From f270f72badd9c61d48f290c3396004802841b9df Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- .../Chapter_1.ipynb | 533 ++++++++++ .../Chapter_10.ipynb | 635 ++++++++++++ .../Chapter_2.ipynb | 438 +++++++++ .../Chapter_3.ipynb | 896 +++++++++++++++++ .../Chapter_4.ipynb | 1022 +++++++++++++++++++ .../Chapter_5.ipynb | 767 +++++++++++++++ .../Chapter_6.ipynb | 712 ++++++++++++++ .../Chapter_7.ipynb | 1029 ++++++++++++++++++++ .../Chapter_8.ipynb | 440 +++++++++ .../Chapter_9.ipynb | 101 ++ Physical_Chemsitry_by_William_F_Sheehan/README.txt | 10 + .../screenshots/chap3.png | Bin 0 -> 62750 bytes .../screenshots/chap4.png | Bin 0 -> 67939 bytes .../screenshots/chap6.png | Bin 0 -> 63857 bytes 14 files changed, 6583 insertions(+) create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_1.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_10.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_2.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_3.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_4.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_5.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_6.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_7.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_8.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/Chapter_9.ipynb create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/README.txt create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap3.png create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap4.png create mode 100755 Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap6.png (limited to 'Physical_Chemsitry_by_William_F_Sheehan') diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_1.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_1.ipynb new file mode 100755 index 00000000..ca5fb73c --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_1.ipynb @@ -0,0 +1,533 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1833f0f72d4fcfdfc05d274c870f8929bea706e80b14f9268d3407df8540de4d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 - Kinetic theory of gases and equations of state" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - Pg 5" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the final volume of the gas\n", + "#initialisation of variables\n", + "V= 22.394 #l\n", + "m= 32 #gm\n", + "T= 0 #C\n", + "T1= 50. #C\n", + "p= .8 #atm\n", + "#CALCULATIONS\n", + "V1= (T1+273.16)*V/(T+273.16)\n", + "V2= (1./p)*V1\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Volume = ',V2,'lt')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Volume = 33.116 lt\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - Pg 7" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate gthe argon temperature\n", + "#initialisation of variables\n", + "P= 1 #atm\n", + "T= 0 #C\n", + "#CALCULATIONS\n", + "T1= 10*(T+273.2)\n", + "#RESULTS\n", + "print '%s %.1f %s' %(' Argon temperature =',T1,' K')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Argon temperature = 2732.0 K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - Pg 9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Atomic Weight\n", + "#initialisation of variables\n", + "x= 0.0820544\n", + "T= 0 #C\n", + "l= 1.7826 #gl^-1atm^-1\n", + "#CALCULATIONS\n", + "M= x*(273.16+T)*l\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Atomic Weight =',M,'gm mole^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Atomic Weight = 39.955 gm mole^-1\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - Pg 11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Molecular weight and molecular formula\n", + "#initialisation of variables\n", + "g=.270 #g\n", + "R=0.08205\n", + "T=296.4 #K\n", + "P=754.6/760.0 #atm\n", + "V=0.03576 #lt\n", + "m1= 12\n", + "m2= 19\n", + "m3= 35.46\n", + "yx=.57\n", + "#CALCULATIONS\n", + "M1=g*R*T/(P*V)\n", + "y=round(yx*M1/m3)\n", + "n=round((M1-m3*y+m2)/(2*m2+m1))\n", + "x=2*n-1\n", + "M= n*m1+x*m2+y*m3\n", + "#RESULTS\n", + "print '%s %.2f %s' %('Approximate molecular weight = ',M1,\"gms\")\n", + "print '%s %.2f %s' % (' Molecular weight =',M,' gms')\n", + "print '%s %d %s %d %s %d' %('Molecular formula is C',n,'F',x,'Cl',y)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Approximate molecular weight = 184.94 gms\n", + " Molecular weight = 187.38 gms\n", + "Molecular formula is C 2 F 3 Cl 3\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - Pg 14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the pressure in both cases\n", + "#initialisation of variables\n", + "n= 10 #moles\n", + "R= 0.08205 #atml/molK\n", + "T= 300 #K\n", + "V= 4.86 #l\n", + "b= 0.0643 #ml mol**-1\n", + "a= 5.44 #l**2\n", + "#CALCULATIONS\n", + "P= n*R*T/V\n", + "P1= (n*R*T/(V-n*b))-(a*n**2/V**2)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Pressure in case of perfect gas law=',P,' atm')\n", + "print '%s %.1f %s' % (' \\n Pressure in case of vanderwaals equation =',P1,' atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure in case of perfect gas law= 50.6 atm\n", + " \n", + " Pressure in case of vanderwaals equation = 35.3 atm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6 - Pg 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the pressure of the gas\n", + "#initialisation of variables\n", + "n= 10 #moles\n", + "T= 300 #K\n", + "V= 4.86 #l\n", + "R= 0.08205 #atml/molK\n", + "v= 0.1417 #l\n", + "T1= 305.7 #K\n", + "#CALCULATIONS\n", + "b= v/2\n", + "a= 2*v*R*T1\n", + "P= ((n*R*T)/(V-n*b))*2.71**(-a*n/(V*R*T))\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Pressure =',P,' atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure = 32.8 atm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - Pg 23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the root mean square velocity\n", + "#initialisation of variables\n", + "import math\n", + "from math import sqrt\n", + "T= 0 #C\n", + "T1= 100 #C\n", + "R= 8.314 #atm lit/mol K\n", + "n= 3\n", + "M= 2.016 #gm\n", + "M1= 28.02 #gm\n", + "M2= 146.1 #gm\n", + "#CALCULATIONS\n", + "u= sqrt(n*R*10**7*(T+273.2)/M)\n", + "u1= sqrt(n*R*10**7*(T+273.2)/M1)\n", + "u2= sqrt(n*R*10**7*(T+273.2)/M2)\n", + "u3= sqrt(n*R*10**7*(T1+273.2)/M)\n", + "u4= sqrt(n*R*10**7*(T1+273.2)/M1)\n", + "u5= sqrt(n*R*10**7*(T1+273.2)/M2)\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' root mean square velocity of H2 at 0 C =',u*10**-4,' cm/sec')\n", + "print '%s %.3f %s' % (' \\n root mean square velocity of N2 at 0 C=',u1*10**-4,' cm/sec')\n", + "print '%s %.3f %s' % (' \\n root mean square velocity of SF6 at 0 C =',u2*10**-4,'cm/sec')\n", + "print '%s %.2f %s' % (' \\n root mean square velocity of H2 at 100 C =',u3*10**-4,' cm/sec')\n", + "print '%s %.3f %s' % (' \\n root mean square velocity of N2 at 100 C =',u4*10**-4,' cm/sec')\n", + "print '%s %.3f %s' % (' \\n root mean square velocity of SF6 at 100 C =',u5*10**-4,' cm/sec')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " root mean square velocity of H2 at 0 C = 18.38 cm/sec\n", + " \n", + " root mean square velocity of N2 at 0 C= 4.931 cm/sec\n", + " \n", + " root mean square velocity of SF6 at 0 C = 2.160 cm/sec\n", + " \n", + " root mean square velocity of H2 at 100 C = 21.49 cm/sec\n", + " \n", + " root mean square velocity of N2 at 100 C = 5.764 cm/sec\n", + " \n", + " root mean square velocity of SF6 at 100 C = 2.524 cm/sec\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - Pg 34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the no. of collisions in He and N2\n", + "#Initialisation of variables\n", + "import math\n", + "from math import sqrt\n", + "P= 1 #at,\n", + "T= 300 #K\n", + "R= 82.05 #atm l/mol K\n", + "R1= 8.314\n", + "s= 4*10**-8 #cm\n", + "s1= 2*10**-8 #cm\n", + "m= 4 #gm\n", + "m1= 28 #gm\n", + "#CALCULATIONS\n", + "N= P*6.02*10**23/(R*T)\n", + "n= 2*s1**2*N**2*sqrt(math.pi*R1*10**7*T/m)\n", + "n1= 2*s**2*N**2*sqrt(math.pi*R1*10**7*T/m1)\n", + "#RESULTS\n", + "print '%s %.e %s' % (' no of collisions =',n,'collisions sec^-1 mol^-1')\n", + "print '%s %.2e %s' % (' \\n no of collisions =',n1,' collisions sec^-1 mol^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " no of collisions = 7e+28 collisions sec^-1 mol^-1\n", + " \n", + " no of collisions = 1.01e+29 collisions sec^-1 mol^-1\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - Pg 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the viscosity of N2\n", + "#initialisation of variables\n", + "import math\n", + "from math import sqrt\n", + "M= 28 #gm\n", + "R= 8.314*10**7 #atm l/mol K\n", + "N= 6.023*10**23\n", + "T= 300 #K\n", + "s= 4*10**-8#cm\n", + "#CALCULATIONS\n", + "m= M/N\n", + "k= R/N\n", + "n= (5./16.)*sqrt(math.pi*m*k*T)/(math.pi*s**2)\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' viscosity =',n,'poise')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " viscosity = 1.53e-04 poise\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12 - Pg 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Increase in energy per degree for 1 mole of gas\n", + "#initialisation of variables\n", + "n= 3\n", + "R= 2 #cal mol^-1 deg^-1\n", + "#CALCULATIONS\n", + "I= n*R\n", + "#RESULTS\n", + "print '%s %.1f %s' %(' Increase in energy =',I,'cal mol^-1 deg^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Increase in energy = 6.0 cal mol^-1 deg^-1\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - Pg 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Dipole moment and percentage of ionic character\n", + "#initialisation of variables\n", + "import math\n", + "k= 1.38*10**-16\n", + "N= 6*10**23 #molecules\n", + "a= 105 #degrees\n", + "l= 0.957 #A\n", + "e= 4.8*10**-10 #ev\n", + "#CALCULATIONS\n", + "u= math.sqrt(9*k*2.08*10**4/(4*math.pi*N))\n", + "uh= u/(2*math.cos(a*math.pi/180/2.))\n", + "z= uh/(l*e*10**-8) \n", + "#RESULTS\n", + "print '%s %.2e %s' % (' Dipole moment of H2O=',u,'e.s.u.cm')\n", + "print '%s %.2e %s' % (' \\n Dipole moment of OH bond =',uh,'e.s.u.cm')\n", + "print '%s %.2f' % (' \\n fraction of ionic character =',z)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Dipole moment of H2O= 1.85e-18 e.s.u.cm\n", + " \n", + " Dipole moment of OH bond = 1.52e-18 e.s.u.cm\n", + " \n", + " fraction of ionic character = 0.33\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14 - Pg 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the dielectric constant\n", + "#initialisation of variables\n", + "import math\n", + "u= 1.44*10**-18 #e.s.u\n", + "k= 3.8*10**-16 \n", + "T= 273. #k\n", + "N= 6.023*10**23 #molecules\n", + "v= 6. #cc\n", + "Vm= 44.8*10**3 #cc\n", + "#CALCULATIONS\n", + "Pm= v+(4*math.pi*N*u**2/(3*3*k*T))\n", + "r= Pm/Vm\n", + "k= (2*r+1)/(1-r)\n", + "#RESULTS\n", + "print '%s %.5f' % (' dielectric constant =',k)\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " dielectric constant = 1.00153\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_10.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_10.ipynb new file mode 100755 index 00000000..4ee0e2f1 --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_10.ipynb @@ -0,0 +1,635 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a8663b753e53365cf46aa7f5948fd23b365bccc405a9c5b9305fa79e49b0f6dc" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 - Chemical Kinetics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 543" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Pressure \n", + "#initialisation of variables\n", + "t= 3 #sec\n", + "P0= 200 #mm\n", + "k= 17.3 #mm/sec\n", + "P1= 104 #mm\n", + "#CALCULATIONS\n", + "P= P0-k*t\n", + "P2= P+P1\n", + "#RESULTS\n", + "print '%s %d %s' % (' Pressure=',P2,' mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure= 252 mm of Hg\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 545" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Half time\n", + "#initialisation of variables\n", + "k= 2.63*10**-3 #min^-1\n", + "#CALCULATIONS\n", + "t1= 0.693/k\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Half time=',t1,'min')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Half time= 263.5 min\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 546" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Partial Pressure of the reactant\n", + "#initialisation of variables\n", + "P= 200. #mm\n", + "t= 30. #min\n", + "k= 2.5*10**-4 #sec^-1\n", + "#CALCULATIONS\n", + "P0= P/(10**(k*t*60/2.303))\n", + "P1= P-P0\n", + "#RESULTS\n", + "print '%s %d %s' % (' Partial Pressure of reactant=',P1,'mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Partial Pressure of reactant= 72 mm of Hg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - pg 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the no of atoms\n", + "#initialisation of variables\n", + "t= 5600*365*24*60.