MNe = 0.020 #Molecular wt of Ne, kg/mol\n",
- "MKr = 0.083 #Molecular wt of Kr, kg/mol\n",
- "\n",
- "#Calculations\n",
- "vmpNe = sqrt(2*R*T/MNe)\n",
- "vmpKr = sqrt(2*R*T/MKr)\n",
- "\n",
- "#Results\n",
- "print 'Most probable speed of Ne and Krypton at 298 K are %4.0f, %4.0f m/s'%(vmpNe,vmpKr)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Most probable speed of Ne and Krypton at 298 K are 498, 244\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.2, Page Number 401"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperatureof Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "\n",
- "\n",
- "#Calculations\n",
- "vmp = sqrt(2*R*T/M)\n",
- "vave = sqrt(8*R*T/(M*pi))\n",
- "vrms = sqrt(3*R*T/M)\n",
- "\n",
- "#Results\n",
- "print 'Maximum, average, root mean square speed of Ar\\nat 298 K are %4.0f, %4.0f, %4.0f m/s'%(vmp,vave,vrms)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum, average, root mean square speed of Ar\n",
- "at 298 K are 352, 397, 431 m/s\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.4, Page Numbe 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "P = 101325 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "V = 1.0 #Volume of Container, L\n",
- "\n",
- "#Calculations\n",
- "Zc = P*NA/sqrt(2*pi*R*T*M)\n",
- "Nc = Zc*A\n",
- "#Results\n",
- "print 'Number of Collisions %4.2e per s'%(Nc)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of Collisions 2.45e+27 per s\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.5, Page Number 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.040 #Molecular wt of Ar, kg/mol\n",
- "P0 = 1013.25 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "V = 1.0 #Volume of Container, L\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "t = 3600 #time of effusion, s\n",
- "A = 0.01 #Area, um2\n",
- "\n",
- "#Calculations\n",
- "A = A*1e-12\n",
- "V = V*1e-3\n",
- "expo = (A*t/V)*(k*T/(2*pi*M/NA))\n",
- "P = P0*exp(-expo)\n",
- "#Results\n",
- "print 'Pressure after 1 hr of effusion is %4.3e Pa'%(P/101325)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure after 1 hr of effusion is 1.00e-02 Pa\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.6, Page Number 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "M = 0.044 #Molecular wt of CO2, kg/mol\n",
- "P = 101325 #Pressure, N/m2\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "sigm = 5.2e-19 #m2\n",
- "\n",
- "#Calculations\n",
- "zCO2 = (P*NA/(R*T))*sigm*sqrt(2)*sqrt(8*R*T/(pi*M)) \n",
- "#Results\n",
- "print 'Single particle collisional frequency is %4.1e per s'%(zCO2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Single particle collisional frequency is 6.9e+09 per s\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 16.7, Page Number 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt, pi\n",
- "\n",
- "#Variable Declaration\n",
- "R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
- "T = 298 #Temperature of Gas, K\n",
- "MAr = 0.04 #Molecular wt of Ar, kg/mol\n",
- "MKr = 0.084 #Molecular wt of Kr, kg/mol\n",
- "pAr = 360 #Partial Pressure Ar, torr\n",
- "pKr = 400 #Partial Pressure Kr, torr\n",
- "rAr = 0.17e-9 #Hard sphere radius of Ar, m\n",
- "rKr = 0.20e-9 #Hard sphere radius of Kr, m\n",
- "NA = 6.022e23 #Number of particles per mol\n",
- "k = 1.38e-23 #Boltzmann constant, J/K\n",
- "\n",
- "#Calculations\n",
- "pAr = pAr*101325/760\n",
- "pKr = pKr*101325/760\n",
- "p1 = pAr*NA/(R*T)\n",
- "p2 = pKr*NA/(R*T)\n",
- "sigm = pi*(rAr+rKr)**2\n",
- "mu = MAr*MKr/((MAr+MKr)*NA)\n",
- "p3 = sqrt(8*k*T/(pi*mu)) \n",
- "zArKr = p1*p2*sigm*p3\n",
- "\n",
- "#Results\n",
- "print 'Collisional frequency is %4.2e m-3s-1'%(zArKr)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Collisional frequency is 3.14e+34 m-3s-1\n"
- ]
- }
- ],
- "prompt_number": 12
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17.ipynb deleted file mode 100755 index 1a941897..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17.ipynb +++ /dev/null @@ -1,509 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:28b1b127871a242526fc780b7f95bfe7e107178dbbf2ff194eb0ee46817a771a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Transport Phenomena"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.1, Page Number 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from scipy import constants\n",
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "M = 0.040 #Molecualar wt of Argon, kh/mol\n",
- "P, T = 101325.0, 298.0 #Pressure and Temperature, Pa, K\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "N_A = 6.02214129e+23 #mol^-1\n",
- "#Calculations\n",
- "DAr = (1./3)*sqrt(8*R*T/(pi*M))*(R*T/(P*N_A*sqrt(2)*sigm))\n",
- "\n",
- "#Results\n",
- "print 'Diffusion coefficient of Argon %3.1e m2/s'%DAr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diffusion coefficient of Argon 1.1e-05 m2/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.2, Page Number 418"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from scipy import constants\n",
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "DHebyAr = 4.0 \n",
- "MAr, MHe = 39.9, 4.0 #Molecualar wt of Argon and Neon, kg/mol\n",
- "P, T = 101325.0, 298.0 #Pressure and Temperature, Pa, K\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "N_A = 6.02214129e+23 #mol^-1\n",
- "#Calculations\n",
- "sigHebyAr = (1./DHebyAr)*sqrt(MAr/MHe)\n",
- "\n",
- "#Results\n",
- "print 'Ratio of collision cross sections of Helium to Argon %4.3f'%sigHebyAr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of collision cross sections of Helium to Argon 0.790\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.3, Page Number 420"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "D = 1.0e-5 #Diffusion coefficient, m2/s \n",
- "t1 = 1000 #Time, s\n",
- "t10 = 10000 #Time, s\n",
- "\n",
- "#Calculations\n",
- "xrms1 = sqrt(2*D*t1)\n",
- "xrms10 = sqrt(2*D*t10)\n",
- "\n",
- "#Results\n",
- "print 'rms displacement at %4d and %4d is %4.3f and %4.3f m respectively'%(t1,t10,xrms1,xrms10)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "rms displacement at 1000 and 10000 is 0.141 and 0.447 m respectively\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.4, Page Number 421"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "D = 2.2e-5 #Diffusion coefficient of benzene, cm2/s \n",
- "x0 = 0.3 #molecular diameter of benzene, nm\n",
- "\n",
- "#Calculations\n",
- "t = (x0*1e-9)**2/(2*D*1e-4)\n",
- "\n",
- "#Results\n",
- "print 'Time per random walk is %4.3e s or %4.2f ps'%(t,t/1e-12)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time per random walk is 2.045e-11 s or 20.45 ps\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.5, Page Number 424"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "P = 101325 #Pressure, Pa\n",
- "kt = 0.0177 #Thermal conductivity, J/(K.m.s)\n",
- "T = 300.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 39.9 #Molecualar wt of Argon and Neon, kg/mol\n",
- "#Calculations\n",
- "CvmbyNA = 3.*k/2\n",
- "nuavg = sqrt(8*R*T/(pi*M*1e-3))\n",
- "N = NA*P/(R*T)\n",
- "labda = 3*kt/(CvmbyNA*nuavg*N)\n",
- "sigm = 1/(sqrt(2)*N*labda)\n",
- "\n",
- "#Results\n",
- "print 'Mean free path %4.3e m and collisional cross section %4.2e m2'%(labda, sigm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mean free path 2.627e-07 m and collisional cross section 1.10e-19 m2\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.6, Page Number 427"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "eta = 227. #Viscosity of Ar, muP\n",
- "P = 101325 #Pressure, Pa\n",
- "kt = 0.0177 #Thermal conductivity, J/(K.m.s)\n",
- "T = 300.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 39.9 #Molecualar wt of Argon and Neon, kg/mol\n",
- "\n",
- "#Calculations\n",
- "nuavg = sqrt(8*R*T/(pi*M*1e-3))\n",
- "N = NA*P/(R*T)\n",
- "m = M*1e-3/NA\n",
- "labda = 3.*eta*1e-7/(nuavg*N*m) #viscosity in kg m s units\n",
- "sigm = 1./(sqrt(2)*N*labda)\n",
- "\n",
- "#Results\n",
- "print 'Collisional cross section %4.2e m2'%(sigm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Collisional cross section 2.74e-19 m2\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.7, Page Number 429"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "m = 22.7 #Mass of CO2, kg\n",
- "T = 293.0 #Temperature, K\n",
- "L = 1.0 #length of the tube, m\n",
- "d = 0.75 #Diameter of the tube, mm\n",
- "eta = 146 #Viscosity of CO2, muP\n",
- "p1 = 1.05 #Inlet pressure, atm\n",
- "p2 = 1.00 #Outlet pressure, atm\n",
- "atm2pa = 101325 #Conversion for pressure from atm to Pa \n",
- "M = 0.044 #Molecular wt of CO2, kg/mol\n",
- "R = 8.314 #Molar Gas constant, J mol^-1 K^-1\n",
- "\n",
- "#Calculations\n",
- "p1 = p1*atm2pa\n",
- "p2 = p2*atm2pa\n",
- "F = pi*(d*1e-3/2)**4*(p1**2-p2**2)/(16.*eta/1.e7*L*p2)\n",
- "nCO2 = m/M\n",
- "v = nCO2*R*T/((p1+p2)/2)\n",
- "t = v/F\n",
- "\n",
- "#Results\n",
- "print 'Flow rate is %4.3e m3/s'%(F)\n",
- "print 'Cylinder can be used for %4.3e s nearly %3.1f days'%(t, t/(24*3600))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Flow rate is 2.762e-06 m3/s\n",