\n", + "x= 5 #atoms\n", + "#CALCULATIONS\n", + "k= 0.693/t\n", + "N= x/k\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' No of atoms=',N, 'atoms')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " No of atoms= 2.12e+10 atoms\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - pg 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the time passed\n", + "#initialisation of variables\n", + "import math\n", + "t= 5600 #sec\n", + "r= 0.256\n", + "#CALCULATIONS\n", + "t1= (t/0.693)*2.303*math.log10(1/r)\n", + "#RESULTS\n", + "print '%s %d %s' % (' Time=',t1,'years ago')\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Time= 11012 years ago\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6 - pg 549" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the first order rate constant and half life\n", + "#initialisation of variables\n", + "import math\n", + "t= 25.1 #hr\n", + "C= 0.004366 \n", + "C1= 0.002192\n", + "C2= 0.006649\n", + "#CALCULATIONS\n", + "r= (C-C1)/(C2-C1)\n", + "k= 2.303*math.log10(1/r)/t\n", + "t1= 0.693/k\n", + "#RESULTS\n", + "print '%s %.1f %s' %(' Time=',t1,' hr')\n", + "print '%s %.2e %s' %(' Time=',k,' hr')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Time= 24.2 hr\n", + " Time= 2.86e-02 hr\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - pg 552" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Rate constant\n", + "#initialisation of variables\n", + "s= 18.6*10**4 #mm of hg\n", + "#CALCULATIONS\n", + "k= 1./s\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' Rate constant=',k,' (mm Hg)^-1 sec^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Rate constant= 5.38e-06 (mm Hg)^-1 sec^-1\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 552" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the requried Pressure\n", + "#initialisation of variables\n", + "k= 1.14*10**-4 #sec^-1\n", + "k1= 5.38*10**-6 #sec^-1\n", + "#CALCULATIONS\n", + "P= k/k1\n", + "P2=0.01*P\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Pressure=',P2,'mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure= 0.212 mm of Hg\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the no of molecules\n", + "#initialisation of variables\n", + "T= 600 #K\n", + "P= 1 #atm\n", + "R= 0.082 #atm lit/mol K\n", + "#CALCULATIONS\n", + "C= P/(R*T)\n", + "r= C**2*4*10**-6 \n", + "r1= 6*10**23*r\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' No of molecules=',r1,'molecules l^-1 sec^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " No of molecules= 9.9e+14 molecules l^-1 sec^-1\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - pg 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the time required\n", + "#initialisation of variables\n", + "k= 6.3*10**2 #ml mole^-1 sec^-1\n", + "P= 400. #mm\n", + "T= 600. #K\n", + "R= 82.06\n", + "#CALCULATIONS\n", + "C= (P/760.)/(R*T)\n", + "t= 1/(9.*C*k)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' time=',t,' sec')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " time= 16.5 sec\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - pg 556" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the pressure of No2 in both cases\n", + "#initialisation of variables\n", + "pf2= 2.00 #mm Hg\n", + "y= 0.96 #mm Hg\n", + "Pn= 5 #mm Hg\n", + "#CALCULATIONS\n", + "pF2= pf2-y\n", + "pNO2= Pn-2*y\n", + "pNO2F= 2*y\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' pressure of NO2=',pNO2,'mm of Hg')\n", + "print '%s %.2f %s' % (' \\n pressure of NO2 after 30 sec=',pNO2F,'mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " pressure of NO2= 3.08 mm of Hg\n", + " \n", + " pressure of NO2 after 30 sec= 1.92 mm of Hg\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - pg 561" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Rate constant\n", + "#initialisation of variables\n", + "k= 4*10**-6 #mol^-1 sec^-1\n", + "Kc= 73\n", + "#CALCULATIONS\n", + "K1= k*Kc/2\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' Rate constant=',K1,'l mol^-1 sec^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Rate constant= 1.46e-04 l mol^-1 sec^-1\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14 - pg 568" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the activation energy\n", + "#initialisation of variables\n", + "import math\n", + "R= 1.987 #atm lit/mol K\n", + "T= 573.2 #K\n", + "T1= 594.6 #K\n", + "k= 3.95*10**-6 #mol^-1 sec^-1\n", + "k1= 1.07*10**-6 #mol^-1 sec^-1\n", + "#CALCULATIONS\n", + "H= R*T*T1*2.303*math.log10((k/k1))/(T1-T)\n", + "#RESULTS\n", + "print '%s %d %s' %(' activation energy=',H,'calmol^-1')\n", + "print 'The answers in the texbook are a bit different due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " activation energy= 41338 calmol^-1\n", + "The answers in the texbook are a bit different due to rounding off error\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15 - pg 568" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the time required\n", + "#initialisation of variables\n", + "import math\n", + "H= 41300. #cal\n", + "T= 673. #K\n", + "T1= 595. #K\n", + "R= 1.987 #cal/mol K\n", + "K= 3.95*10**-6\n", + "P= 1 #atm\n", + "R1= 0.08205 #j/mol K\n", + "#CALCULATIONS\n", + "k2= math.e**(H*(T-T1)/(R*T*T1))*K\n", + "C= P/(R1*T)\n", + "t= 44.8/C\n", + "t2=R1*T*10**-2 /k2\n", + "#RESULTS\n", + "print '%s %d %s' %(' time =',t,'sec')\n", + "print '%s %d %s' %('Time required in case 2 = ',t2,'sec')\n", + "print 'The answers in the texbook are a bit different due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " time = 2473 sec\n", + "Time required in case 2 = 2438 sec\n", + "The answers in the texbook are a bit different due to rounding off error\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16 - pg 569" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the collision diameter\n", + "#initialisation of variables\n", + "import math\n", + "H= 41300.\n", + "R= 1.987 #atm lit/mol K\n", + "T= 595. #K\n", + "M= 128. #gm\n", + "R1= 8.314*10**7 #atm lit/mol K\n", + "N= 6.02*10**23 #moleccules\n", + "k= 3.95*10**-6 #sec**-1\n", + "#CALCULATIONS\n", + "s= math.sqrt((k*10**3/(4*N))*(128/(math.pi*R1*T))**0.5*math.e**(H/(R*T)))\n", + "#RESULTS\n", + "print '%s %.3e %s' % (' collision diameter=',s,' cm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " collision diameter= 8.356e-09 cm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18 - pg 577" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Concentration of A and B\n", + "#initialisation of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "p= 20.3 #percent\n", + "p1= 1.77 #percent\n", + "I= 100.\n", + "n= 2.\n", + "l= 300. #l mol^-1 cm^-1\n", + "l1= 30. #l mol^-1 cm^-1\n", + "l2= 10. #l mol^-1 cm^-1\n", + "l3= 200. #l mol^-1 cm^-1\n", + "#CALCULATIONS\n", + "A= ([[n*l, n*l1],[n*l2, n*l3]])\n", + "b= ([[math.log10(I/p1)],[math.log10(I/p)]])\n", + "c= numpy.dot(numpy.linalg.inv(A),b)\n", + "R1=c[0]\n", + "R2=c[1]\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' Concentration of A =',R1,' mole l^-1')\n", + "print '%s %.2e %s' % (' \\n Concentration of B =',R2,' mole l^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Concentration of A = 2.76e-03 mole l^-1\n", + " \n", + " Concentration of B = 1.59e-03 mole l^-1\n" + ] + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_2.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_2.ipynb new file mode 100755 index 00000000..0dbb438a --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_2.ipynb @@ -0,0 +1,438 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9cfae4f7a986d325f2eab7e6f5652e4123a3be0877578166dbb4520edb5ed59f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 - Structures of Condensed Phases" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the size of cubic unit cell\n", + "#initialisation of variables\n", + "import math\n", + "l= 1.5418 #A\n", + "a= 19.076 #degrees\n", + "d2= 1.444 #A\n", + "#CALCULATIONS\n", + "d= l/(2*math.sin(a*math.pi/180.))\n", + "a= math.sqrt(8*d2*d2)\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' size of cubic unit cell =',a,'A')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " size of cubic unit cell = 4.0842 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Density of silver\n", + "#initialisation of variables\n", + "M= 107.88 #gm\n", + "z= 4\n", + "v= 4.086 #A\n", + "N= 6.023*10**23\n", + "#CALCULATIONS\n", + "d= z*M/(v**3*10**-24*N)\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' Density of silver =',d,'gm cm^-3')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Density of silver = 10.5025 gm cm^-3\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the molecular weight\n", + "#initialisation of variables\n", + "d= 1.287 #g cm**-3\n", + "a= 123 #A\n", + "z= 4\n", + "#CALCULATIONS\n", + "M= d*6.023*10**23*a**3*10**-24/z\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' molecular weight =',M,'gm ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " molecular weight = 3.6e+05 gm \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - pg 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the radius of silver atom\n", + "import math\n", + "#initialisation of variables\n", + "a= 4.086 #A\n", + "#CALCULATIONS\n", + "d= a*math.sqrt(2)\n", + "r= d/4.\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' radius of silver atom=',r,' A ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " radius of silver atom= 1.445 A \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - pg 99" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the surface tension\n", + "import math\n", + "#initialisation of variables\n", + "M= 38.3 #mg cm^-1\n", + "d= 13.55 #g cm^-3\n", + "p= 0.9982 #g cm^-3\n", + "g= 980.7 #cm/sec^2\n", + "l= 4.96 #cm\n", + "#CALCULATIONS\n", + "r= math.sqrt(M*10**-3/(d*math.pi))\n", + "R= r*p*g*l/2\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' surface tension =',R,' ergs cm^-2 ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " surface tension = 72.8 ergs cm^-2 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6 - pg 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the dipole moment of water\n", + "#initialisation of variables\n", + "import math\n", + "r= 1.333\n", + "d= 0.9982 #g cm**-3\n", + "m= 18.02 #gm\n", + "Pm= 74.22 #cc\n", + "k= 8.314*10**7 \n", + "N= 6.023*10**23\n", + "T= 293 #k\n", + "#CALCULATIONS\n", + "Rm= ((r**2-1)/(r**2+2))*m/d\n", + "u= math.sqrt(9*k*T*(Pm-Rm)/(4*math.pi*N**2))\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' dipole moment of water =',u,'e.s.u ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " dipole moment of water = 1.84e-18 e.s.u \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - pg 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the radius of argon atom\n", + "#initialisation of variables\n", + "a= 1.66*10**-24 #cm**3\n", + "#CALCULATIONS\n", + "r= a**(1/3.)/10**-8\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' radius =',r,'A ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " radius = 1.18 A \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the index of refraction\n", + "import math\n", + "#initialisation of variables\n", + "N= 6.023*10**23 #molecules\n", + "a= 10**-24\n", + "k= 0.89\n", + "cl= 3.60\n", + "M= 74.56 #gms\n", + "d= 1.989 #g/cm^3\n", + "#CACLULATIONS\n", + "Rm= 4*math.pi*N*(k+cl)*a/3\n", + "r= Rm*d/M\n", + "n= math.sqrt((2*r+1)/(1-r))\n", + "#RESULTS\n", + "print '%s %.3f' % (' index of refraction= ',n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " index of refraction= 1.516\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the radius of K and Cl atoms\n", + "#initialisation of variables\n", + "v= 3.6 #cc\n", + "v1= 0.89 #cc\n", + "s= 3.146 #A\n", + "#CALCULATIONS\n", + "r= (v/v1)**(1/3.)\n", + "r1 = s/(1+r)\n", + "r2 = s-r1\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' radius of k+=',r1,'A ')\n", + "print '%s %.3f %s' % (' \\n radius of cl-=',r2,'A ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " radius of k+= 1.213 A \n", + " \n", + " radius of cl-= 1.933 A \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - pg 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the angle of rotation\n", + "#initialisation of variables\n", + "g= 10 #gm\n", + "d= 1.038 #gm/mol\n", + "M= 100 #gm\n", + "x= 66.412\n", + "y= 0.127\n", + "z= 0.038\n", + "l= 20 #cm\n", + "#CALCULATIONS\n", + "p= g/(M/d)\n", + "X= x+y-z\n", + "ar= X*l*p/10.\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' angle of rotation=',ar,'degrees ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " angle of rotation= 13.81 degrees \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - pg 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the viscosity of toluene\n", + "#initialisation of variables\n", + "t= 68.9 #sec\n", + "t1= 102.2 #sec\n", + "p1= 0.866 #g/cm^3\n", + "p2= 0.998 #gm/cm^3\n", + "n= 0.01009 #dynesc/cm^2\n", + "#CALCULATIONS\n", + "N= n*t*p1/(t1*p2)\n", + "#RESULTS\n", + "print '%s %.5f %s' % (' viscosity of toluene=',N,'dyne sec/cm^2 ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " viscosity of toluene= 0.00590 dyne sec/cm^2 \n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_3.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_3.ipynb new file mode 100755 index 00000000..51fa8398 --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_3.ipynb @@ -0,0 +1,896 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e52ffa5d23d3ddfa48a481e2732abf0ed79242403a53e7634d8954f9e01773b0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 - First law of Thermodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 129" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the increase in energy\n", + "#initialisation of variables\n", + "P= 0.0060 #atm\n", + "M=18. #gm\n", + "L=80 #cal/gm\n", + "H=596.1 #cal/gm\n", + "#calculations\n", + "Hs=M*L+M*H\n", + "#results\n", + "print '%s %d %s' %('Net increase in energy = ',Hs,'cal')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net increase in energy = 12169 cal\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the increase in energy\n", + "#initialisation of variables\n", + "P= 0.0060 #atm\n", + "V1= 0.0181 #l\n", + "H= -10730 #cal\n", + "V2= 22.4 #l\n", + "#CALCULATIONS\n", + "W= (V2-P*V1)*(1.987/.08205)\n", + "E= H+W\n", + "#RESULTS\n", + "print '%s %d %s' % (' increase in energy=',E,' cal ')\n", + "print 'The answer differs a bit from the textbook due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " increase in energy= -10187 cal \n", + "The answer differs a bit from the textbook due to rounding off error\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 132" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the increase in energy\n", + "#initialisation of variables\n", + "T1= 70 #C\n", + "T2= 10 #C\n", + "Cp= 18 #cal mole^-1 deg^-1\n", + "P= 1 #atm\n", + "m= 18. #g\n", + "d= 0.9778 #g/ml\n", + "d1= 0.9997 #g/ml\n", + "e= 1.987 #cal\n", + "x= 82.05 #ml atm\n", + "#CALCULATIONS\n", + "H= Cp*(T1-T2)\n", + "E= H-(e/x)*P*((m/d)-(m/d1))\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' increase in energy=',E,'cal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " increase in energy= 1080.0 cal \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - pg 132" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the conversion factor\n", + "#initialisation of variables\n", + "i= 1 #amp\n", + "r= 2 #ohms\n", + "t= 10 #min\n", + "dT= 2.73 #C\n", + "x= 0.1 #cal/deg\n", + "x1= 100 #cal/deg\n", + "x2= 5 #cal/deg\n", + "#CALCULATIONS\n", + "w= i**2*r*t*60\n", + "H= (x+x1+x2)*dT\n", + "E= w/H\n", + "#RESULTS\n", + "print '%s %.2f %s' % ('conversion factor =',E,'cal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "conversion factor = 4.18 cal \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6 - pg 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the heat at constant pressure and volume\n", + "#initialisation of variables\n", + "Cp= 6.0954 #cal /mol deg\n", + "Cp1= 3.2533*10**-3 #cal /mol deg\n", + "Cp2= 1.071*10**-6 #cal /mol deg\n", + "T= 100 #C\n", + "T1= 0 #C\n", + "R= 1.987 #atml/cal K\n", + "#CALULATIONS\n", + "H= Cp*(T-T1)+(Cp1/2)*((T+273.2)**2-(T1+273.2)**2)-(Cp2/3)*((T+273.2)**3-(T1+273.2)**3)\n", + "q= H-R*(T-T1)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Heat at constant pressure=',H,'cal ')\n", + "print '%s %.1f %s' % (' \\n Heat at constant volume=',q,'cal ')\n", + "print 'The answer differs a bit from the textbook due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Heat at constant pressure= 703.4 cal \n", + " \n", + " Heat at constant volume= 504.7 cal \n", + "The answer differs a bit from the textbook due to rounding off error\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - pg 140" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the work done in the process\n", + "#initialisation of variables\n", + "vl= 0.019 #l\n", + "vg= 16.07 #l\n", + "h= 1489. #mm of Hg\n", + "#CALCULATIONS\n", + "w= -(h/760)*(vl-vg)*(1.987/0.08206)\n", + "#RESULTS\n", + "print '%s %d %s' % (' Work done=',w,'cal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Work done= 761 cal \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the minimum work in both cases\n", + "#initialisation of variables\n", + "import math\n", + "n= 2 #moles\n", + "R= 0.08206 #J/mol K\n", + "T= 25 #C\n", + "b= 0.0428 #lmole^-1\n", + "a= 3.61 #atm l^2 mole^-1\n", + "V1= 20. #l\n", + "V2= 1. #l\n", + "#CALCULATIONS\n", + "w1= n*1.987*(273.2+T)*math.log10(V1/V2) *2.303\n", + "w= (n*R*(273.2+T)*2.303*math.log10((V1-n*b)/(V2-n*b))-a*n**2*((1/V2)-(1/V1)))*(1.987/0.08206)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' minimum work=',w1,'cal ')\n", + "print '%s %.1f %s' % (' \\n minimum work=',w,'cal ')\n", + "print 'The answer differs a bit from the textbook due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " minimum work= 3550.7 cal \n", + " \n", + " minimum work= 3319.5 cal \n", + "The answer differs a bit from the textbook due to rounding off error\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 144" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the final volume and temperature of the gas. Also calculate the work done in the process\n", + "#initialisation of variables\n", + "cv = 5.00 #cal mole^-1 deg^-1\n", + "R= 1.99 #cal mole^-1 deg^-1\n", + "p= 1 #atm\n", + "p1= 100. #atm\n", + "V= 75. #l\n", + "n= 3. #moles\n", + "R1= 0.08206 #cal/mol K\n", + "#CALCULATIONS\n", + "cp= cv+R\n", + "r= cp/cv\n", + "V1= V/(p1/p)**(1/r)\n", + "T2= p1*V1/(n*R1)\n", + "w= (p1*V1-p*V)*R/((r-1)*R1)\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' final volume of gas =',V1,'l ')\n", + "print '%s %d %s' % (' \\n final temperature of gas =',T2,'K ')\n", + "print '%s %d %s' % (' \\n Work done =',w,'cal ')\n", + "print 'The answer differs a bit from the textbook due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " final volume of gas = 2.78 l \n", + " \n", + " final temperature of gas = 1130 K \n", + " \n", + " Work done = 12384 cal \n", + "The answer differs a bit from the textbook due to rounding off error\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - pg 144" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the change in energy and enthalpy\n", + "#initialisation of variables\n", + "cv= 5 #cal mole^-1\n", + "P= 100 #atm\n", + "T= 1130 #K\n", + "T1= 812 #K\n", + "n= 3 #moles\n", + "R= 1.99 #cal/mole K\n", + "#CALCULTIONS\n", + "E= n*cv*(T1-T)\n", + "H= E+n*R*(T1-T)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' change in energy =',E,'cal ')\n", + "print '%s %.1f %s' % (' \\n change in enthalpy=',H,' cal ')\n", + "print 'The answer differs a bit from the textbook due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " change in energy = -4770.0 cal \n", + " \n", + " change in enthalpy= -6668.5 cal \n", + "The answer differs a bit from the textbook due to rounding off error\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - pg 145" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the work done and final pressure\n", + "#initialisation of variables\n", + "import math\n", + "k= 1.435 \n", + "k1= 17.845*10**-3 #K**-1\n", + "k2= -4.165*10**-6 #K**-2\n", + "T= 200. #C\n", + "T1= 0. #C\n", + "P= 10. #atm\n", + "R= 1.987 #cal/mol K\n", + "k3= 3.422\n", + "#CALCULATIONS\n", + "W= k*(T-T1)+(k1/2)*((273+T)**2-(273+T1)**2)+(k2/3)*((273+T)**3-(273+T1)**3)\n", + "P2= (P/math.e**((k*math.log((273+T1)/(273+T))+k1*(T1-T)+(k2/2)*((273+T1)**2-(273+T)**2))/R))/100.\n", + "#RESULTS\n", + "print '%s %d %s' % (' work done by methane =',W,'cal ')\n", + "print '%s %.2f %s' % (' \\n final pressure=',P2,'atm ')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " work done by methane = 1499 cal \n", + " \n", + " final pressure= 0.77 atm \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12 - pg 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the fraction of liquid\n", + "#initialisation of variables\n", + "P= 100 #atm\n", + "P1= 1 #atm\n", + "R= 1.99 #cal/mol**-1 K**-1\n", + "k= 0.3 #atm**-1\n", + "E= 1600 #cal\n", + "T= -183 #C\n", + "T1= 0 #C\n", + "#CALCULATIONS\n", + "X= (k*3.5*R*(P-P1))/(3.5*R*(T1-T)+E)\n", + "#RESULTS\n", + "print '%s %.3f' % (' fraction of liquid = ',X)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " fraction of liquid = 0.072\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - pg 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the enthalpy change of the reaction\n", + "#initialisation of variables\n", + "H= -21.8 #kcal\n", + "H1= 3.3 #kcal\n", + "#CALCULATIONS\n", + "H2= H-H1\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Enthalpy =',H2,'kcal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy = -25.1 kcal \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14 - pg 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the heat of hydrogenation\n", + "#initialisation of variables\n", + "H= -68.317 #kcal\n", + "H1= -310.615 #kcal\n", + "H2= -337.234 #kcal\n", + "R= 1.987 #cal/mol^-1 K^-1\n", + "T= 298.2 #K\n", + "n= 1 #mole\n", + "n1= 1 #mole\n", + "n2= 1 #mole\n", + "#CALCULATIONS\n", + "E= H+H1-H2-(n-n1-n2)*R*T*10**-3\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Heat of hydrogenation =',E,'kcal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Heat of hydrogenation = -41.105 kcal \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15 - pg 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the enthalpy of the process\n", + "#initialisation of variables\n", + "Hf= -196.5 #kcal\n", + "H= -399.14 #kcal\n", + "#CALCULATIONS\n", + "H1= (H-Hf)*1000\n", + "#RESULTS\n", + "print '%s %d %s' % (' Enthalpy =',H1,' kcal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy = -202640 kcal \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16 - pg 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Enthalpy change\n", + "#initialisation of variables\n", + "H= -350.2 #kcal\n", + "H1= -128.67 #kcal\n", + "H2= -216.90 #kcal\n", + "#CALCULATIONS\n", + "H3= H-(H1+H2)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Enthalpy =',H3,'kcal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy = -4.6 kcal \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17 - pg 158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the enthalpy of the process\n", + "#initialisation of variables\n", + "H= -40.023 #kcal\n", + "H1= -22.063 #kcal\n", + "#CALCULATIONS\n", + "H2= H-H1\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Enthalpy =',H2,' kcal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy = -17.960 kcal \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18 - pg 162" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the enthalpy change in the process\n", + "#initialisation of variables\n", + "H= -112.148 #k cal\n", + "H1= 101.99 #k cal\n", + "Hx=-112.148 #kcal\n", + "Hy=-111.015 #kcal\n", + "Hz=-.64\n", + "Hsol=-9.02\n", + "#CALCULATIONS\n", + "H2= H+H1\n", + "H3=2*Hx-2*Hy\n", + "H4=-10*Hz\n", + "H5=Hsol-5*Hz\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' Enthalpy in case 1=',H2,'k cal ')\n", + "print '%s %.3f %s' % (' Enthalpy in case 2=',H3,'k cal ')\n", + "print '%s %.1f %s' % (' Enthalpy in case 3=',H4,'k cal ')\n", + "print '%s %.2f %s' % (' Enthalpy in case 4=',H5,'k cal ')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy in case 1= -10.16 k cal \n", + " Enthalpy in case 2= -2.266 k cal \n", + " Enthalpy in case 3= 6.4 k cal \n", + " Enthalpy in case 4= -5.82 k cal \n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19 - pg 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the dE and dH in the process\n", + "#initialisation of variables\n", + "cp=18.