- "Cylinder can be used for 4.381e+06 s nearly 50.7 days\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.8, Page Number 431"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "eta = 0.891 #Viscosity of hemoglobin in water, cP\n",
- "T = 298.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "D = 6.9e-11 #Diffusion coefficient, m2/s \n",
- "\n",
- "#Calculations\n",
- "r = k*T/(6*pi*eta*1e-3*D)\n",
- "\n",
- "#Results\n",
- "print 'Radius of protein is %4.3f nm'%(r/1e-9)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of protein is 3.550 nm\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.9, Page Number 432"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "s = 1.91e-13 #Sedimentation constant, s\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 14100.0 #Molecualr wt of lysozyme, g/mol\n",
- "rho = 0.998 #Density of water, kg/m3\n",
- "eta = 1.002 #Viscosity lysozyme in water, cP\n",
- "T = 293.15 #Temperature, K\n",
- "vbar = 0.703 #Specific volume of cm3/g\n",
- "\n",
- "#Calculations\n",
- "m = M/NA\n",
- "f = m*(1.-vbar*rho)/s\n",
- "r = f/(6*pi*eta)\n",
- "\n",
- "#Results\n",
- "print 'Radius of Lysozyme particle is %4.3f nm'%(r/1e-9)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of Lysozyme particle is 1.937 nm\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.10, Page Number 433"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg,log, exp\n",
- "from pylab import plot,show\n",
- "\n",
- "\n",
- "#Variable Declaration\n",
- "t = array([0.0,30.0,60.0,90.0,120.0,150.0]) #Time, min\n",
- "xb = array([6.00,6.07,6.14,6.21,6.28,6.35]) #Location of boundary layer, cm\n",
- "rpm = 55000. #RPM of centrifuge \n",
- "\n",
- "#Calculations\n",
- "nx = xb/xb[0]\n",
- "lnx = log(nx)\n",
- "A = array([ t, ones(size(t))])\n",
- "# linearly generated sequence\n",
- "[slope, intercept] = linalg.lstsq(A.T,lnx)[0] # obtaining the parameters\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = slope*t+intercept # regression line\n",
- "#Results\n",
- "plot(t,line,'-',t,lnx,'o')\n",
- "xlabel('Time, min')\n",
- "ylabel('log(xb/xb0)')\n",
- "show()\n",
- "sbar = (slope/60)/(rpm*2*pi/60)**2\n",
- "print 'Slope is %6.2e 1/min or %4.3e 1/s '%(slope, slope/60)\n",
- "print 'Sedimentation factor is %4.3e s'%(sbar)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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Yp/v734df/hL22CPpRCLSmpkZ7l77DE9aYu003H2bmRUDM4G2wL3uvszMLoz2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\n9v8BCPiGPVHALbQAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x850c3d0>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope is 3.78e-04 1/min or 6.299e-06 1/s \n",
- "Sedimentation factor is 1.899e-13 s\n"
- ]
- }
- ],
- "prompt_number": 67
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.11, Page Number 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "LMg = 0.0106 #Ionic conductance for Mg, S.m2/mol\n",
- "LCl = 0.0076 #Ionic conductance for Cl, S.m2/mol\n",
- "nMg, nCl = 1, 2 #Coefficients of Mg and Cl \n",
- "\n",
- "\n",
- "#Calculations\n",
- "LMgCl2 = nMg*LMg + nCl*LCl\n",
- "\n",
- "#Results\n",
- "print 'Molar conductivity of MgCl2 on infinite dilution is %5.4f S.m2/mol'%(LMgCl2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molar conductivity of MgCl2 on infinite dilution is 0.0258 S.m2/mol\n"
- ]
- }
- ],
- "prompt_number": 59
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17_1.ipynb deleted file mode 100755 index 67736314..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17_1.ipynb +++ /dev/null @@ -1,504 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:aa0a13ac0b2f590a7935298520fa0293eee8e76499442eb9a4a0b49a824ca2fb"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Transport Phenomena"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.1, Page Number 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from scipy import constants\n",
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "M = 0.040 #Molecualar wt of Argon, kh/mol\n",
- "P, T = 101325.0, 298.0 #Pressure and Temperature, Pa, K\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "N_A = 6.02214129e+23 #mol^-1\n",
- "#Calculations\n",
- "DAr = (1./3)*sqrt(8*R*T/(pi*M))*(R*T/(P*N_A*sqrt(2)*sigm))\n",
- "\n",
- "#Results\n",
- "print 'Diffusion coefficient of Argon %3.1e m2/s'%DAr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diffusion coefficient of Argon 1.1e-05 m2/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.2, Page Number 418"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "DHebyAr = 4.0 \n",
- "MAr, MHe = 39.9, 4.0 #Molecualar wt of Argon and Neon, kg/mol\n",
- "P, T = 101325.0, 298.0 #Pressure and Temperature, Pa, K\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "N_A = 6.02214129e+23 #mol^-1\n",
- "#Calculations\n",
- "sigHebyAr = (1./DHebyAr)*sqrt(MAr/MHe)\n",
- "\n",
- "#Results\n",
- "print 'Ratio of collision cross sections of Helium to Argon %4.3f'%sigHebyAr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of collision cross sections of Helium to Argon 0.790\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.3, Page Number 420"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "D = 1.0e-5 #Diffusion coefficient, m2/s \n",
- "t1 = 1000 #Time, s\n",
- "t10 = 10000 #Time, s\n",
- "\n",
- "#Calculations\n",
- "xrms1 = sqrt(2*D*t1)\n",
- "xrms10 = sqrt(2*D*t10)\n",
- "\n",
- "#Results\n",
- "print 'rms displacement at %4d and %4d is %4.3f and %4.3f m respectively'%(t1,t10,xrms1,xrms10)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "rms displacement at 1000 and 10000 is 0.141 and 0.447 m respectively\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.4, Page Number 421"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "D = 2.2e-5 #Diffusion coefficient of benzene, cm2/s \n",
- "x0 = 0.3 #molecular diameter of benzene, nm\n",
- "\n",
- "#Calculations\n",
- "t = (x0*1e-9)**2/(2*D*1e-4)\n",
- "\n",
- "#Results\n",
- "print 'Time per random walk is %4.3e s or %4.2f ps'%(t,t/1e-12)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time per random walk is 2.045e-11 s or 20.45 ps\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.5, Page Number 424"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "P = 101325 #Pressure, Pa\n",
- "kt = 0.0177 #Thermal conductivity, J/(K.m.s)\n",
- "T = 300.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 39.9 #Molecualar wt of Argon and Neon, kg/mol\n",
- "\n",
- "#Calculations\n",
- "CvmbyNA = 3.*k/2\n",
- "nuavg = sqrt(8*R*T/(pi*M*1e-3))\n",
- "N = NA*P/(R*T)\n",
- "labda = 3*kt/(CvmbyNA*nuavg*N)\n",
- "sigm = 1/(sqrt(2)*N*labda)\n",
- "\n",
- "#Results\n",
- "print 'Mean free path %4.3e m and collisional cross section %4.2e m2'%(labda, sigm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mean free path 2.627e-07 m and collisional cross section 1.10e-19 m2\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.6, Page Number 427"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "eta = 227. #Viscosity of Ar, muP\n",
- "P = 101325 #Pressure, Pa\n",
- "kt = 0.0177 #Thermal conductivity, J/(K.m.s)\n",
- "T = 300.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 39.9 #Molecualar wt of Argon and Neon, kg/mol\n",
- "\n",
- "#Calculations\n",
- "nuavg = sqrt(8*R*T/(pi*M*1e-3))\n",
- "N = NA*P/(R*T)\n",
- "m = M*1e-3/NA\n",
- "labda = 3.*eta*1e-7/(nuavg*N*m) #viscosity in kg m s units\n",
- "sigm = 1./(sqrt(2)*N*labda)\n",
- "\n",
- "#Results\n",
- "print 'Collisional cross section %4.2e m2'%(sigm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Collisional cross section 2.74e-19 m2\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.7, Page Number 429"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "m = 22.7 #Mass of CO2, kg\n",
- "T = 293.0 #Temperature, K\n",
- "L = 1.0 #length of the tube, m\n",
- "d = 0.75 #Diameter of the tube, mm\n",
- "eta = 146 #Viscosity of CO2, muP\n",
- "p1 = 1.05 #Inlet pressure, atm\n",
- "p2 = 1.00 #Outlet pressure, atm\n",
- "atm2pa = 101325 #Conversion for pressure from atm to Pa \n",
- "M = 0.044 #Molecular wt of CO2, kg/mol\n",
- "R = 8.314 #Molar Gas constant, J mol^-1 K^-1\n",
- "\n",
- "#Calculations\n",
- "p1 = p1*atm2pa\n",
- "p2 = p2*atm2pa\n",
- "F = pi*(d*1e-3/2)**4*(p1**2-p2**2)/(16.*eta/1.e7*L*p2)\n",
- "nCO2 = m/M\n",
- "v = nCO2*R*T/((p1+p2)/2)\n",
- "t = v/F\n",
- "\n",
- "#Results\n",
- "print 'Flow rate is %4.3e m3/s'%(F)\n",
- "print 'Cylinder can be used for %4.3e s nearly %3.1f days'%(t, t/(24*3600))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Flow rate is 2.762e-06 m3/s\n",
- "Cylinder can be used for 4.381e+06 s nearly 50.7 days\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.8, Page Number 431"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "eta = 0.891 #Viscosity of hemoglobin in water, cP\n",
- "T = 298.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "D = 6.9e-11 #Diffusion coefficient, m2/s \n",
- "\n",
- "#Calculations\n",
- "r = k*T/(6*pi*eta*1e-3*D)\n",
- "\n",
- "#Results\n",
- "print 'Radius of protein is %4.3f nm'%(r/1e-9)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of protein is 3.550 nm\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.9, Page Number 432"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "s = 1.91e-13 #Sedimentation constant, s\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 14100.0 #Molecualr wt of lysozyme, g/mol\n",
- "rho = 0.998 #Density of water, kg/m3\n",
- "eta = 1.002 #Viscosity lysozyme in water, cP\n",
- "T = 293.15 #Temperature, K\n",
- "vbar = 0.703 #Specific volume of cm3/g\n",
- "\n",
- "#Calculations\n",
- "m = M/NA\n",
- "f = m*(1.-vbar*rho)/s\n",
- "r = f/(6*pi*eta)\n",
- "\n",
- "#Results\n",
- "print 'Radius of Lysozyme particle is %4.3f nm'%(r/1e-9)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of Lysozyme particle is 1.937 nm\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.10, Page Number 433"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg,log, exp\n",
- "from pylab import plot,show\n",
- "\n",
- "#Variable Declaration\n",
- "t = array([0.0,30.0,60.0,90.0,120.0,150.0]) #Time, min\n",
- "xb = array([6.00,6.07,6.14,6.21,6.28,6.35]) #Location of boundary layer, cm\n",
- "rpm = 55000. #RPM of centrifuge \n",
- "\n",
- "#Calculations\n",
- "nx = xb/xb[0]\n",
- "lnx = log(nx)\n",
- "A = array([ t, ones(size(t))])\n",
- "# linearly generated sequence\n",
- "[slope, intercept] = linalg.lstsq(A.T,lnx)[0] # obtaining the parameters\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = slope*t+intercept # regression line\n",
- "\n",
- "#Results\n",
- "plot(t,line,'-',t,lnx,'o')\n",
- "xlabel('Time, min')\n",
- "ylabel('log(xb/xb0)')\n",
- "show()\n",
- "sbar = (slope/60)/(rpm*2*pi/60)**2\n",
- "print 'Slope is %6.2e 1/min or %4.3e 1/s '%(slope, slope/60)\n",
- "print 'Sedimentation factor is %4.3e s'%(sbar)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x850c3d0>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope is 3.78e-04 1/min or 6.299e-06 1/s \n",
- "Sedimentation factor is 1.899e-13 s\n"
- ]
- }
- ],
- "prompt_number": 67
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.11, Page Number 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "LMg = 0.0106 #Ionic conductance for Mg, S.m2/mol\n",
- "LCl = 0.0076 #Ionic conductance for Cl, S.m2/mol\n",
- "nMg, nCl = 1, 2 #Coefficients of Mg and Cl \n",
- "\n",
- "#Calculations\n",
- "LMgCl2 = nMg*LMg + nCl*LCl\n",
- "\n",
- "#Results\n",
- "print 'Molar conductivity of MgCl2 on infinite dilution is %5.4f S.m2/mol'%(LMgCl2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molar conductivity of MgCl2 on infinite dilution is 0.0258 S.m2/mol\n"
- ]
- }
- ],
- "prompt_number": 59
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17_2.ipynb deleted file mode 100755 index 7cd1557e..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter17_2.ipynb +++ /dev/null @@ -1,506 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:eb5ec1473fe382bc4c20517f981244e27fb89ea55598df7473fff0c3c26d19fc"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Transport Phenomena"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.1, Page Number 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from scipy import constants\n",
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "M = 0.040 #Molecualar wt of Argon, kh/mol\n",
- "P, T = 101325.0, 298.0 #Pressure and Temperature, Pa, K\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "N_A = 6.02214129e+23 #mol^-1\n",
- "#Calculations\n",
- "DAr = (1./3)*sqrt(8*R*T/(pi*M))*(R*T/(P*N_A*sqrt(2)*sigm))\n",
- "\n",
- "#Results\n",
- "print 'Diffusion coefficient of Argon %3.1e m2/s'%DAr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diffusion coefficient of Argon 1.1e-05 m2/s\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.2, Page Number 418"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "DHebyAr = 4.0 \n",
- "MAr, MHe = 39.9, 4.0 #Molecualar wt of Argon and Neon, kg/mol\n",
- "P, T = 101325.0, 298.0 #Pressure and Temperature, Pa, K\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "N_A = 6.02214129e+23 #mol^-1\n",
- "#Calculations\n",
- "sigHebyAr = (1./DHebyAr)*sqrt(MAr/MHe)\n",
- "\n",
- "#Results\n",
- "print 'Ratio of collision cross sections of Helium to Argon %4.3f'%sigHebyAr"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of collision cross sections of Helium to Argon 0.790\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.3, Page Number 420"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt\n",
- "\n",
- "#Variable Declaration\n",
- "D = 1.0e-5 #Diffusion coefficient, m2/s \n",
- "t1 = 1000 #Time, s\n",
- "t10 = 10000 #Time, s\n",
- "\n",
- "#Calculations\n",
- "xrms1 = sqrt(2*D*t1)\n",
- "xrms10 = sqrt(2*D*t10)\n",
- "\n",
- "#Results\n",
- "print 'rms displacement at %4d and %4d is %4.3f and %4.3f m respectively'%(t1,t10,xrms1,xrms10)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "rms displacement at 1000 and 10000 is 0.141 and 0.447 m respectively\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.4, Page Number 421"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "D = 2.2e-5 #Diffusion coefficient of benzene, cm2/s \n",
- "x0 = 0.3 #molecular diameter of benzene, nm\n",
- "\n",
- "#Calculations\n",
- "t = (x0*1e-9)**2/(2*D*1e-4)\n",
- "\n",
- "#Results\n",
- "print 'Time per random walk is %4.3e s or %4.2f ps'%(t,t/1e-12)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time per random walk is 2.045e-11 s or 20.45 ps\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.5, Page Number 424"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "P = 101325 #Pressure, Pa\n",
- "kt = 0.0177 #Thermal conductivity, J/(K.m.s)\n",
- "T = 300.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "sigm = 3.6e-19 #\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 39.9 #Molecualar wt of Argon and Neon, kg/mol\n",
- "\n",
- "#Calculations\n",
- "CvmbyNA = 3.*k/2\n",
- "nuavg = sqrt(8*R*T/(pi*M*1e-3))\n",
- "N = NA*P/(R*T)\n",
- "labda = 3*kt/(CvmbyNA*nuavg*N)\n",
- "sigm = 1/(sqrt(2)*N*labda)\n",
- "\n",
- "#Results\n",
- "print 'Mean free path %4.3e m and collisional cross section %4.2e m2'%(labda, sigm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mean free path 2.627e-07 m and collisional cross section 1.10e-19 m2\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.6, Page Number 427"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "eta = 227. #Viscosity of Ar, muP\n",
- "P = 101325 #Pressure, Pa\n",
- "kt = 0.0177 #Thermal conductivity, J/(K.m.s)\n",
- "T = 300.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 39.9 #Molecualar wt of Argon and Neon, kg/mol\n",
- "\n",
- "#Calculations\n",
- "nuavg = sqrt(8*R*T/(pi*M*1e-3))\n",
- "N = NA*P/(R*T)\n",
- "m = M*1e-3/NA\n",
- "labda = 3.*eta*1e-7/(nuavg*N*m) #viscosity in kg m s units\n",
- "sigm = 1./(sqrt(2)*N*labda)\n",
- "\n",
- "#Results\n",
- "print 'Collisional cross section %4.2e m2'%(sigm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Collisional cross section 2.74e-19 m2\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.7, Page Number 429"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "m = 22.7 #Mass of CO2, kg\n",
- "T = 293.0 #Temperature, K\n",
- "L = 1.0 #length of the tube, m\n",
- "d = 0.75 #Diameter of the tube, mm\n",
- "eta = 146 #Viscosity of CO2, muP\n",
- "p1 = 1.05 #Inlet pressure, atm\n",
- "p2 = 1.00 #Outlet pressure, atm\n",
- "atm2pa = 101325 #Conversion for pressure from atm to Pa \n",
- "M = 0.044 #Molecular wt of CO2, kg/mol\n",
- "R = 8.314 #Molar Gas constant, J mol^-1 K^-1\n",
- "\n",
- "#Calculations\n",
- "p1 = p1*atm2pa\n",
- "p2 = p2*atm2pa\n",
- "F = pi*(d*1e-3/2)**4*(p1**2-p2**2)/(16.*eta/1.e7*L*p2)\n",
- "nCO2 = m/M\n",
- "v = nCO2*R*T/((p1+p2)/2)\n",
- "t = v/F\n",
- "\n",
- "#Results\n",
- "print 'Flow rate is %4.3e m3/s'%(F)\n",
- "print 'Cylinder can be used for %4.3e s nearly %3.1f days'%(t, t/(24*3600))"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Flow rate is 2.762e-06 m3/s\n",
- "Cylinder can be used for 4.381e+06 s nearly 50.7 days\n"
- ]
- }
- ],
- "prompt_number": 80
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.8, Page Number 431"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "eta = 0.891 #Viscosity of hemoglobin in water, cP\n",
- "T = 298.0 #Temperature, K\n",
- "k = 1.3806488e-23 #Boltzmanconstant,J K^-1\n",
- "R = 8.314 #Molar Gas constant, mol^-1 K^-1\n",
- "D = 6.9e-11 #Diffusion coefficient, m2/s \n",
- "\n",
- "#Calculations\n",
- "r = k*T/(6*pi*eta*1e-3*D)\n",
- "\n",
- "#Results\n",
- "print 'Radius of protein is %4.3f nm'%(r/1e-9)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of protein is 3.550 nm\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.9, Page Number 432"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import sqrt,pi\n",
- "\n",
- "#Variable Declaration\n",
- "s = 1.91e-13 #Sedimentation constant, s\n",
- "NA = 6.02214129e+23 #mol^-1\n",
- "M = 14100.0 #Molecualr wt of lysozyme, g/mol\n",
- "rho = 0.998 #Density of water, kg/m3\n",
- "eta = 1.002 #Viscosity lysozyme in water, cP\n",
- "T = 293.15 #Temperature, K\n",
- "vbar = 0.703 #Specific volume of cm3/g\n",
- "\n",
- "#Calculations\n",
- "m = M/NA\n",
- "f = m*(1.-vbar*rho)/s\n",
- "r = f/(6*pi*eta)\n",
- "\n",