\n", + "T2=373 #K\n", + "T1=298 #K\n", + "T3=403.2 #K\n", + "hvap=9713 #cal\n", + "H4= 0 #cal\n", + "E4= 0 #cal\n", + "a=7.1873\n", + "b=2.3733e-3\n", + "c=.2084e-6\n", + "R=1.987\n", + "#RESULTS\n", + "H1=cp*(T2-T1)\n", + "H2=hvap\n", + "H3=a*(T3-T2) + b/2 *(T3**2-T2**2) + c/3 *(T3**3-T2**3)\n", + "E1=H1\n", + "E2=H2-R*T2\n", + "E3=H3-R*(T3-T2)\n", + "H= H1+H2+H3+H4\n", + "E= E1+E2+E3+E4\n", + "#RESULTS\n", + "print '%s %d %s' % (' Enthalpy=',H,'cal ')\n", + "print '%s %d %s' % (' \\n Energy=',E,' cal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy= 11308 cal \n", + " \n", + " Energy= 10507 cal \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20 - pg 171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the enthalpy change \n", + "#initialisation of variables\n", + "H= -114009.8 #cal\n", + "x= -5.6146 #K**-1\n", + "y= 0.9466*10**-3 #K**-2\n", + "z= 0.1578*10**-6 #K**-3\n", + "T= 1000\n", + "#CALCULATIONS\n", + "H1= H+x*T+y*T**2+z*T**3\n", + "#RESULTS\n", + "print '%s %.1f %s' %(' Enthalpy =',H1,'cal ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy = -118520.0 cal \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21 - pg 173" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#caculate the temperature achieved\n", + "#Initialization of variables\n", + "import numpy\n", + "a=72.3639\n", + "b=36.2399e-3\n", + "c=3.7621e-6\n", + "H=214920\n", + "#calculations\n", + "vec=([-c/3,b/2,a,-H])\n", + "vec2=numpy.roots(vec)\n", + "vec22=(vec2[2])\n", + "#results\n", + "print '%s %.1f %s' %('The required temperature observed is', vec22,'K')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required temperature observed is 2059.4 K\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22 - pg 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the change in enthalpy\n", + "#initialisation of variables\n", + "T= 298 #K\n", + "R= 1.987 #atmcc/mol K\n", + "x= 128.16\n", + "y= 0.9241\n", + "H= -8739 #cal\n", + "H1=-771.1 #cal\n", + "H22=-196.5 #cal\n", + "H33=-30.4 #cal\n", + "n1= 10 #mol\n", + "n2= 12 #mol\n", + "M1=122.12\n", + "M11=128.16\n", + "M2=55.85\n", + "dT=2.51\n", + "x1=.8\n", + "H2=1947.4\n", + "x2=1.\n", + "H3=1944.2\n", + "dT2=3.62\n", + "#CALCULATIONS\n", + "E1=H1+1.5*R*T/1000.\n", + "E2=H22+1.5*R*T/1000.\n", + "E3=(H33+3.5*R*T/1000.)/2.\n", + "C=(-E1*.9619/M1-E2/2/M2*.0002 -.47/1000. *E3)*1000/dT - x1-H2\n", + "E11=(-E2/2/M2*.0002-.76/1000. *E3 - (x2+H3+C)*dT2)*M11/.9241\n", + "H= (E11+R*T*(n1-n2))/1000\n", + "#RESULTS\n", + "print '%s %.1f %s' %(' Enthalpy =',H,'kcal mole^-1 ')\n", + "print 'The answer is a bit different from textbook due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy = -1214.5 kcal mole^-1 \n", + "The answer is a bit different from textbook due to rounding off error\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_4.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_4.ipynb new file mode 100755 index 00000000..b8239152 --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_4.ipynb @@ -0,0 +1,1022 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f5ac26c556550c99333fe9a615118f88638c4e49d92a9c806704f151f26a3795" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 - Second and Third laws of thermodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the maximum efficiency in all cases\n", + "#initialisation of variables\n", + "T = 100. #C\n", + "T1= 25. #C\n", + "T2= 150. #C\n", + "T3= 357. #C\n", + "T4= 500. #C\n", + "T5= 2000. #C\n", + "T6= 5*10**6\n", + "T7= 1000. #C\n", + "#CALCULATIONS\n", + "e= (T-T1)/(T+273)\n", + "e1= (T2-T1)/(273+T2)\n", + "e2= (T3-T)/(273+T3)\n", + "e3= (T5-T4)/(T5+273)\n", + "e4= (T6-T7)/T6\n", + "#RESULTS\n", + "print '%s %.2f' % (' maximum efficiency in case 1= ',e)\n", + "print '%s %.2f' % (' \\n maximum efficiency in case 2 = ',e1)\n", + "print '%s %.2f' % (' \\n maximum efficiency in case 3 = ',e2)\n", + "print '%s %.2f' % (' \\n maximum efficiency in case 4 = ',e3)\n", + "print '%s %.2f' % (' \\n maximum efficiency in case 5 = ',e4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " maximum efficiency in case 1= 0.20\n", + " \n", + " maximum efficiency in case 2 = 0.30\n", + " \n", + " maximum efficiency in case 3 = 0.41\n", + " \n", + " maximum efficiency in case 4 = 0.66\n", + " \n", + " maximum efficiency in case 5 = 1.00\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 194" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the maximum efficiency and minimum work\n", + "#initialisation of variables\n", + "T= 20. #C\n", + "T1= -10. #C\n", + "q= 1000. #cal\n", + "#CALCULATIONS\n", + "e= (273+T1)/(T-T1)\n", + "w= (T-T1)*q/(273+T1)\n", + "#RESULTS\n", + "print '%s %.1f' % (' maximum efficiency = ',e)\n", + "print '%s %d %s' % (' \\n minimum work =',w,'cal')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " maximum efficiency = 8.8\n", + " \n", + " minimum work = 114 cal\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the net work done on and by the gas\n", + "#initialisation of variables\n", + "T= 1000 #K\n", + "T1= 400 #/K\n", + "w= 1000 #cal\n", + "E= 0 #cal\n", + "gam=7/5.\n", + "#CALCULATIONS\n", + "q= w-E\n", + "W= q*(T-T1)/T\n", + "q1= W-q\n", + "W1= -q1\n", + "dE=5/2.*(T1-T)\n", + "dH=7/2. *(T1-T)\n", + "w2=dE-E\n", + "w3=(T-T1)/(gam-1)\n", + "#RESULTS\n", + "print '%s %.1f %s' % ('net work done by gas=',W,' cal')\n", + "print '%s %.1f %s' % ('net work done on gas =',W1,'cal')\n", + "print '%s %.1f %s' %('Change in Internal energy = ',dE,'R cal')\n", + "print '%s %.1f %s' %('Change in Enthalpy = ',dH,'R cal')\n", + "print '%s %.1f %s' %('Work for adiabatic compression =',w3,'R cal')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "net work done by gas= 600.0 cal\n", + "net work done on gas = 400.0 cal\n", + "Change in Internal energy = -1500.0 R cal\n", + "Change in Enthalpy = -2100.0 R cal\n", + "Work for adiabatic compression = 1500.0 R cal\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - pg 199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the entropy of substances\n", + "#initialisation of variables\n", + "import numpy as np\n", + "Hv= np.array([1960.,1560.,4880.,37000.,5500.,27400.,60700.,9720.,30900.]) #cal mole^-1\n", + "Tb= np.array([112.,87.3,239.,1806.,259.,1180.,2466.,373.,1029.]) #K\n", + "#CALCULATIONS\n", + "Sv= np.round(Hv/Tb,1)\n", + "#RESULTS\n", + "print '%s' % (' Entropy (cal mole deg^-1)')\n", + "print Sv\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Entropy (cal mole deg^-1)\n", + "[ 17.5 17.9 20.4 20.5 21.2 23.2 24.6 26.1 30. ]\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - pg 201" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the entropy at constant pressure and volume\n", + "#initialisation of variables\n", + "import math\n", + "T= 300. #K\n", + "T1= 400. #K\n", + "k= 6.0954\n", + "k1= 3.2533*10**-3\n", + "k2= -1.0171*10**-6\n", + "R= 1.98719 #cal/mol K\n", + "#CALCULATIONS\n", + "S= 2*(k*math.log(T1/T)+k1*(T1-T)+k2*(T1**2-T**2)/2)\n", + "S1= S-2*R*math.log(T1/T)\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' Entropy=',S,' cal deg^-1')\n", + "print '%s %.4f %s' % (' \\n Entropy =',S1,'cal deg^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Entropy= 4.0865 cal deg^-1\n", + " \n", + " Entropy = 2.9432 cal deg^-1\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the final temperature \n", + "#initialisation of variables\n", + "T1= 273.16 #K\n", + "R= 1.987 #cal /mol K\n", + "R1= 0.08205 #J /mol K\n", + "n= 10 #moles\n", + "V1= 22.4 #lit\n", + "a= 1.36\n", + "Cv= 4.9\n", + "#CALCULATIONS\n", + "T2= T1-(R*a*(n-1)/(R1*n*Cv*V1))\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' temperature=',T2,' K')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " temperature= 272.89 K\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the inversion Temperature\n", + "#initialisation of variables\n", + "a= 1.360 #l^2 atm mole^-1\n", + "b= 0.0317 #l mole^-1\n", + "R= 0.08205 #J/mol K\n", + "#CALCULATIONS\n", + "T= 2*a/(b*R)\n", + "#RESULTS\n", + "print '%s %.1f %s' % ('Inversion Temperature=',T,'K')\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Inversion Temperature= 1045.8 K\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - pg 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the joule thomson coefficient\n", + "#initialisation of variables\n", + "a= 1.360 #l^2 atm mole^-1\n", + "b= 0.0317 #l mole^-1\n", + "R= 0.08205 #J/mol K\n", + "R1= 1.987 #cal/mole K\n", + "Cp= 6.9 #cal mole^-1 deg^-1\n", + "T= 273.2 #K\n", + "#CALCULATIONS\n", + "u= ((2*a/(R*T))-b)/(Cp*(R/R1))\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Joule thomson coefficient=',u,' atm^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Joule thomson coefficient= 0.315 atm^-1\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12 - pg 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the increase in entropy\n", + "#initialisation of variables\n", + "import math\n", + "p= 4/3. #atm\n", + "p1= 1 #atm\n", + "R= 1.9872 #cal /mole K\n", + "#CALCULATIONS\n", + "S= 2*R*math.log(p/p1)\n", + "#RESULTS\n", + "print '%s %.4f %s' %(' increase in entropy=',S,'cal deg^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " increase in entropy= 1.1434 cal deg^-1\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - pg 222" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the increase in entropy\n", + "#initialisation of variables\n", + "import math\n", + "p1= 1 #atm\n", + "R= 1.9872 #cal /mole K\n", + "#CALCULATIONS\n", + "S= 0 #Initial and final states are alike\n", + "#RESULTS\n", + "print '%s %.4f %s' %(' increase in entropy=',S,'cal deg^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " increase in entropy= 0.0000 cal deg^-1\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14 - pg 222" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate rhe increase in entropy in both cases\n", + "#initialisation of variables\n", + "import math\n", + "T= 25. #C\n", + "T1= 100. #C\n", + "R= 1.9872 #cal /mole K\n", + "p= 1 #atm\n", + "p1= 10. #atm\n", + "#CALCULATIONS\n", + "S= 3.5*R*math.log((T1+273)/(T+273))\n", + "S1= S+R*math.log(p/p1)\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' increase in entropy in case 1 =',S,'cal deg^-1')\n", + "print '%s %.2f %s' % (' \\n increase in entropy in case 2 =',S1,'cal deg^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " increase in entropy in case 1 = 1.56 cal deg^-1\n", + " \n", + " increase in entropy in case 2 = -3.01 cal deg^-1\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15 - pg 222" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the absolute entropy\n", + "#initialisation of variables\n", + "import math\n", + "S= 45.77 #cal deg^-1\n", + "T= 25. #C\n", + "T1= 100. #C\n", + "R= 1.9872 #cal /mole K\n", + "#CALCULATIONS\n", + "S0= S+ 3.5*R*math.log((T1+273)/(T+273))\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' absolute entropy=',S0,'cal deg^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " absolute entropy= 47.33 cal deg^-1\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16 - pg 226" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the change in entropy\n", + "#initialisation of variables\n", + "import math\n", + "Cp= 18. #cal deg^-1\n", + "T= 0. #C\n", + "T1= -5. #C\n", + "H2= -1440. #cal\n", + "Cp1= 9. #cal deg^-1\n", + "H= 0.