- "#Results\n",
- "print 'Radius of Lysozyme particle is %4.3f nm'%(r/1e-9)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radius of Lysozyme particle is 1.937 nm\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.10, Page Number 433"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg,log, exp\n",
- "from matplotlib.pylab import plot,show\n",
- "\n",
- "%matplotlib inline\n",
- "\n",
- "#Variable Declaration\n",
- "t = array([0.0,30.0,60.0,90.0,120.0,150.0]) #Time, min\n",
- "xb = array([6.00,6.07,6.14,6.21,6.28,6.35]) #Location of boundary layer, cm\n",
- "rpm = 55000. #RPM of centrifuge \n",
- "\n",
- "#Calculations\n",
- "nx = xb/xb[0]\n",
- "lnx = log(nx)\n",
- "A = array([ t, ones(size(t))])\n",
- "# linearly generated sequence\n",
- "[slope, intercept] = linalg.lstsq(A.T,lnx)[0] # obtaining the parameters\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = slope*t+intercept # regression line\n",
- "\n",
- "#Results\n",
- "plot(t,line,'-',t,lnx,'o')\n",
- "xlabel('$ Time, min $')\n",
- "ylabel('$ \\log(x_b/x_{b0}) $')\n",
- "show()\n",
- "sbar = (slope/60)/(rpm*2*pi/60)**2\n",
- "print 'Slope is %6.2e 1/min or %4.3e 1/s '%(slope, slope/60)\n",
- "print 'Sedimentation factor is %4.3e s'%(sbar)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5925330>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope is 3.78e-04 1/min or 6.299e-06 1/s \n",
- "Sedimentation factor is 1.899e-13 s\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 17.11, Page Number 439"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "LMg = 0.0106 #Ionic conductance for Mg, S.m2/mol\n",
- "LCl = 0.0076 #Ionic conductance for Cl, S.m2/mol\n",
- "nMg, nCl = 1, 2 #Coefficients of Mg and Cl \n",
- "\n",
- "#Calculations\n",
- "LMgCl2 = nMg*LMg + nCl*LCl\n",
- "\n",
- "#Results\n",
- "print 'Molar conductivity of MgCl2 on infinite dilution is %5.4f S.m2/mol'%(LMgCl2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molar conductivity of MgCl2 on infinite dilution is 0.0258 S.m2/mol\n"
- ]
- }
- ],
- "prompt_number": 59
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18.ipynb deleted file mode 100755 index f1741f74..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18.ipynb +++ /dev/null @@ -1,407 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:739f6a7aa866a8931c051faa7d51c479c27f099d05036b55ada1468f0458424b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Elementary Chemical Kinetics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.2, Page Number 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "Ca0 = [2.3e-4,4.6e-4,9.2e-4] #Initial Concentration of A, M\n",
- "Cb0 = [3.1e-5,6.2e-5,6.2e-5] #Initial Concentration of B, M\n",
- "Ri = [5.25e-4,4.2e-3,1.68e-2] #Initial rate of reaction, M\n",
- "\n",
- "#Calculations\n",
- "alp = log(Ri[1]/Ri[2])/log(Ca0[1]/Ca0[2])\n",
- "beta = (log(Ri[0]/Ri[1]) - 2*log((Ca0[0]/Ca0[1])))/(log(Cb0[0]/Cb0[1]))\n",
- "k = Ri[2]/(Ca0[2]**2*Cb0[2]**beta)\n",
- "\n",
- "#REsults\n",
- "print 'Order of reaction with respect to reactant A: %3.2f'%alp\n",
- "print 'Order of reaction with respect to reactant A: %3.2f'%beta\n",
- "print 'Rate constant of the reaction: %4.3e 1./(M.s)'%k"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Order of reaction with respect to reactant A: 2.00\n",
- "Order of reaction with respect to reactant A: 1.00\n",
- "Rate constant of the reaction: 3.201e+08 1./(M.s)\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.3, Page Number 457"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "t1by2 = 2.05e4 #Half life for first order decomposition of N2O5, s\n",
- "x = 60. #percentage decay of N2O5\n",
- "\n",
- "#Calculations\n",
- "k = log(2)/t1by2\n",
- "t = -log(x/100)/k\n",
- "\n",
- "#REsults\n",
- "print 'Rate constant of the reaction: %4.3e 1/s'%k\n",
- "print 'Timerequire for 60 percent decay of N2O5: %4.3e s'%t"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rate constant of the reaction: 3.381e-05 1/s\n",
- "Timerequire for 60 percent decay of N2O5: 1.511e+04 s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.4, Page Number 457 Incomplete"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "t1by2 = 5760 #Half life for C14, years\n",
- "\n",
- "\n",
- "#Calculations\n",
- "k = log(2)/t1by2\n",
- "t = -log(x/100)/k\n",
- "\n",
- "#REsults\n",
- "print 'Rate constant of the reaction: %4.3e 1/s'%k\n",
- "print 'Timerequire for 60 percent decay of N2O5: %4.3e s'%t"
- ],
- "language": "python",
- "metadata": {},
- "outputs": []
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.5, Page Number 463"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "kAbykI = 2.0 #Ratio of rate constants\n",
- "kA = 0.1 #First order rate constant for rxn 1, 1/s \n",
- "kI = 0.05 #First order rate constant for rxn 2, 1/s \n",
- "#Calculations\n",
- "tmax = 1/(kA-kI)*log(kA/kI)\n",
- "\n",
- "#Results\n",
- "print 'Time required for maximum concentration of A: %4.2f s'%tmax"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required for maximum concentration of A: 13.86 s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.7, Page Number 467"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "T = 22.0 #Temperature of the reaction,\u00b0C\n",
- "k1 = 7.0e-4 #Rate constants for rxn 1, 1/s\n",
- "k2 = 4.1e-3 #Rate constant for rxn 2, 1/s \n",
- "k3 = 5.7e-3 #Rate constant for rxn 3, 1/s \n",
- "#Calculations\n",
- "phiP1 = k1/(k1+k2+k3)\n",
- "\n",
- "#Results\n",
- "print 'Percentage of Benzyl Penicillin that under acid catalyzed reaction by path 1: %4.2f '%(phiP1*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Percentage of Benzyl Penicillin that under acid catalyzed reaction by path 1: 6.67 \n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.8, Page Number 468"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg,log, exp\n",
- "from pylab import plot,show\n",
- "\n",
- "\n",
- "#Variable Declaration\n",
- "T = array([22.7,27.2,33.7,38.0])\n",
- "k1 = array([7.e-4,9.8e-4,1.6e-3,2.e-3])\n",
- "R = 8.314 \n",
- "\n",
- "#Calculations\n",
- "T = T +273.15\n",
- "x = 1./T\n",
- "y = log(k1)\n",
- "A = array([ x, ones(size(x))])\n",
- "# linearly generated sequence\n",
- "[slope, intercept] = linalg.lstsq(A.T,y)[0] # obtaining the parameters\n",
- "\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = w[0]*x+w[1] # regression line\n",
- "#Results\n",
- "plot(x,line,'-',x,y,'o')\n",
- "xlabel('1/T, $K^{-1}$')\n",
- "ylabel('log(k)')\n",
- "show()\n",
- "Ea = -slope*R\n",
- "A = exp(intercept)\n",
- "print 'Slope and intercept are, %6.1f and %4.2f'%(slope, intercept)\n",
- "print 'Pre-exponential factor and Activation energy are %4.2f kJ/mol and %4.2e 1/s'%(Ea/1e3, A)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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REZFEKdGIiEiilGhERCRRSjQiIpIoJRoREUmUEo2IiCRKiUZERBKlRCMiIolS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\nAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x8281150>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are, -6419.8 and 14.45\n",
- "Pre-exponential factor and Activation energy are 53.37 kJ/mol and 1.88e+06 1/s\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.9, Page Number 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration\n",
- "Ea = 42.e3 #Activation energy for reaction, J/mol\n",
- "A = 1.e12 #Pre-exponential factor for reaction, 1/s\n",
- "T = 298.0 #Temeprature, K\n",
- "Kc = 1.0e4 #Equilibrium constant for reaction\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "#Calculations\n",
- "kB = A*exp(-Ea/(R*T))\n",
- "kA = kB*Kc\n",
- "kApp = kA + kB\n",
- "\n",
- "#Results\n",
- "print 'Forward Rate constant is %4.2e 1/s'%kA\n",
- "print 'Backward Rate constant is %4.2e 1/s'%kB\n",
- "print 'Apperent Rate constant is %4.2e 1/s'%kApp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forward Rate constant is 4.34e+08 1/s\n",
- "Backward Rate constant is 4.34e+04 1/s\n",
- "Apperent Rate constant is 4.34e+08 1/s\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.10, Page Number 480"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi\n",
- "#Variable Declaration\n",
- "Dh = 7.6e-7 #Diffusion coefficient of Hemoglobin, cm2/s\n",
- "Do2 = 2.2e-5 #Diffusion coefficient of oxygen, cm2/s\n",
- "rh = 35. #Radius of Hemoglobin, \u00b0A\n",
- "ro2 = 2.0 #Radius of Oxygen, \u00b0A\n",
- "k = 4e7 #Rate constant for binding of O2 to Hemoglobin, 1/(M.s)\n",
- "NA =6.022e23 #Avagadro Number\n",
- "#Calculations\n",
- "DA = Dh + Do2\n",
- "kd = 4*pi*NA*(rh+ro2)*1e-8*DA\n",
- "\n",
- "#Results\n",
- "print 'Estimated rate %4.1e 1/(M.s) is far grater than experimental value of %4.1e 1/(M.s), \\nhence the reaction is not diffusion controlled'%(kd,k)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Estimated rate 6.4e+13 1/(M.s) is far grater than experimental value of 4.0e+07 1/(M.s), \n",
- "hence the reaction is not diffusion controlled\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.11, Page Number 484"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, e\n",
- "#Variable Declaration\n",
- "Ea = 104e3 #Activation energy for reaction, J/mol\n",