\n", + "#CALCULATIONS\n", + "T2= (-Cp*(T-T1)-H2+Cp1*(273.16+T))/Cp1\n", + "S= Cp*math.log((273.16+T)/(273.16+T1))-(Cp*(T-T1)/(T+273.16))\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' Change in entropy=',S,'cal deg^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Change in entropy= 0.0031 cal deg^-1\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18 - pg 231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Gibbs free energy\n", + "#initialisation of variables\n", + "H= -57.7979 #cal\n", + "H1= -68.3174 #cal\n", + "S= 45.106 #cal deg^-1\n", + "S1= 16.716 #cal deg^-1\n", + "T= 25 #C\n", + "#CALCULATIONS\n", + "H2= (H-H1)*1000\n", + "S2= S-S1\n", + "G= H2-(273.16+T)*S2\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Gibbs free energy=',G,'cal')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gibbs free energy= 2054.7 cal\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19 - pg 231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Gibbs free energy\n", + "#initialisation of variables\n", + "H= -68317.4 #cal\n", + "S= 16.716 #cal\n", + "S1= 49.003 #cal\n", + "S2= 31.211 #cal\n", + "T= 25 #C\n", + "#CALCULATIONS\n", + "H1= 2*H\n", + "S3= 2*S-(S1+2*S2)\n", + "G= H1-(T+273.16)*S3\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Gibbs free energy=',G,'cal')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gibbs free energy= -113380.4 cal\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20 - pg 232" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the gibbs free energy change\n", + "#initialisation of variables\n", + "H= -57.7979 #kcal\n", + "H1= -196.5 #kcal\n", + "S1=45.106\n", + "S2=6.49\n", + "S3=21.5\n", + "S4=31.211\n", + "T=298.16\n", + "#calculations\n", + "dH=3*H-H1\n", + "dS=3*S1+2*S2-S3-3*S4\n", + "dG=dH*1000-T*dS\n", + "#results\n", + "print '%s %d %s' %(\"Gibbs free energy change =\",dG,\"cal\")\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Gibbs free energy change = 13217 cal\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22 - pg 240" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Gibbs free energy and value of A in both cases\n", + "#initialisation of variables\n", + "import math\n", + "p= 1. #atm\n", + "p1= 3. #atm\n", + "R= 1.987 #cal/mole K\n", + "T= 27. #C\n", + "b= 0.0428 #l mole^-1\n", + "a= 3.61 #l^2 atm mole^-1\n", + "#CALCULATIONS\n", + "G= R*(273+T)*math.log(p/p1)\n", + "A= R*(273+T)*math.log(p/p1)\n", + "G1= R*(273+T)*math.log(p/p1)+(b-(a/(0.08205*(T+273))))*(p-p1)*(R/0.08205)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Gibs free energy=',G,'cal')\n", + "print '%s %.1f %s' % (' \\n Value of dA=',A,'cal')\n", + "print '%s %.1f %s' % (' \\n Gibs free energy=',G1,'cal')\n", + "print '%s %.1f %s' % (' \\n Value of dA=',A,'cal')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gibs free energy= -654.9 cal\n", + " \n", + " Value of dA= -654.9 cal\n", + " \n", + " Gibs free energy= -649.9 cal\n", + " \n", + " Value of dA= -654.9 cal\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24 - pg 244" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the fugacities of both components\n", + "#initialisation of variables\n", + "import math\n", + "b= 0.0386 #l**2 atm mole**-1\n", + "a= 1.348 #l mole**-1\n", + "R= 0.08205 #cal /mole K\n", + "T= 25 #C\n", + "a1= 3.61 #l**2 atm mole**-1\n", + "b1= 0.0428 #l mole**-1\n", + "P= 50 #atm\n", + "P1= 1 #atm\n", + "#CALCULATIONS\n", + "Bn= b-(a/(R*(273.2+T)))\n", + "Bc= b1-(a1/(R*(273.2+T))) \n", + "Fn= P1*math.e**(Bn*P1/(R*(273.2+T)))\n", + "Fc= P1*math.e**(Bc*P1/(R*(273.2+T)))\n", + "Fn1= P*math.e**(Bn*P/(R*(273.2+T)))\n", + "Fc1= P*math.e**(Bc*P/(R*(273.2+T)))\n", + "#RESULTS\n", + "print '%s %.3f %.2f %s' % (' Fugacity of N2 at 1 and 50 atm are respectively =',Fn,Fn1,'atm')\n", + "print '%s %.3f %.2f %s' % (' \\n Fugacity of CO2 at 1 and 50 atm are respectively =',Fc,Fc1,'atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Fugacity of N2 at 1 and 50 atm are respectively = 0.999 48.34 atm\n", + " \n", + " Fugacity of CO2 at 1 and 50 atm are respectively = 0.996 40.37 atm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25 - pg 245" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Increase in pressure in all cases\n", + "#initialisation of variables\n", + "import math\n", + "P1= 23.756 #atm\n", + "T= 25. #C\n", + "P2= 1. #atm\n", + "P3= 10. #atm\n", + "P4= 100. #atm\n", + "R= 82.02 #J/mole K\n", + "v= 18.07 #ml\n", + "#CALCULATIONS\n", + "p1= P1/760.\n", + "p2= 10**(math.log10(P1)+(v*(P2-p1)/(2.303*R*(273.2+T))))\n", + "p3= 10**(math.log10(P1)+(v*(P3-p1)/(2.303*R*(273.2+T))))\n", + "p4= 10**(math.log10(P1)+(v*(P4-p1)/(2.303*R*(273.2+T))))\n", + "x= -(P1-p2)*100/P1\n", + "x1= -(P1-p3)*100/P1\n", + "x2= -(P1-p4)*100/P1\n", + "#RESULTS\n", + "print '%s %.2f %s' % ('Increase in pressure=',x,'percent')\n", + "print '%s %.2f %s' % ('Increase in pressure=',x1,' percent')\n", + "print '%s %.1f %s' % ('Increase in pressure=',x2,' percent')\n", + "print '%s %.3f %s' %('Vapor pressure at 1 atm',p2,'mm Hg')\n", + "print '%s %.3f %s' %('Vapor pressure at 10 atm',p3,'mm Hg')\n", + "print '%s %.3f %s' %('Vapor pressure at 100 atm',p4,'mm Hg')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Increase in pressure= 0.07 percent\n", + "Increase in pressure= 0.74 percent\n", + "Increase in pressure= 7.7 percent\n", + "Vapor pressure at 1 atm 23.773 mm Hg\n", + "Vapor pressure at 10 atm 23.932 mm Hg\n", + "Vapor pressure at 100 atm 25.577 mm Hg\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 26 - pg 247" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the pressure required\n", + "#initialisation of variables\n", + "H= 1436.3 #cal mole^-1\n", + "d= 0.9999 #g ml^-1\n", + "d1= 0.9168 #g ml^-1\n", + "P= 1. #atm\n", + "m= 18.02 #gm\n", + "R= 1.987 #cal/mole K\n", + "T= 2 #C\n", + "#CALCULATIONS\n", + "V= (P/d)-(P/d1)\n", + "H1= H*82.05/(m*R) \n", + "P1= H1*(-T)/(273*V)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' pressure required to decrease=',P1,'atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " pressure required to decrease= 266.0 atm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 27 - pg 249" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the heat of vapourisation\n", + "#initialisation of variables\n", + "H= 540. #cal gram ^-1\n", + "T= 95. #C\n", + "T1= 100. #C\n", + "m= 18. #gms\n", + "R= 1.987 #cal /mole K\n", + "P= 760. #mm of Hg\n", + "#CALCULATIONS\n", + "H1= m*H\n", + "P1= P/(10**(H1*(T1-T)/(2.303*R*(273+T)*(273+T1))))\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' heat of vapourisation=',P1,'mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " heat of vapourisation= 636.0 mm of Hg\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 28 - pg 249" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the vapor pressure of water\n", + "#initialisation of variables\n", + "H= 9720 #cal mole^-1\n", + "P= 1 #atm\n", + "R= 1.987 #cal /mole K\n", + "T= 100 #C\n", + "T1= 95 #C\n", + "#CALCULATIONS\n", + "r= P*H/(R*(273+T)**2)\n", + "dP= r*(T1-T)\n", + "P1= (P+dP)*626/0.824\n", + "#RESULTS\n", + "print '%s %d %s' % (' vapour pressure=',P1,'mm Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " vapour pressure= 626 mm Hg\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 29 - pg 250" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the vapour pressure\n", + "#initialisation of variables\n", + "G= 145 #cal\n", + "R= 1.987 #cal/mole K\n", + "T= 95 #C\n", + "#CALCULATIONS\n", + "P= 10**(-G/(2.303*R*(273+T)))*(624/0.820)\n", + "#RESULTS\n", + "print '%s %d %s' % (' vapour pressure=',P,'mm Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " vapour pressure= 624 mm Hg\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 30 - pg 250" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the molar heat of vapourisation\n", + "#initialisation of variables\n", + "import math\n", + "R= 1.987 #cal/mole K\n", + "T1= 25 #C\n", + "T2= 76.8 #C\n", + "P2= 760. #mm\n", + "P1= 115. #mm\n", + "#CALCULATIONS\n", + "H= 2.303*R*(273.2+T1)*(273.2+T2)*math.log10(P2/P1)/(T2-T1)\n", + "#RESULTS\n", + "print '%s %d %s' % (' molar heat of vapourisation=',H,'cal mole^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " molar heat of vapourisation= 7561 cal mole^-1\n" + ] + } + ], + "prompt_number": 17 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_5.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_5.ipynb new file mode 100755 index 00000000..98aa3041 --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_5.ipynb @@ -0,0 +1,767 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:18c54e3b106846f46428edc2ce784211e8ed1cb16969a115044ec2bc914626ae" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 - The phase rule and solutions" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the molality of the mixture\n", + "#initialisation of variables\n", + "m= 98.08 #gms\n", + "d= 1.102 #g ml^-1\n", + "m1= 165.3 #gm\n", + "v= 1000 #ml\n", + "wt=.15\n", + "#CALCULATIONS\n", + "form=d*v*wt/m\n", + "M= d*v-m1\n", + "norm=2*form\n", + "m2= m1*v/(m*M)\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' molality = ',m2,'molal')\n", + "print '%s %.3f %s' %('Formality = ',form,'gm formula wt/l')\n", + "print '%s %.3f %s' %('Normality = ',norm,'N')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " molality = 1.799 molal\n", + "Formality = 1.685 gm formula wt/l\n", + "Normality = 3.371 N\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Increase in enthalpy\n", + "#initialisation of variables\n", + "T= -40 #C\n", + "v= 217.4 #cm^3\n", + "r= 8.8 # atm deg^-1\n", + "m= 18 #gms\n", + "#CALCULATIONS\n", + "H= (273+T)*(-v*m/1000)*r*(1.987/82.05)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Increase in enthalpy =',H,'cal mole^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Increase in enthalpy = -194.3 cal mole^-1\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 279" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the density\n", + "#initialisation of variables\n", + "T= 27 #C\n", + "R= 0.08206 #cal/mol T\n", + "W= 28.6 #gms\n", + "#CALCULATIONS\n", + "d= W/((273.2+T)*R)\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' density =',d,' g l^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " density = 1.161 g l^-1\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - pg 289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the mole fraction and total pressure\n", + "#initialisation of variables\n", + "P= 408. #mm of Hg\n", + "P1= 141. # mm of Hg\n", + "p= 60.\n", + "#CALCULATIONS\n", + "P2= P*(100-p)/100.\n", + "P3= P1*p/100.\n", + "N= P2/(P2+P3)\n", + "P4= P2+P3\n", + "#RESULTS\n", + "print '%s %.3f' % (' mole fraction = ',N)\n", + "print '%s %.1f %s' % (' \\n total pressure =',P4,' mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " mole fraction = 0.659\n", + " \n", + " total pressure = 247.8 mm of Hg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - pg 289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the molality\n", + "#initialisation of variables\n", + "P2= 760. #mm of Hg\n", + "m2= 2.18*10**-3\n", + "v= 23.5 #ml\n", + "p= 21.\n", + "p1= 79.\n", + "#CALCULATIONS\n", + "K= P2*55.5/m2\n", + "K1= 760*55.5*22.4*10**3/v\n", + "m= 55.5*(p*760/(100*K))+55.5*(p1*760/(100*K1))\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' molality =',m,'molal')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " molality = 1.29e-03 molal\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - pg 297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the percentage of Br in the vapors in steam distillation\n", + "#initialisation of variables\n", + "Ph= 643. #mm of Hg\n", + "Mh= 18. #gms\n", + "Po= 117. #mm of Hg\n", + "Mo= 157. #gms\n", + "#CALCULATIONS\n", + "r= Ph*Mh/(Po*Mo)\n", + "P= 100*(1/(1+r))\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' percentage =',P,'percent')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " percentage = 61.3 percent\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 306" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the amounts of phases present at 375 and 370 C\n", + "#initialisation of variables\n", + "n= 1 \n", + "n1= 0.5\n", + "n3= 0.36\n", + "n4= 0.67\n", + "n5= 0.34\n", + "r= 3\n", + "#CALCULATIONS\n", + "A= (n-n1)/(n1-n3)\n", + "A1= r*(n4-n1)/(n1-n5)\n", + "#RESULTS\n", + "print '%s %.1f' % (' amount of phase at 375 C = ',A)\n", + "print '%s %.1f' % (' \\n amount of phase at 370 C = ',A1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " amount of phase at 375 C = 3.6\n", + " \n", + " amount of phase at 370 C = 3.2\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the vapour pressure\n", + "#initialisation of variables\n", + "m= 100 #gms\n", + "m1= 1 #gms\n", + "m2= 2 #gms\n", + "P= 23.756 #mm of Hg\n", + "n= 18.02 \n", + "n1= 60.06\n", + "n2= 342.3 \n", + "#CALCULATIONS\n", + "r= ((m1/n1)+(m2/n2))/((m1/n1)+(m2/n2)+(m/n))\n", + "dp= P*r\n", + "P1= P-dp\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' vapour pressure =',P1,' mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " vapour pressure = 23.660 mm of Hg\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - pg 315" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the boiling point of solution\n", + "#initialisation of variables\n", + "kf= 0.514 #K/molal\n", + "m= 0.225 #molal\n", + "#CALCULATIONS\n", + "dT= kf*m\n", + "T2=dT+100.