- "A = 1.e13 #Pre-exponential factor for reaction, 1/s\n",
- "T = 300.0 #Temeprature, K\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "h = 6.626e-34 #Plnak constant, Js\n",
- "c = 1.0 #Std. State concentration, M\n",
- "k = 1.38e-23 #,J/K\n",
- "\n",
- "#Calculations\n",
- "dH = Ea - 2*R*T\n",
- "dS = R*log(A*h*c/(k*T*e**2))\n",
- "\n",
- "#Results\n",
- "print 'Forward Rate constant is %4.2e 1/s'%dH\n",
- "print 'Backward Rate constant is %4.2f 1/s'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forward Rate constant is 9.90e+04 1/s\n",
- "Backward Rate constant is -12.72 1/s\n"
- ]
- }
- ],
- "prompt_number": 72
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18_1.ipynb deleted file mode 100755 index d26f90a9..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18_1.ipynb +++ /dev/null @@ -1,407 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:445a31bf90e68e213ebec0ec2a3db9f1c5fd3f96b2af63c47524b2b781345068"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Elementary Chemical Kinetics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.2, Page Number 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "Ca0 = [2.3e-4,4.6e-4,9.2e-4] #Initial Concentration of A, M\n",
- "Cb0 = [3.1e-5,6.2e-5,6.2e-5] #Initial Concentration of B, M\n",
- "Ri = [5.25e-4,4.2e-3,1.68e-2] #Initial rate of reaction, M\n",
- "\n",
- "#Calculations\n",
- "alp = log(Ri[1]/Ri[2])/log(Ca0[1]/Ca0[2])\n",
- "beta = (log(Ri[0]/Ri[1]) - 2*log((Ca0[0]/Ca0[1])))/(log(Cb0[0]/Cb0[1]))\n",
- "k = Ri[2]/(Ca0[2]**2*Cb0[2]**beta)\n",
- "\n",
- "#REsults\n",
- "print 'Order of reaction with respect to reactant A: %3.2f'%alp\n",
- "print 'Order of reaction with respect to reactant A: %3.2f'%beta\n",
- "print 'Rate constant of the reaction: %4.3e 1./(M.s)'%k"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Order of reaction with respect to reactant A: 2.00\n",
- "Order of reaction with respect to reactant A: 1.00\n",
- "Rate constant of the reaction: 3.201e+08 1./(M.s)\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.3, Page Number 457"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "t1by2 = 2.05e4 #Half life for first order decomposition of N2O5, s\n",
- "x = 60. #percentage decay of N2O5\n",
- "\n",
- "#Calculations\n",
- "k = log(2)/t1by2\n",
- "t = -log(x/100)/k\n",
- "\n",
- "#REsults\n",
- "print 'Rate constant of the reaction: %4.3e 1/s'%k\n",
- "print 'Timerequire for 60 percent decay of N2O5: %4.3e s'%t"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rate constant of the reaction: 3.381e-05 1/s\n",
- "Timerequire for 60 percent decay of N2O5: 1.511e+04 s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.4, Page Number 457 Incomplete"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "t1by2 = 5760 #Half life for C14, years\n",
- "\n",
- "\n",
- "#Calculations\n",
- "k = log(2)/t1by2\n",
- "t = -log(x/100)/k\n",
- "\n",
- "#REsults\n",
- "print 'Rate constant of the reaction: %4.3e 1/s'%k\n",
- "print 'Timerequire for 60 percent decay of N2O5: %4.3e s'%t"
- ],
- "language": "python",
- "metadata": {},
- "outputs": []
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.5, Page Number 463"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "kAbykI = 2.0 #Ratio of rate constants\n",
- "kA = 0.1 #First order rate constant for rxn 1, 1/s \n",
- "kI = 0.05 #First order rate constant for rxn 2, 1/s \n",
- "#Calculations\n",
- "tmax = 1/(kA-kI)*log(kA/kI)\n",
- "\n",
- "#Results\n",
- "print 'Time required for maximum concentration of A: %4.2f s'%tmax"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required for maximum concentration of A: 13.86 s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.7, Page Number 467"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "T = 22.0 #Temperature of the reaction,\u00b0C\n",
- "k1 = 7.0e-4 #Rate constants for rxn 1, 1/s\n",
- "k2 = 4.1e-3 #Rate constant for rxn 2, 1/s \n",
- "k3 = 5.7e-3 #Rate constant for rxn 3, 1/s \n",
- "#Calculations\n",
- "phiP1 = k1/(k1+k2+k3)\n",
- "\n",
- "#Results\n",
- "print 'Percentage of Benzyl Penicillin that under acid catalyzed reaction by path 1: %4.2f '%(phiP1*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Percentage of Benzyl Penicillin that under acid catalyzed reaction by path 1: 6.67 \n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.8, Page Number 468"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg,log, exp\n",
- "from pylab import plot,show\n",
- "\n",
- "\n",
- "#Variable Declaration\n",
- "T = array([22.7,27.2,33.7,38.0])\n",
- "k1 = array([7.e-4,9.8e-4,1.6e-3,2.e-3])\n",
- "R = 8.314 \n",
- "\n",
- "#Calculations\n",
- "T = T +273.15\n",
- "x = 1./T\n",
- "y = log(k1)\n",
- "A = array([ x, ones(size(x))])\n",
- "# linearly generated sequence\n",
- "[slope, intercept] = linalg.lstsq(A.T,y)[0] # obtaining the parameters\n",
- "\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = w[0]*x+w[1] # regression line\n",
- "#Results\n",
- "plot(x,line,'-',x,y,'o')\n",
- "xlabel('1/T, $K^{-1}$')\n",
- "ylabel('log(k)')\n",
- "show()\n",
- "Ea = -slope*R\n",
- "A = exp(intercept)\n",
- "print 'Slope and intercept are, %6.1f and %4.2f'%(slope, intercept)\n",
- "print 'Pre-exponential factor and Activation energy are %4.2f kJ/mol and %4.2e 1/s'%(Ea/1e3, A)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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REZFEKdGIiEiilGhERCRRSjQiIpIoJRoREUmUEo2IiCRKiUZERBKlRCMiIolS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\nAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x8281150>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are, -6419.8 and 14.45\n",
- "Pre-exponential factor and Activation energy are 53.37 kJ/mol and 1.88e+06 1/s\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.9, Page Number 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "Ea = 42.e3 #Activation energy for reaction, J/mol\n",
- "A = 1.e12 #Pre-exponential factor for reaction, 1/s\n",
- "T = 298.0 #Temeprature, K\n",
- "Kc = 1.0e4 #Equilibrium constant for reaction\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "#Calculations\n",
- "kB = A*exp(-Ea/(R*T))\n",
- "kA = kB*Kc\n",
- "kApp = kA + kB\n",
- "\n",
- "#Results\n",
- "print 'Forward Rate constant is %4.2e 1/s'%kA\n",
- "print 'Backward Rate constant is %4.2e 1/s'%kB\n",
- "print 'Apperent Rate constant is %4.2e 1/s'%kApp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forward Rate constant is 4.34e+08 1/s\n",
- "Backward Rate constant is 4.34e+04 1/s\n",
- "Apperent Rate constant is 4.34e+08 1/s\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.10, Page Number 480"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi\n",
- "#Variable Declaration\n",
- "Dh = 7.6e-7 #Diffusion coefficient of Hemoglobin, cm2/s\n",
- "Do2 = 2.2e-5 #Diffusion coefficient of oxygen, cm2/s\n",
- "rh = 35. #Radius of Hemoglobin, \u00b0A\n",
- "ro2 = 2.0 #Radius of Oxygen, \u00b0A\n",
- "k = 4e7 #Rate constant for binding of O2 to Hemoglobin, 1/(M.s)\n",
- "NA =6.022e23 #Avagadro Number\n",
- "#Calculations\n",
- "DA = Dh + Do2\n",
- "kd = 4*pi*NA*(rh+ro2)*1e-8*DA\n",
- "\n",
- "#Results\n",
- "print 'Estimated rate %4.1e 1/(M.s) is far grater than experimental value of %4.1e 1/(M.s), \\nhence the reaction is not diffusion controlled'%(kd,k)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Estimated rate 6.4e+13 1/(M.s) is far grater than experimental value of 4.0e+07 1/(M.s), \n",
- "hence the reaction is not diffusion controlled\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.11, Page Number 484"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, e\n",
- "#Variable Declaration\n",
- "Ea = 104e3 #Activation energy for reaction, J/mol\n",
- "A = 1.e13 #Pre-exponential factor for reaction, 1/s\n",
- "T = 300.0 #Temeprature, K\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "h = 6.626e-34 #Plnak constant, Js\n",
- "c = 1.0 #Std. State concentration, M\n",
- "k = 1.38e-23 #,J/K\n",
- "\n",
- "#Calculations\n",
- "dH = Ea - 2*R*T\n",
- "dS = R*log(A*h*c/(k*T*e**2))\n",
- "\n",
- "#Results\n",
- "print 'Forward Rate constant is %4.2e 1/s'%dH\n",
- "print 'Backward Rate constant is %4.2f 1/s'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forward Rate constant is 9.90e+04 1/s\n",
- "Backward Rate constant is -12.72 1/s\n"
- ]
- }
- ],
- "prompt_number": 72
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18_2.ipynb deleted file mode 100755 index 703987ca..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter18_2.ipynb +++ /dev/null @@ -1,407 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:16e7490a05ae8ab95bdce6309e35e472520df00964c8812973f2cdbef573b0c0"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Elementary Chemical Kinetics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.2, Page Number 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "Ca0 = [2.3e-4,4.6e-4,9.2e-4] #Initial Concentration of A, M\n",
- "Cb0 = [3.1e-5,6.2e-5,6.2e-5] #Initial Concentration of B, M\n",
- "Ri = [5.25e-4,4.2e-3,1.68e-2] #Initial rate of reaction, M\n",
- "\n",
- "#Calculations\n",
- "alp = log(Ri[1]/Ri[2])/log(Ca0[1]/Ca0[2])\n",