\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' boiling point =',T2,' C')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " boiling point = 100.116 C\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12 - pg 315" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the molecular weight of the solute\n", + "#initialisation of variables\n", + "kb= 2.64 #C gm\n", + "dT= 0.083 #C\n", + "m= 120 #gms\n", + "W2= 0.764 #gms\n", + "#CALCULATIONS\n", + "m2= dT/kb\n", + "M2= W2*1000/(m2*m)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' molecular weight of solute =',M2,'gms')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " molecular weight of solute = 202.5 gms\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - pg 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the value of n\n", + "#initialisation of variables\n", + "T= 176.5 #C\n", + "T1= 158.8 #C\n", + "Kf= 37.7\n", + "W1= 0.522 #gms\n", + "W2= 0.0386 #gms\n", + "m= 12 #gms\n", + "m1= 1 #gm\n", + "#CALCULATIONS\n", + "m3= (T-T1)/Kf\n", + "M2= W2*1000/(m3*W1)\n", + "r= M2/(m+m1)\n", + "#RESULTS\n", + "print '%s %d' % ('value of n = ',r)\n", + "print '%s %d %s' %('Molecular weight = ',M2,'gm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "value of n = 12\n", + "Molecular weight = 157 gm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14 - pg 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the triple point of the system\n", + "#initialisation of variables\n", + "T= 273.2 #K\n", + "P= 0.0060 #atm\n", + "P1= 1 #atm\n", + "H= 3290 #cal\n", + "dV= -0.0907 #cc\n", + "#CALCULATIONS\n", + "dT= T*dV*(P-P1)/H\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' triple point =',dT,'C') \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " triple point = 0.0075 C\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16 - pg 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the fraction of impurity in both cases\n", + "#initialisation of variables\n", + "n= 100.\n", + "K= 2.\n", + "V= 100. #ml\n", + "V2= 1000. #ml\n", + "n= 10.\n", + "n1= 100.\n", + "#CALCULATIONS\n", + "x= (K*V/(K*V+(V2/n)))**n\n", + "y= (K*V/(K*V+(V2/n1)))**n1\n", + "#RESULTS\n", + "print '%s %.4f' % (' fraction of impurity = ',x)\n", + "print '%s %.4f' % (' \\n fraction of impurity = ',y)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " fraction of impurity = 0.0173\n", + " \n", + " fraction of impurity = 0.0076\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17 - pg 328" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the molecular weight of the protein\n", + "#initialisation of variables\n", + "T= 27 #C\n", + "m= 0.635 #gms\n", + "V= 100 #ml\n", + "R= 0.08205 #cal/mol K\n", + "p= 2.35 #cm\n", + "#CALCULATIONS\n", + "M= 13.6*76*m*R*(T+273)*1000/(p*V)\n", + "#RESULTS\n", + "print '%s %d %s' % (' molecular weight =',M,'gms')\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " molecular weight = 68747 gms\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18 - pg 328" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the osmotic pressure\n", + "#initialisation of variables\n", + "import math\n", + "R= 0.08205 #cal/mol K\n", + "v1= 0.0180#cc\n", + "N= 0.9820\n", + "T= 273.2\n", + "#CALCULATIONS\n", + "P= -R*T*math.log(N)/v1\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' osmotic pressure =',P,'atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " osmotic pressure = 22.6 atm\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19 - pg 331" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the osmotic pressure\n", + "#initialisation of variables\n", + "kf= 1.86\n", + "dT= 0.402 #K\n", + "T= 310 #K\n", + "R= 0.08205 #cal/mol K\n", + "#CALCULATIONS\n", + "P= dT*T*R/kf\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' osmotic pressure =',P,'atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " osmotic pressure = 5.50 atm\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20 - pg 333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Degrees of ionisation\n", + "#initialisation of variables\n", + "m= 0.100 #gms\n", + "kf= 1.86 #K/gms\n", + "dT= 0.300 #K\n", + "v= 2\n", + "#CALCULATIONS\n", + "T= kf*m\n", + "i= dT/T\n", + "a= (i-1)/(v-1)\n", + "#RESULTS\n", + "print '%s %.2f' % (' Degrees of ionisation = ',a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Degrees of ionisation = 0.61\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21 - pg 335" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the lowering of the freezing point\n", + "#initialisation of variables\n", + "W= 0.0020 #M\n", + "W1= 0.0010 #M\n", + "W2= 0.0040 #M\n", + "T= 1.86 #C\n", + "n= 1 #moles\n", + "n1= 1 #moles\n", + "n2= 2 #moles\n", + "a= 1.122\n", + "#CALCULATIONS\n", + "dT= T*(W+W1+W2)\n", + "I= 0.5*(n**2*W+n1**2*W2+n2**2*W1)\n", + "g= 1-(2*a*I**1.5/(3*(W+W1+W2)))\n", + "dT1= g*dT\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' lowering the freezing point =',dT1,'C ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " lowering the freezing point = 0.0125 C \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22 - pg 338" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the activity coefficient of acetone and water\n", + "#initialisation of variables\n", + "p= 1820 #mm\n", + "n= 2.5 #mole percent\n", + "f= 0.470\n", + "P= 420 #mm\n", + "n1= 97.5 #percent\n", + "#CALCULATIONS\n", + "P1= p*n/(100*760)\n", + "F= f/P1\n", + "F1= (1-f)*760.*100/(P*n1)\n", + "#RESULTS\n", + "print '%s %.2f' % (' activity coefficient of acetone = ',F)\n", + "print '%s %.2f' % (' \\n activity coefficient of water = ',F1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " activity coefficient of acetone = 7.85\n", + " \n", + " activity coefficient of water = 0.98\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_6.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_6.ipynb new file mode 100755 index 00000000..88034fe1 --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_6.ipynb @@ -0,0 +1,712 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e8a56bec98c8dfc26716196236590a8e6e7a9f919a9f30f91ea12d37d82b5d68" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 - Chemical Equilibrium" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the value of Kc\n", + "#initialisation of variables\n", + "d= 3.880 #g l^-1\n", + "M= 208.3 #gm\n", + "P= 1 #atm\n", + "R= 0.08205 #cal/mol K\n", + "T= 473.1 #K\n", + "#CALCULATIONS\n", + "d1= M*P/(R*T)\n", + "d2= (d1-d)/d\n", + "Kp= d2**2/(1-d2**2)\n", + "Kc= Kp/(R*T)\n", + "#RESULTS\n", + "print '%s %.3e %s' %(' Kc =',Kc,'moles l^-1')\n", + "print '%s %.4f %s' %(' Kp =',Kp,'atm')\n", + "print '%s %.4f' %('Fraction dissociated = ',d2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Kc = 4.429e-03 moles l^-1\n", + " Kp = 0.1719 atm\n", + "Fraction dissociated = 0.3830\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the percentage of Pcl5 dissociated\n", + "#initialisation of variables\n", + "import math\n", + "P= 10 #atm\n", + "Kp= 0.1719\n", + "#CALCULATIONS\n", + "a= math.sqrt(Kp/(10+Kp))*100\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' percentage =',a,'percent')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " percentage = 13.000 percent\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the value of Kp\n", + "#initialisation of variables\n", + "P= 0.3429 #atm\n", + "p0= 0.3153 #atm\n", + "#CALCULATIONS\n", + "Kp= (2*(P-p0))**2/(2*p0-P)\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' Kp =',Kp,'atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Kp = 1.06e-02 atm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - pg 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the pressure required\n", + "#initialisation of variables\n", + "Kp= 1.06*10**-2 #atm\n", + "a= 0.990\n", + "#CALCULATIONS\n", + "P= Kp*(1-a**2)/(4*a**2)\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' pressure =',P,' atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " pressure = 5.38e-05 atm\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - pg 357" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the pressure required\n", + "#initialisation of variables\n", + "G= 2054.7 #cal\n", + "R= 1.9872 #cal/mol K\n", + "T= 298.16 #K\n", + "#CALCULATIONS\n", + "P= 10**(-G/(2.303*T*R))\n", + "#RESULTS\n", + "print '%s %.5f %s' % (' pressure =',P,'atm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " pressure = 0.03120 atm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - pg 359" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the solubility product constant\n", + "#initialisation of variables\n", + "T= 25 #C\n", + "H= 25.31 #cal\n", + "H1= -40.02 #cal\n", + "H2= -30.36 #cal\n", + "S1= 17.67 #cal deg^-1\n", + "S2= 13.17 #cal deg^-1\n", + "S3= -22.97 #cal deg^-1\n", + "R= 1.987 #cal/mol K\n", + "#CALCULATIONS\n", + "H3= (H+H1-H2)*1000\n", + "S4= S1+S2+S3\n", + "G= H3-(273.2+T)*S4\n", + "Ka= 10**(-G/(2.303*R*(273.2+T)))\n", + "#RESULTS\n", + "print '%s %.1e' %(' solubility product constant = ',Ka)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " solubility product constant = 1.8e-10\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 359" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the increase in free energy\n", + "#initialisation of variables\n", + "import math\n", + "T= 25 #C\n", + "H= -36430 #cal\n", + "S= -4.19 #cal deg^-1\n", + "a= 0.1\n", + "f= 0.2\n", + "R= 1.987 #cal/mol K\n", + "#CALCULATIONS\n", + "G= H-(273.2+T)*S\n", + "Q= a*f/a**2\n", + "G1= G+R*(273.2+T)*math.log(Q)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' increase in free energy =',G1, 'cal')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " increase in free energy = -34769.8 cal\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 361" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the free energy of formation\n", + "#initialisation of variables\n", + "H= 21600. #cal\n", + "S= 50.339 #cal\n", + "S1= 49.003 #cal\n", + "S2= 45.767 #cal\n", + "T= 298.2 #K\n", + "#CALCULATIONS\n", + "H1= 2*H\n", + "S1= 2*S-S1-S2\n", + "G= H1-T*S1\n", + "Gj= G/(2*1000)\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' free energy of formation =',Gj,'kcal')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " free energy of formation = 20.719 kcal\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - pg 361" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the range of humidity\n", + "#initialisation of variables\n", + "R= 1.987 #cal/mol K\n", + "T= 25. #C\n", + "G1= -193.8 #cal\n", + "G2= -54.6 #cal\n", + "G3= -253.1 #cal\n", + "G4= -253.1 #cal\n", + "G5= -54.6 #cal\n", + "G6= -309.7 #cal\n", + "#CALCULATIONS\n", + "G= G1+G2-G3\n", + "Ph= 10**(-G*10**3/(2.303*R*(273.2+T)))\n", + "G0= G4+G5-G6\n", + "Ph1= 10**(-G0*10**3/(2.303*R*(273.2+T)))\n", + "p= Ph*100./Ph1\n", + "#RESULTS\n", + "print '%s %.2f %s' %(' range of humidity =',p,'percent')\n", + "print 'The answers are a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " range of humidity = 1.05 percent\n", + "The answers are a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - pg 362" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Ksp\n", + "#initialisation of variables\n", + "m= 10**-2\n", + "m1= 10**-22\n", + "G= -22.15 #kcal\n", + "G1= -5.81 #kcal\n", + "G2= 20.6 #kcal\n", + "T= 25 #C\n", + "R= 1.987 #cal/mol K\n", + "#CALCULATIONS\n", + "G3= G-(G1+G2)\n", + "Ksp= 10**(G3*10**3/(2.303*R*(273+T)))\n", + "#RESULTS\n", + "print '%s %.0e' %(' Ksp = ',Ksp)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Ksp = 8e-28\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12 - pg 366" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the heat of dissociation and standard free energy of iodine\n", + "%matplotlib inline\n", + "import numpy\n", + "import math\n", + "import matplotlib\n", + "from matplotlib import pyplot\n", + "import warnings\n", + "#initialization of variables\n", + "R=1.987\n", + "T=1000 #K\n", + "T=numpy.array([973.,1073.,1173.,1274.])