- "beta = (log(Ri[0]/Ri[1]) - 2*log((Ca0[0]/Ca0[1])))/(log(Cb0[0]/Cb0[1]))\n",
- "k = Ri[2]/(Ca0[2]**2*Cb0[2]**beta)\n",
- "\n",
- "#REsults\n",
- "print 'Order of reaction with respect to reactant A: %3.2f'%alp\n",
- "print 'Order of reaction with respect to reactant A: %3.2f'%beta\n",
- "print 'Rate constant of the reaction: %4.3e 1./(M.s)'%k"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Order of reaction with respect to reactant A: 2.00\n",
- "Order of reaction with respect to reactant A: 1.00\n",
- "Rate constant of the reaction: 3.201e+08 1./(M.s)\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.3, Page Number 457"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "from sympy import *\n",
- "\n",
- "#Variable Declaration\n",
- "t1by2 = 2.05e4 #Half life for first order decomposition of N2O5, s\n",
- "x = 60. #percentage decay of N2O5\n",
- "\n",
- "#Calculations\n",
- "k = log(2)/t1by2\n",
- "t = -log(x/100)/k\n",
- "\n",
- "#REsults\n",
- "print 'Rate constant of the reaction: %4.3e 1/s'%k\n",
- "print 'Timerequire for 60 percent decay of N2O5: %4.3e s'%t"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rate constant of the reaction: 3.381e-05 1/s\n",
- "Timerequire for 60 percent decay of N2O5: 1.511e+04 s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.4, Page Number 457 Incomplete"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "t1by2 = 5760 #Half life for C14, years\n",
- "\n",
- "\n",
- "#Calculations\n",
- "k = log(2)/t1by2\n",
- "t = -log(x/100)/k\n",
- "\n",
- "#REsults\n",
- "print 'Rate constant of the reaction: %4.3e 1/s'%k\n",
- "print 'Timerequire for 60 percent decay of N2O5: %4.3e s'%t"
- ],
- "language": "python",
- "metadata": {},
- "outputs": []
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.5, Page Number 463"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "kAbykI = 2.0 #Ratio of rate constants\n",
- "kA = 0.1 #First order rate constant for rxn 1, 1/s \n",
- "kI = 0.05 #First order rate constant for rxn 2, 1/s \n",
- "#Calculations\n",
- "tmax = 1/(kA-kI)*log(kA/kI)\n",
- "\n",
- "#Results\n",
- "print 'Time required for maximum concentration of A: %4.2f s'%tmax"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time required for maximum concentration of A: 13.86 s\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.7, Page Number 467"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log\n",
- "\n",
- "#Variable Declaration\n",
- "T = 22.0 #Temperature of the reaction,\u00b0C\n",
- "k1 = 7.0e-4 #Rate constants for rxn 1, 1/s\n",
- "k2 = 4.1e-3 #Rate constant for rxn 2, 1/s \n",
- "k3 = 5.7e-3 #Rate constant for rxn 3, 1/s \n",
- "#Calculations\n",
- "phiP1 = k1/(k1+k2+k3)\n",
- "\n",
- "#Results\n",
- "print 'Percentage of Benzyl Penicillin that under acid catalyzed reaction by path 1: %4.2f '%(phiP1*100)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Percentage of Benzyl Penicillin that under acid catalyzed reaction by path 1: 6.67 \n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.8, Page Number 468"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg,log, exp\n",
- "from matplotlib.pylab import plot, show\n",
- "%matplotlib inline\n",
- "\n",
- "#Variable Declaration\n",
- "T = array([22.7,27.2,33.7,38.0])\n",
- "k1 = array([7.e-4,9.8e-4,1.6e-3,2.e-3])\n",
- "R = 8.314 \n",
- "\n",
- "#Calculations\n",
- "T = T +273.15\n",
- "x = 1./T\n",
- "y = log(k1)\n",
- "A = array([ x, ones(size(x))])\n",
- "# linearly generated sequence\n",
- "[slope, intercept] = linalg.lstsq(A.T,y)[0] # obtaining the parameters\n",
- "\n",
- "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
- "# plotting the line\n",
- "line = slope*x+intercept # regression line\n",
- "#Results\n",
- "plot(x,line,'-',x,y,'o')\n",
- "xlabel('$ 1/T, K^{-1} $')\n",
- "ylabel('$ log(k) $')\n",
- "show()\n",
- "Ea = -slope*R\n",
- "A = exp(intercept)\n",
- "print 'Slope and intercept are, %6.1f and %4.2f'%(slope, intercept)\n",
- "print 'Pre-exponential factor and Activation energy are %4.2f kJ/mol and %4.2e 1/s'%(Ea/1e3, A)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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6VnSkelx1WcyCUwISEZGP0CQEEREpaUpAIiIShRKQiIhEoQQkIiJRKAGJiEgU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\nJSAREYlCCUhERKJQAhIRkSj+P8yT25TURYlKAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x8fa5f10>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are, -6419.8 and 14.45\n",
- "Pre-exponential factor and Activation energy are 53.37 kJ/mol and 1.88e+06 1/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.9, Page Number 473"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "\n",
- "#Variable Declaration\n",
- "Ea = 42.e3 #Activation energy for reaction, J/mol\n",
- "A = 1.e12 #Pre-exponential factor for reaction, 1/s\n",
- "T = 298.0 #Temeprature, K\n",
- "Kc = 1.0e4 #Equilibrium constant for reaction\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "#Calculations\n",
- "kB = A*exp(-Ea/(R*T))\n",
- "kA = kB*Kc\n",
- "kApp = kA + kB\n",
- "\n",
- "#Results\n",
- "print 'Forward Rate constant is %4.2e 1/s'%kA\n",
- "print 'Backward Rate constant is %4.2e 1/s'%kB\n",
- "print 'Apperent Rate constant is %4.2e 1/s'%kApp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forward Rate constant is 4.34e+08 1/s\n",
- "Backward Rate constant is 4.34e+04 1/s\n",
- "Apperent Rate constant is 4.34e+08 1/s\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.10, Page Number 480"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import pi\n",
- "#Variable Declaration\n",
- "Dh = 7.6e-7 #Diffusion coefficient of Hemoglobin, cm2/s\n",
- "Do2 = 2.2e-5 #Diffusion coefficient of oxygen, cm2/s\n",
- "rh = 35. #Radius of Hemoglobin, \u00b0A\n",
- "ro2 = 2.0 #Radius of Oxygen, \u00b0A\n",
- "k = 4e7 #Rate constant for binding of O2 to Hemoglobin, 1/(M.s)\n",
- "NA =6.022e23 #Avagadro Number\n",
- "#Calculations\n",
- "DA = Dh + Do2\n",
- "kd = 4*pi*NA*(rh+ro2)*1e-8*DA\n",
- "\n",
- "#Results\n",
- "print 'Estimated rate %4.1e 1/(M.s) is far grater than experimental value of %4.1e 1/(M.s), \\nhence the reaction is not diffusion controlled'%(kd,k)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Estimated rate 6.4e+13 1/(M.s) is far grater than experimental value of 4.0e+07 1/(M.s), \n",
- "hence the reaction is not diffusion controlled\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem 18.11, Page Number 484"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import log, e\n",
- "#Variable Declaration\n",
- "Ea = 104e3 #Activation energy for reaction, J/mol\n",
- "A = 1.e13 #Pre-exponential factor for reaction, 1/s\n",
- "T = 300.0 #Temeprature, K\n",
- "R = 8.314 #Ideal gas constant, J/(mol.K)\n",
- "h = 6.626e-34 #Plnak constant, Js\n",
- "c = 1.0 #Std. State concentration, M\n",
- "k = 1.38e-23 #,J/K\n",
- "\n",
- "#Calculations\n",
- "dH = Ea - 2*R*T\n",
- "dS = R*log(A*h*c/(k*T*e**2))\n",
- "\n",
- "#Results\n",
- "print 'Forward Rate constant is %4.2e 1/s'%dH\n",
- "print 'Backward Rate constant is %4.2f 1/s'%dS"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Forward Rate constant is 9.90e+04 1/s\n",
- "Backward Rate constant is -12.72 1/s\n"
- ]
- }
- ],
- "prompt_number": 72
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19.ipynb deleted file mode 100755 index 8e67d995..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19.ipynb +++ /dev/null @@ -1,340 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:426928f27add269a4fb950a40e7b077ed56eb54390661454909c07f1497ce752"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Complex Reaction Mechanism "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.1, Page Number 501"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import numpy as np\n",
- "from numpy import arange,array,ones,linalg\n",
- "from pylab import plot,show\n",
- "#Variable declaration\n",
- "Ce = 2.3e-9 #Initial value of enzyme concentration, M\n",
- "r = array([2.78e-5,5.e-5,8.33e-5,1.67e-4])\n",
- "CCO2 = array([1.25e-3,2.5e-3,5.e-3,20.e-3])\n",
- "\n",
- "#Calculations\n",
- "rinv = 1./r\n",
- "CCO2inv = 1./CCO2\n",
- "xlim(0,850)\n",
- "ylim(0,38000)\n",
- "xi = CCO2inv\n",
- "A = array([ CCO2inv, ones(size(CCO2inv))])\n",
- "# linearly generated sequence\n",
- "w = linalg.lstsq(A.T,rinv)[0] # obtaining the parameters\n",
- "slope = w[0]\n",
- "intercept = w[1]\n",
- "\n",
- "line = w[0]*CCO2inv+w[1] # regression line\n",
- "plot(CCO2inv,line,'r-',CCO2inv,rinv,'o')\n",
- "show()\n",
- "rmax = 1./intercept\n",
- "k2 = rmax/Ce\n",
- "Km = slope*rmax\n",
- "\n",
- "#Results\n",
- "print 'Km and k2 are %4.1f mM and %3.1e s-1'%(Km*1e3,k2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
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- "text": [
- "<matplotlib.figure.Figure at 0x8d31770>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Km and k2 are 10.0 mM and 1.1e+05 s-1\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.2, Page Number 507"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg\n",
- "from pylab import plot,show\n",
- "\n",
- "#Variable declaration\n",
- "Vads = array([5.98,7.76,10.1,12.35,16.45,18.05,19.72,21.1]) #Adsorption data at 193.5K\n",
- "P = array([2.45,3.5,5.2,7.2,11.2,12.8,14.6,16.1]) #Pressure, torr\n",
- "\n",
- "#Calculations\n",
- "Vinv = 1./Vads\n",
- "Pinv =1./P\n",
- "xlim(0,0.5)\n",
- "ylim(0,0.2)\n",
- "A = array([ Pinv, ones(size(Pinv))])\n",
- "# linearly generated sequence\n",
- "w = linalg.lstsq(A.T,Vinv)[0] # obtaining the parameters\n",
- "m = w[0]\n",
- "c = w[1]\n",
- "line = m*Pinv+c # regression line\n",
- "plot(Pinv,line,'r-',Pinv,Vinv,'o')\n",
- "show()\n",
- "Vm = 1./c\n",
- "K = 1./(m*Vm)\n",
- "\n",
- "#Results\n",
- "print 'Slope and intercept are %5.4f torr.g/cm3 and %5.4f g/cm3'%(m,c)\n",
- "print 'K and Vm are %4.2e Torr^-1 and %3.1f cm3/g'%(K,Vm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x8295610>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are 0.3449 torr.g/cm3 and 0.0293 g/cm3\n",
- "K and Vm are 8.48e-02 Torr^-1 and 34.2 cm3/g\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.3, Page Number 510"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.4, Page Number 520"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg\n",
- "from pylab import plot,show\n",
- "\n",
- "#Variable declaration\n",
- "CBr = array([0.0005,0.001,0.002,0.003,0.005]) #C6Br6 concentration, M\n",
- "tf = array([2.66e-7,1.87e-7,1.17e-7,8.50e-8,5.51e-8]) #Fluroscence life time, s\n",
- "\n",
- "#Calculations\n",
- "Tfinv = 1./tf\n",
- "xlim(0,0.006)\n",
- "ylim(0,2.e7)\n",
- "A = array([ CBr, ones(size(CBr))])\n",
- "# linearly generated sequence\n",
- "[m,c] = linalg.lstsq(A.T,Tfinv)[0] # obtaining the parameters\n",
- "\n",
- "line = m*CBr+c # regression line\n",
- "plot(CBr,line,'r-',CBr,Tfinv,'o')\n",
- "show()\n",
- "\n",
- "#Results\n",
- "print 'Slope and intercept are kq = %5.4e per s and kf = %5.4e per s'%(m,c)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x860a730>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are kq = 3.1995e+09 per s and kf = 2.1545e+06 per s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.5, Page Number 523"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from scipy.optimize import root\n",
- "\n",
- "#Variable Declaration\n",
- "r = 11. #Distance of residue separation, \u00b0A\n",
- "r0 = 9. #Initial Distance of residue separation, \u00b0A\n",
- "EffD = 0.2 #Fraction decrease in eff\n",
- "\n",
- "#Calculations\n",
- "Effi = r0**6/(r0**6+r**6)\n",
- "Eff = Effi*(1-EffD)\n",
- "f = lambda r: r0**6/(r0**6+r**6) - Eff\n",
- "sol = root(f, 12)\n",
- "rn = sol.x[0]\n",
- "\n",
- "#Results\n",
- "print 'Separation Distance at decreased efficiency %4.2f'%rn"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Separation Distance at decreased efficiency 11.53\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.6, Page Number 525"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "mr = 2.5e-3 #Moles reacted, mol\n",
- "P = 100.0 #Irradiation Power, J/s\n",
- "t = 27 #Time of irradiation, s\n",
- "h = 6.626e-34 #Planks constant, Js\n",
- "c = 3.0e8 #Speed of light, m/s\n",
- "labda = 280e-9 #Wavelength of light, m\n",
- "\n",
- "#Calculation\n",
- "Eabs = P*t\n",
- "Eph = h*c/labda\n",
- "nph = Eabs/Eph #moles of photone\n",
- "phi = mr/6.31e-3\n",
- "\n",
- "#Results\n",
- "print 'Total photon energy absorbed by sample %3.1e J'%Eabs\n",
- "print 'Photon energy absorbed at 280 nm is %3.1e J'%Eph\n",
- "print 'Total number of photon absorbed by sample %3.1e photones'%nph\n",
- "print 'Overall quantum yield %4.2f'%phi"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total photon energy absorbed by sample 2.7e+03 J\n",
- "Photon energy absorbed at 280 nm is 7.1e-19 J\n",
- "Total number of photon absorbed by sample 3.8e+21 photones\n",
- "Overall quantum yield 0.40\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.7, Page Number 530"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "#Variable Declarations\n",
- "r = 2.0e9 #Rate constant for electron transfer, per s\n",
- "labda = 1.2 #Gibss energy change, eV\n",
- "DG = -1.93 #Gibss energy change for 2-naphthoquinoyl, eV\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "T = 298.0 #Temeprature, K\n",
- "#Calculation\n",
- "DGS = (DG+labda)**2/(4*labda)\n",
- "k193 = r*exp(-DGS*1.6e-19/(k*T))\n",
- "#Results\n",
- "print 'DGS = %5.3f eV'%DGS\n",
- "print 'Rate constant with barrier to electron transfer %3.2e per s'%k193"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "DGS = 0.111 eV\n",
- "Rate constant with barrier to electron transfer 2.66e+07 per s\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19_1.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19_1.ipynb deleted file mode 100755 index fa06dbe1..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19_1.ipynb +++ /dev/null @@ -1,332 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c61898b3497bb5b0aa45fad66ec37778e660d1c5f95436ee3dc7430758def5e0"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Complex Reaction Mechanism "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.1, Page Number 501"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import numpy as np\n",
- "from numpy import arange,array,ones,linalg\n",
- "from pylab import plot,show\n",
- "#Variable declaration\n",
- "Ce = 2.3e-9 #Initial value of enzyme concentration, M\n",
- "r = array([2.78e-5,5.e-5,8.33e-5,1.67e-4])\n",
- "CCO2 = array([1.25e-3,2.5e-3,5.e-3,20.e-3])\n",
- "\n",
- "#Calculations\n",
- "rinv = 1./r\n",
- "CCO2inv = 1./CCO2\n",
- "xlim(0,850)\n",
- "ylim(0,38000)\n",
- "xi = CCO2inv\n",
- "A = array([ CCO2inv, ones(size(CCO2inv))])\n",
- "# linearly generated sequence\n",
- "w = linalg.lstsq(A.T,rinv)[0] # obtaining the parameters\n",
- "slope = w[0]\n",
- "intercept = w[1]\n",
- "\n",
- "line = w[0]*CCO2inv+w[1] # regression line\n",
- "plot(CCO2inv,line,'r-',CCO2inv,rinv,'o')\n",
- "show()\n",
- "rmax = 1./intercept\n",
- "k2 = rmax/Ce\n",
- "Km = slope*rmax\n",
- "\n",
- "#Results\n",
- "print 'Km and k2 are %4.1f mM and %3.1e s-1'%(Km*1e3,k2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x8d31770>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Km and k2 are 10.0 mM and 1.1e+05 s-1\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.2, Page Number 507"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg\n",
- "from pylab import plot,show\n",
- "\n",
- "#Variable declaration\n",
- "Vads = array([5.98,7.76,10.1,12.35,16.45,18.05,19.72,21.1]) #Adsorption data at 193.5K\n",
- "P = array([2.45,3.5,5.2,7.2,11.2,12.8,14.6,16.1]) #Pressure, torr\n",
- "\n",
- "#Calculations\n",
- "Vinv = 1./Vads\n",
- "Pinv =1./P\n",
- "xlim(0,0.5)\n",
- "ylim(0,0.2)\n",
- "A = array([ Pinv, ones(size(Pinv))])\n",
- "# linearly generated sequence\n",
- "w = linalg.lstsq(A.T,Vinv)[0] # obtaining the parameters\n",
- "m = w[0]\n",
- "c = w[1]\n",
- "line = m*Pinv+c # regression line\n",
- "plot(Pinv,line,'r-',Pinv,Vinv,'o')\n",
- "show()\n",
- "Vm = 1./c\n",
- "K = 1./(m*Vm)\n",
- "\n",
- "#Results\n",
- "print 'Slope and intercept are %5.4f torr.g/cm3 and %5.4f g/cm3'%(m,c)\n",
- "print 'K and Vm are %4.2e Torr^-1 and %3.1f cm3/g'%(K,Vm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x8295610>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are 0.3449 torr.g/cm3 and 0.0293 g/cm3\n",
- "K and Vm are 8.48e-02 Torr^-1 and 34.2 cm3/g\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.4, Page Number 520"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg\n",
- "from pylab import plot,show\n",
- "\n",
- "#Variable declaration\n",
- "CBr = array([0.0005,0.001,0.002,0.003,0.005]) #C6Br6 concentration, M\n",
- "tf = array([2.66e-7,1.87e-7,1.17e-7,8.50e-8,5.51e-8]) #Fluroscence life time, s\n",
- "\n",
- "#Calculations\n",
- "Tfinv = 1./tf\n",
- "xlim(0,0.006)\n",
- "ylim(0,2.e7)\n",
- "A = array([ CBr, ones(size(CBr))])\n",
- "# linearly generated sequence\n",
- "[m,c] = linalg.lstsq(A.T,Tfinv)[0] # obtaining the parameters\n",
- "\n",
- "line = m*CBr+c # regression line\n",
- "plot(CBr,line,'r-',CBr,Tfinv,'o')\n",
- "show()\n",
- "\n",
- "#Results\n",
- "print 'Slope and intercept are kq = %5.4e per s and kf = %5.4e per s'%(m,c)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x860a730>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are kq = 3.1995e+09 per s and kf = 2.1545e+06 per s\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.5, Page Number 523"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from scipy.optimize import root\n",
- "\n",
- "#Variable Declaration\n",
- "r = 11. #Distance of residue separation, \u00b0A\n",
- "r0 = 9. #Initial Distance of residue separation, \u00b0A\n",
- "EffD = 0.2 #Fraction decrease in eff\n",
- "\n",
- "#Calculations\n",
- "Effi = r0**6/(r0**6+r**6)\n",
- "Eff = Effi*(1-EffD)\n",
- "f = lambda r: r0**6/(r0**6+r**6) - Eff\n",
- "sol = root(f, 12)\n",
- "rn = sol.x[0]\n",
- "\n",
- "#Results\n",
- "print 'Separation Distance at decreased efficiency %4.2f'%rn"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Separation Distance at decreased efficiency 11.53\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.6, Page Number 525"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "mr = 2.5e-3 #Moles reacted, mol\n",
- "P = 100.0 #Irradiation Power, J/s\n",
- "t = 27 #Time of irradiation, s\n",
- "h = 6.626e-34 #Planks constant, Js\n",
- "c = 3.0e8 #Speed of light, m/s\n",
- "labda = 280e-9 #Wavelength of light, m\n",
- "\n",
- "#Calculation\n",
- "Eabs = P*t\n",
- "Eph = h*c/labda\n",
- "nph = Eabs/Eph #moles of photone\n",
- "phi = mr/6.31e-3\n",
- "\n",
- "#Results\n",
- "print 'Total photon energy absorbed by sample %3.1e J'%Eabs\n",
- "print 'Photon energy absorbed at 280 nm is %3.1e J'%Eph\n",
- "print 'Total number of photon absorbed by sample %3.1e photones'%nph\n",
- "print 'Overall quantum yield %4.2f'%phi"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total photon energy absorbed by sample 2.7e+03 J\n",
- "Photon energy absorbed at 280 nm is 7.1e-19 J\n",
- "Total number of photon absorbed by sample 3.8e+21 photones\n",
- "Overall quantum yield 0.40\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.7, Page Number 530"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "#Variable Declarations\n",
- "r = 2.0e9 #Rate constant for electron transfer, per s\n",
- "labda = 1.2 #Gibss energy change, eV\n",
- "DG = -1.93 #Gibss energy change for 2-naphthoquinoyl, eV\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "T = 298.0 #Temeprature, K\n",
- "#Calculation\n",
- "DGS = (DG+labda)**2/(4*labda)\n",
- "k193 = r*exp(-DGS*1.6e-19/(k*T))\n",
- "#Results\n",
- "print 'DGS = %5.3f eV'%DGS\n",
- "print 'Rate constant with barrier to electron transfer %3.2e per s'%k193"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "DGS = 0.111 eV\n",
- "Rate constant with barrier to electron transfer 2.66e+07 per s\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19_2.ipynb b/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19_2.ipynb deleted file mode 100755 index 78526d43..00000000 --- a/Thermodynamics,_Statistical_Thermodynamics,_&_Kinetics/Chapter19_2.ipynb +++ /dev/null @@ -1,342 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:4e717dccd3881fc251151e670a02cca0fba87118374ab9969adbd84fab322b22"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Complex Reaction Mechanism "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.1, Page Number 501"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import numpy as np\n",
- "from numpy import arange,array,ones,linalg\n",
- "from matplotlib.pylab import plot,show\n",
- "%matplotlib inline\n",
- "\n",
- "#Variable declaration\n",
- "Ce = 2.3e-9 #Initial value of enzyme concentration, M\n",
- "r = array([2.78e-5,5.e-5,8.33e-5,1.67e-4])\n",
- "CCO2 = array([1.25e-3,2.5e-3,5.e-3,20.e-3])\n",
- "\n",
- "#Calculations\n",
- "rinv = 1./r\n",
- "CCO2inv = 1./CCO2\n",
- "xlim(0,850)\n",
- "ylim(0,38000)\n",
- "xi = CCO2inv\n",
- "A = array([ CCO2inv, ones(size(CCO2inv))])\n",
- "# linearly generated sequence\n",
- "w = linalg.lstsq(A.T,rinv)[0] # obtaining the parameters\n",
- "slope = w[0]\n",
- "intercept = w[1]\n",
- "\n",
- "line = w[0]*CCO2inv+w[1] # regression line\n",
- "plot(CCO2inv,line,'r-',CCO2inv,rinv,'o')\n",
- "xlabel('$ {C_{CO}}_2, mM^{-1} $')\n",
- "ylabel('$ Rate^{-1}, s/M^{-1} $')\n",
- "show()\n",
- "rmax = 1./intercept\n",
- "k2 = rmax/Ce\n",
- "Km = slope*rmax\n",
- "\n",
- "#Results\n",
- "print 'Km and k2 are %4.1f mM and %3.1e s-1'%(Km*1e3,k2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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NwEZP3kv6UkQ8VuE8ufDkvbU4CxbAmWfCo4+m3srXv+4eitVLua5j8SkhZk3dBx+k04a/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\nKOw6oO96bcOA8RG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- "text": [
- "<matplotlib.figure.Figure at 0x22b5270>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Km and k2 are 10.0 mM and 1.1e+05 s-1\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.2, Page Number 507"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg\n",
- "from matplotlib.pylab import plot,show\n",
- "%matplotlib inline\n",
- "\n",
- "#Variable declaration\n",
- "Vads = array([5.98,7.76,10.1,12.35,16.45,18.05,19.72,21.1]) #Adsorption data at 193.5K\n",
- "P = array([2.45,3.5,5.2,7.2,11.2,12.8,14.6,16.1]) #Pressure, torr\n",
- "\n",
- "#Calculations\n",
- "Vinv = 1./Vads\n",
- "Pinv =1./P\n",
- "xlim(0,0.5)\n",
- "ylim(0,0.2)\n",
- "A = array([ Pinv, ones(size(Pinv))])\n",
- "# linearly generated sequence\n",
- "w = linalg.lstsq(A.T,Vinv)[0] # obtaining the parameters\n",
- "m = w[0]\n",
- "c = w[1]\n",
- "line = m*Pinv+c # regression line\n",
- "plot(Pinv,line,'r-',Pinv,Vinv,'o')\n",
- "xlabel('$ 1/P, Torr^{-1} $')\n",
- "ylabel('$ 1/V_{abs}, cm^{-1}g $')\n",
- "show()\n",
- "Vm = 1./c\n",
- "K = 1./(m*Vm)\n",
- "\n",
- "#Results\n",
- "print 'Slope and intercept are %5.4f torr.g/cm3 and %5.4f g/cm3'%(m,c)\n",
- "print 'K and Vm are %4.2e Torr^-1 and %3.1f cm3/g'%(K,Vm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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- "text": [
- "<matplotlib.figure.Figure at 0x5bde3f0>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are 0.3449 torr.g/cm3 and 0.0293 g/cm3\n",
- "K and Vm are 8.48e-02 Torr^-1 and 34.2 cm3/g\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.4, Page Number 520"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from numpy import arange,array,ones,linalg\n",
- "from matplotlib.pylab import plot,show\n",
- "%matplotlib inline\n",
- "\n",
- "#Variable declaration\n",
- "CBr = array([0.0005,0.001,0.002,0.003,0.005]) #C6Br6 concentration, M\n",
- "tf = array([2.66e-7,1.87e-7,1.17e-7,8.50e-8,5.51e-8]) #Fluroscence life time, s\n",
- "\n",
- "#Calculations\n",
- "Tfinv = 1./tf\n",
- "xlim(0,0.006)\n",
- "ylim(0,2.e7)\n",
- "A = array([ CBr, ones(size(CBr))])\n",
- "# linearly generated sequence\n",
- "[m,c] = linalg.lstsq(A.T,Tfinv)[0] # obtaining the parameters\n",
- "\n",
- "line = m*CBr+c # regression line\n",
- "plot(CBr,line,'r-',CBr,Tfinv,'o')\n",
- "xlabel('$ Br_6C_6, M $')\n",
- "ylabel('$ tau_f $')\n",
- "show()\n",
- "\n",
- "#Results\n",
- "print 'Slope and intercept are kq = %5.4e per s and kf = %5.4e per s'%(m,c)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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VxAQiSaqICUSSVJH/D6UzTURQtsUNAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x5c02050>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Slope and intercept are kq = 3.1995e+09 per s and kf = 2.1545e+06 per s\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.5, Page Number 523"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from scipy.optimize import root\n",
- "\n",
- "#Variable Declaration\n",
- "r = 11. #Distance of residue separation, \u00b0A\n",
- "r0 = 9. #Initial Distance of residue separation, \u00b0A\n",
- "EffD = 0.2 #Fraction decrease in eff\n",
- "\n",
- "#Calculations\n",
- "Effi = r0**6/(r0**6+r**6)\n",
- "Eff = Effi*(1-EffD)\n",
- "f = lambda r: r0**6/(r0**6+r**6) - Eff\n",
- "sol = root(f, 12)\n",
- "rn = sol.x[0]\n",
- "\n",
- "#Results\n",
- "print 'Separation Distance at decreased efficiency %4.2f'%rn"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Separation Distance at decreased efficiency 11.53\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.6, Page Number 525"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declarations\n",
- "mr = 2.5e-3 #Moles reacted, mol\n",
- "P = 100.0 #Irradiation Power, J/s\n",
- "t = 27 #Time of irradiation, s\n",
- "h = 6.626e-34 #Planks constant, Js\n",
- "c = 3.0e8 #Speed of light, m/s\n",
- "labda = 280e-9 #Wavelength of light, m\n",
- "\n",
- "#Calculation\n",
- "Eabs = P*t\n",
- "Eph = h*c/labda\n",
- "nph = Eabs/Eph #moles of photone\n",
- "phi = mr/6.31e-3\n",
- "\n",
- "#Results\n",
- "print 'Total photon energy absorbed by sample %3.1e J'%Eabs\n",
- "print 'Photon energy absorbed at 280 nm is %3.1e J'%Eph\n",
- "print 'Total number of photon absorbed by sample %3.1e photones'%nph\n",
- "print 'Overall quantum yield %4.2f'%phi"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total photon energy absorbed by sample 2.7e+03 J\n",
- "Photon energy absorbed at 280 nm is 7.1e-19 J\n",
- "Total number of photon absorbed by sample 3.8e+21 photones\n",
- "Overall quantum yield 0.40\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example Problem: 19.7, Page Number 530"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from math import exp\n",
- "#Variable Declarations\n",
- "r = 2.0e9 #Rate constant for electron transfer, per s\n",
- "labda = 1.2 #Gibss energy change, eV\n",
- "DG = -1.93 #Gibss energy change for 2-naphthoquinoyl, eV\n",
- "k = 1.38e-23 #Boltzman constant, J/K\n",
- "T = 298.0 #Temeprature, K\n",
- "#Calculation\n",
- "DGS = (DG+labda)**2/(4*labda)\n",
- "k193 = r*exp(-DGS*1.6e-19/(k*T))\n",
- "#Results\n",
- "print 'DGS = %5.3f eV'%DGS\n",
- "print 'Rate constant with barrier to electron transfer %3.2e per s'%k193"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "DGS = 0.111 eV\n",
- "Rate constant with barrier to electron transfer 2.66e+07 per s\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
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