\n", + "kp=numpy.array([.175e-2,1.108e-2,4.87e-2,17.05e-2])\n", + "#calculations\n", + "Tx=1000./T\n", + "logkp=numpy.log10(kp)\n", + "slope, intercept = numpy.polyfit(Tx,logkp,1)\n", + "dH=-slope*2.303*R*1000.\n", + "dH0=dH-1.5*R*T\n", + "dG1=-R*T*logkp[1]*2.303\n", + "dGt=28720 #cal\n", + "dGI=(dGt/1000. + 4.63)/2\n", + "#results\n", + "print '%s %d %s' %('Heat of dissociation = ',dH,'cal')\n", + "print '%s %.2f %s' %('standard free energy of iodine atom = ',dGI,'kcal/mol')\n", + "pyplot.plot(Tx,logkp)\n", + "pyplot.xlabel('1000/T')\n", + "pyplot.ylabel('log Kp')\n", + "pyplot.title('Logarithm of Kp for dissociation of Iodine as a function of reciprocal temperature')\n", + "pyplot.show()\n", + "warnings.filterwarnings(\"ignore\")\n", + "print 'The answers are a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat of dissociation = 37460 cal\n", + "standard free energy of iodine atom = 16.68 kcal/mol\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The answers are a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - pg 361" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Partial pressure\n", + "#initialisation of variables\n", + "import math\n", + "T= 2000 #K\n", + "P= 1 #atm\n", + "G= 41438 #cal\n", + "R= 1.987 #cal/mol K\n", + "T2= 298.2 #K\n", + "T1= 2000 #K\n", + "H= 43200 #cal\n", + "#CALCULATIONS\n", + "Kp= 10**(-G/(2.303*R*T2))\n", + "Kp1= Kp*10**(H*(T-T2)/(2.303*R*T1*T2))\n", + "p= math.sqrt(Kp1*0.8*0.2)\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' Partial pressure of NO =',p,'atm ')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Partial pressure of NO = 7.7e-03 atm \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15 - pg 368" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Temperature required\n", + "#initialisation of variables\n", + "G0 = 0 #cal\n", + "G= 13200. #cal\n", + "T1= 298.2\n", + "H1= 23100. #cal\n", + "#CALCULATIONS\n", + "T= H1/((H1/T1)-(G/T1)+(G0/T1))\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Temperature =',T,' K ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Temperature = 695.8 K \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16 - pg 373" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Equilibrium constant\n", + "#initialisation of variables\n", + "T= 2000 #K\n", + "R= 1.987 #cal /mol K\n", + "G= 31160 #cal\n", + "#CALULATIONS\n", + "Kp= 10**(-G/(2.303*R*T))\n", + "#RESULTS\n", + "print '%s %.2e' % ('Equilibrium constant =',Kp )\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equilibrium constant = 3.94e-04\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17 - pg 373" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the fraction of methane decomposed\n", + "#initialisation of variables\n", + "p= 0.08 #atm\n", + "#CALCULATIONS\n", + "a= (1-p)/(p+1)\n", + "#RESULTS\n", + "print '%s %.2f' % ('fraction = ',a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "fraction = 0.85\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18 - pg 374" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the enthalpy of the reaction\n", + "#initialisation of variables\n", + "H= -57240. #cal\n", + "T= 2257. #C\n", + "Hh= -54.60 #cal\n", + "Ho= -38.56 #cal\n", + "HO= -57.08 #cal\n", + "#CALCULATIONS\n", + "H1= H-T*(2*Hh-2*Ho-HO)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Enthalpy =',H1,'cal')\n", + "print 'The answers in the textbook are a different due to a rounding off error '" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy = -113665.0 cal\n", + "The answers in the textbook are a different due to a rounding off error \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19 - pg 375" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Enthalpy\n", + "#initialisation of variables\n", + "H= -57797 #cal\n", + "T= 25 #C\n", + "Hh= 7.934 #cal\n", + "Ho= -6.788 #cal\n", + "HO= 6.912 #cal\n", + "#CALCULATIONS\n", + "H1= 2*H-(T+273.16)*(2*Hh+2*Ho-HO)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Enthalpy =',H1,'cal ')\n", + "print 'The answers in the textbook are a different due to a rounding off error '" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Enthalpy = -114216.5 cal \n", + "The answers in the textbook are a different due to a rounding off error \n" + ] + } + ], + "prompt_number": 17 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_7.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_7.ipynb new file mode 100755 index 00000000..d0564a44 --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_7.ipynb @@ -0,0 +1,1029 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3529f6dd0800b2bdaab8573a3a6af2c519dc83ab93935a987777c3348bc812ea" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 - Electrochemistry" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 391" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Avagadro number\n", + "#initialisation of variables\n", + "e= 1.6016*10**-19 #coloumb\n", + "F= 96493 #\n", + "#CALCULATIONS\n", + "N= F/e\n", + "#RESULTS\n", + "print '%s %.4e %s' % (' Avagadro number = ',N,'molecules/mol')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Avagadro number = 6.0248e+23 molecules/mol\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 391" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Time required\n", + "#initialisation of variables\n", + "m= 1 #gms\n", + "M= 63.54 #gms\n", + "e= 2 #farady\n", + "F= 96493\n", + "n= 3\n", + "#CALCULATIONS\n", + "t= (m/M)*(e*F/n)\n", + "#RESULTS\n", + "print '%s %d %s' % (' Time =',t,'sec')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Time = 1012 sec\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 396" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the transference number\n", + "#initialisation of variables\n", + "M= 25.01 #gms\n", + "n= 1.0053 #moles\n", + "n1= 6.6*10**-5 #moles\n", + "e= 1.350*10**-3 #coloumbs\n", + "#CALCULATIONS\n", + "x= M/n\n", + "y= n1*x\n", + "nm= y*10**3+e*10**3-(x/10)\n", + "t= nm/(e*10**3)\n", + "#CALCULATIONS\n", + "print '%s %.3f' % (' transference number = ',t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " transference number = 0.373\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - pg 404" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the electrokinetic potential\n", + "#initialisation of variables\n", + "import math\n", + "x= 0.033 #cm\n", + "t= 38.2 #sec\n", + "e= 3.2 #v\n", + "V= 9*10**-3 #dyne sec cm**-2\n", + "k= 78\n", + "#CALCULATIONS\n", + "v= x/t\n", + "u= v/e\n", + "S= -300**2*u*V*4*math.pi/k\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' electrokinetic potential =',S,' volt ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " electrokinetic potential = -0.035 volt \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6 - pg 406" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the specific conductivity\n", + "#initialisation of variables\n", + "o= 0.999505 #mho cm^-1\n", + "k= 0.0128560\n", + "i= 97.36 #ohms\n", + "I= 117.18 #ohms\n", + "#CALCULATIONS\n", + "Lsp= k*o\n", + "L1sp= k*i/I\n", + "#RESULTS\n", + "print '%s %.6f %s' % (' specific conductivity =',L1sp,'mho cm^-1 ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " specific conductivity = 0.010682 mho cm^-1 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - pg 410" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the equivalent conductance of the anion at infinite dilution\n", + "#initialisation of variables\n", + "A= 388.5\n", + "l= 349.8\n", + "a= 0.61\n", + "m= 0.1 #M\n", + "#CALCULATIONS\n", + "L= A-l\n", + "A1= a*A\n", + "Lsp= m*A1/1000.\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' equivalent conductance of the anion at infinite dilution =',Lsp,' mho cm^-1 ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " equivalent conductance of the anion at infinite dilution = 2.37e-02 mho cm^-1 \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 410" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the effective mobility\n", + "#initialisation of variables\n", + "l= 349.82 \n", + "F= 96493.1 #coloumb\n", + "#CALCULATIONS\n", + "u= l/F\n", + "#RESULTS\n", + "print '%s %.3e %s' % (' effective mobility =',u,'cm^2 volt sec^-1 ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " effective mobility = 3.625e-03 cm^2 volt sec^-1 \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 413" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the solubility product constant\n", + "#initialisation of variables\n", + "G1= -7800 #cal\n", + "G2= -24600 #cal\n", + "G3= -39700 #cal\n", + "R= 1.987 #cal/mol K\n", + "T= 25 #C\n", + "#CALCULATIONS\n", + "G= G1+G2-G3\n", + "Ksp= 10**(-G/(2.303*R*(273.2+T)))\n", + "#RESULTS\n", + "print '%s %.1e' % (' solubility product constant = ',Ksp)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " solubility product constant = 4.5e-06\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - pg 417" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the concentration of hydrogen ion\n", + "#initialisation of variables\n", + "import math\n", + "Ka= 6*10**-10\n", + "C= 10**-1 #moles l^-1\n", + "#CALCULATIONS\n", + "C1= math.sqrt(Ka*C)\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' concentration of hydrogen ion =',C1,'moles l^-1 ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " concentration of hydrogen ion = 7.7e-06 moles l^-1 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - pg 419" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the concentration of hydrogen ion\n", + "#initialisation of variables\n", + "Ka= 1.8*10**-5 \n", + "n= 2 #milli moles\n", + "v= 45 #ml\n", + "n1= 0.5#milli moles\n", + "#CALCULATIONS\n", + "x= Ka*v*n1/n\n", + "C= x/v\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' concentration of hydrogen ion =',C,' moles l^-1 ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " concentration of hydrogen ion = 4.5e-06 moles l^-1 \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14 - pg 421" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the pH of the solution and activity coefficient\n", + "#initialisation of variables\n", + "import math\n", + "a= 2.4*10**-4\n", + "Ph= 11.54\n", + "#CALCULATIONS\n", + "Ph1= -math.log10(a)\n", + "a= 10**(-Ph)\n", + "#RESULTS\n", + "print '%s %.2f' % (' pH of solution = ',Ph1)\n", + "print '%s %.1e' % (' \\n activity coefficient = ',a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " pH of solution = 3.62\n", + " \n", + " activity coefficient = 2.9e-12\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15 - pg 426" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Gibbs free energy\n", + "#initialisation of variables\n", + "E= 0.35240 #volts\n", + "F= 96493.1 #coloumb\n", + "n= 2 #electrons\n", + "#CALCULATIONS\n", + "G= -n*F*E\n", + "#RESULTS\n", + "print '%s %d %s' % (' Gibbs free energy =',G,' absolute joules ')\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gibbs free energy = -68008 absolute joules \n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16 - pg 428" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Entropy and Enthalpy of the mixture\n", + "#initialisation of variables\n", + "E= 0.35240 #volts\n", + "E1= 0.35321 #volts\n", + "E2= 0.35140 #volts\n", + "E3=.35252\n", + "T= 25. #C\n", + "T1= 20. #C\n", + "T2= 30. #C\n", + "n= 2. #electrons\n", + "F= 96493.1 #coloumb\n", + "#CALCULATIONS\n", + "r= (E-E1)/(T-T1)\n", + "r1= (E2-E)/(T2-T)\n", + "R= (r+r1)/2\n", + "S= n*F*R\n", + "H= n*F*((273.16+T)*R-E3)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Entropy =',S,'joules deg^-1')\n", + "print '%s %.1f %s' % (' \\n Enthalpy =',H,'joules')\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Entropy = -34.9 joules deg^-1\n", + " \n", + " Enthalpy = -78446.4 joules\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18 - pg 431" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Gibbs free energy\n", + "#initialisation of variables\n", + "import math\n", + "v= 0.11834 #volt\n", + "F= 96493.1 #coloumb\n", + "n= 1 #electron\n", + "R= 8.3144 #J/mol K\n", + "T= 25 #C\n", + "m= 0.1\n", + "m1= 0.9862\n", + "#CALCULATIONS\n", + "G= -n*F*v\n", + "G1= 2*R*(273.16+T)*math.log(m/m1)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Gibbs free energy =',G,'joules')\n", + "print '%s %d %s' % (' \\n Gibbs free energy =',G1,'joules')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gibbs free energy = -11419.0 joules\n", + " \n", + " Gibbs free energy = -11347 joules\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19 - pg 432" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the potential difference\n", + "#initialisation of variables\n", + "import math\n", + "n= 2 #electrons\n", + "R= 8.314 #bJ/mol K\n", + "F= 96493 #coloumb\n", + "T= 25 #C\n", + "N2= 3.17*10**-6\n", + "N1= 6.13*10**-3\n", + "#CALCULATIONS\n", + "E= -(R*(273.16+T)*2.3026/(n*F))*math.log10(N2/N1)\n", + "#RESULTS\n", + "print '%s %.5f %s' % (' potential difference =',E,' volt')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " potential difference = 0.09720 volt\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20 - pg 432" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Equilibrium constant\n", + "#initialisation of variables\n", + "import math\n", + "E= 0.84 #volts\n", + "n= 1 #electron\n", + "F= 96500 #coloumb\n", + "R= 8.314 #J/mol K\n", + "T= 25 #C\n", + "#CALCULATIONS\n", + "K= math.e**(E*n*F/(R*(273+T)))\n", + "#RESULTS\n", + "print '%s %.1e' % (' Equilibrium constant =',K)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Equilibrium constant = 1.6e+14\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21 - pg 432" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Equilibrium constant\n", + "#initialisation of variables\n", + "import math\n", + "E= -0.0029 #volts\n", + "V= 0.1 #volts\n", + "V1= 0.05 #volts\n", + "f= 0.05916 #J/mol coloumb\n", + "T= 25. #C\n", + "F= 96500 #coloumb\n", + "R= 8.314 #J/mol K\n", + "#CALCULATIONS \n", + "e= E+f*math.log10(V*V1/V1)\n", + "K= math.e**(e*F/(R*(273+T)))\n", + "#RESULTS\n", + "print '%s %.1e' % (' Equilibrium constant =',K)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Equilibrium constant = 8.9e-02\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22 - pg 438" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Standard electrode potential\n", + "#initialisation of variables\n", + "import math\n", + "E= 1.0508 #volts\n", + "V= 0.3338 #volts\n", + "a= 0.0796 \n", + "a1= math.sqrt(0.0490)\n", + "f= 0.05916 #J/mol coloumb\n", + "#CALCULATIONS\n", + "V= E+V+f*math.log10(a/a1)\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' Standard electrode potential =',V,'volts')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Standard electrode potential = 1.3583 volts\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 23 - pg 438" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Standard molar free energy\n", + "#initialisation of variables\n", + "V= 1.3595 #volts\n", + "n= 1 #electron\n", + "F= 96493 #coloumb\n", + "#CALCULATIONS\n", + "G= -n*F*V/4.28\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Standard molar free energy =',G,'cal')\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Standard molar free energy = -30650.1 cal\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24 - pg 439" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the ion product\n", + "#Initialization of variables\n", + "import math\n", + "I=0.0050\n", + "E0=.22619\n", + "con=.0602\n", + "E2=1.05080\n", + "R=8.3144\n", + "T=298.16 #K\n", + "#calculations\n", + "E1=E0-con*math.sqrt(I)\n", + "E3=-E2+E1\n", + "Kw=10**(E3*96493/2.3026/R/T)\n", + "#results\n", + "print '%s %.3e' %(\"Ion product = \",Kw)\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ion product = 9.741e-15\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 25 - pg 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Solubility constant\n", + "#initialisation of variables\n", + "V= -0.658 #volt\n", + "V1= -0.3363 #volt\n", + "n= 1 #electron\n", + "F= 96438 #coloumb\n", + "R= 8.314 #j/mol K\n", + "T= 25 #C\n", + "#CLACULATIONS\n", + "V2= V-V1\n", + "Ksp= 10**(V2*n*F/(2.303*R*(273.2+T)))\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' Solubility constant =',Ksp,' volt')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Solubility constant = 3.7e-06 volt\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 26 - pg 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the cell potential\n", + "#initialisation of variables\n", + "import math\n", + "e= 0\n", + "e1= -0.37\n", + "k= -0.05916 #j/mol\n", + "a= 0.02\n", + "a1= 0.01\n", + "a3=.2\n", + "p= 730. #mm of Hg\n", + "#CALCULATIONS\n", + "E= (e-e1)+k*math.log10(a*math.sqrt(p/760.) /a1/a3)\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' cell potential =',E,'volt') \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " cell potential = 0.31 volt\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 27 - pg 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the cell potential\n", + "#initialisation of variables\n", + "V= -0.440 #volt\n", + "V1= 0.771 #volt\n", + "F= 96500 #coloumb\n", + "n=2 #electrons\n", + "n1= 1 #electrons\n", + "n2= 3 #electrons\n", + "#CALCULATIONS\n", + "G= -n*F*V\n", + "G1= -n1*F*V1\n", + "G2= G+G1\n", + "V= G2/(n2*F)\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' cell potential =',-V,'volt') \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " cell potential = -0.0363 volt\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 28 - pg 444" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the cell potential\n", + "#initialisation of variables\n", + "import math\n", + "p1=386.6 #atm\n", + "p2=1 #atm\n", + "f= 2\n", + "k= -0.05916 #j/mol\n", + "#CALCULATIONS\n", + "E= (k/f)*math.log10(p1/p2)\n", + "#RESULTS\n", + "print '%s %.4f %s' % (' cell potential =',E,'volt')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " cell potential = -0.0765 volt\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 29 - pg 445" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the cell potential\n", + "#initialisation of variables\n", + "import math\n", + "c= 10**-7\n", + "c1= 1\n", + "f= 1\n", + "k= -0.05915 #j/mol\n", + "#CALCULATIONS\n", + "E= (k/f)*math.log10(c/c1)\n", + "#RESULTS\n", + "print '%s %.5f %s' % (' cell potential =',E,' volt')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " cell potential = 0.41405 volt\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 30 - pg 448" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the junction potential\n", + "#initialisation of variables\n", + "import math\n", + "c= 391.\n", + "c1= 129.\n", + "f= 1.\n", + "k= -0.05915 #j/mol\n", + "#CALCULATIONS\n", + "E= (k/f)*math.log10(c1/c)\n", + "#RESULS\n", + "print '%s %.4f %s' % (' junction potential =',E,'volt')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " junction potential = 0.0285 volt\n" + ] + } + ], + "prompt_number": 26 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_8.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_8.ipynb new file mode 100755 index 00000000..c1054064 --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_8.ipynb @@ -0,0 +1,440 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e7ff5ec8c26bca61ce95f7f9a6bfe9182508039293520fdb892a47c60afd9608" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 - Quantum chemistry" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 460" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Wavelength\n", + "#initialisation of variables\n", + "v= 299.8 #V\n", + "e= 4.802*10**-10 #ev\n", + "h= 6.624*10**-27 #ergs sec\n", + "c= 3*10**10 #cm/sec\n", + "#CALCULATIONS\n", + "E= e/v\n", + "l= h*c*10**8/(2*E)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Wavelength =',l,'A')\n", + "print 'The answers are a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Wavelength = 6203.3 A\n", + "The answers are a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 462" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the value of numerical coefficient\n", + "#initialisation of variables\n", + "u= 109677.583 #cm**-1\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' value of numerical coefficient =',u,' cm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " value of numerical coefficient = 109677.6 cm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 464" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the wavelength in both cases\n", + "#initialisation of variables\n", + "import math\n", + "h= 6.6234*10**-27 #ergs sec\n", + "m= 2.59 #gms\n", + "v= 3.35*10**4 #cm sec **-1\n", + "e= 4.8*10**-10 #ev\n", + "V= 40000. #volts\n", + "M= 300. #gms\n", + "L= 1836. #A\n", + "N= 6*10**23 #molecules\n", + "#CALCULATIONS\n", + "p= m*v\n", + "l= h/p\n", + "E= V*e/M\n", + "P= math.sqrt(2*E*(1/(L*N)))\n", + "L1= h*10**8/P\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' wavelength =',l,'cm')\n", + "print '%s %.4f %s' % (' \\n wavelength =',L1,'A')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " wavelength = 7.63e-32 cm\n", + " \n", + " wavelength = 0.0614 A\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - pg 471" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the lifetime of this excited state\n", + "#initialisation of variables\n", + "import math\n", + "h= 6.624*10**-27 #ergs sec\n", + "c= 3*10**10 #cm/sec\n", + "u= 5 #cm**-1\n", + "#CALCULATIONS \n", + "T= h/(h*2*math.pi*c*u)\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' lifetime of this excited state =',T,'sec')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " lifetime of this excited state = 1.1e-12 sec\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - pg 471" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the lifetime\n", + "#initialisation of variables\n", + "import math\n", + "V= 2.5*10**4 #m/sec\n", + "m= 30 #gms\n", + "s= 10*10**-16 #cm**2\n", + "N= 6.023*10**23 #molecules\n", + "T= 300 #K\n", + "k= 8.3*10**7\n", + "#CALCULATIONS\n", + "t= math.sqrt((m/(math.pi*k*T)))*(V/(4*s*N))\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' lifetime =',t,' sec')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " lifetime = 2.0e-10 sec\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - pg 494" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the internuclear distances\n", + "#initialisation of variables\n", + "import math\n", + "h= 6.6238*10**-27 #ergssec\n", + "N= 6.0254*10**23 #molecules\n", + "c= 2.9979*10**10\n", + "Be= 60.809\n", + "mh= 1.00812 #gms\n", + "#CALCULATIONS\n", + "u= mh/2.\n", + "Re= math.sqrt(h*N/(c*8*math.pi**2*Be*u))\n", + "#RESULTS\n", + "print '%s %.4e %s' % (' internuclear distances =',Re,'cm ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " internuclear distances = 7.4168e-09 cm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 497" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Resonance energy\n", + "#initialisation of variables\n", + "H= 19.8 #kcal\n", + "H1= -0.8 #kcal\n", + "H2= -29.4 #kcal\n", + "#CALCULATIONS\n", + "H3= -85.8\n", + "H4= -49.2\n", + "H5= -H3+H4\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Resonance energy =',H5,'cal')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Resonance energy = 36.6 cal\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 500" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the no of bonds\n", + "#initialisation of variables\n", + "import math\n", + "R= 1.69 #A\n", + "l= 1.49 #A\n", + "r= 0.706\n", + "#CALCULATIONS\n", + "n= 10**((R-l)/r)\n", + "#RESULTS\n", + "print '%s %.2f' % (' no of bonds = ',n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " no of bonds = 1.92\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - pg 504" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the lattice energy\n", + "#initialisation of variables\n", + "N= 6.*10**23 #molecules\n", + "R= 2.82 #A\n", + "e= 4.8*10**-10 #ev\n", + "n= 9.\n", + "z= 1.748\n", + "#CALCULATIONS\n", + "U= (N*z*e**2*(1-(1/n)))*182.2/(R*10**-8*7.63*10**12)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' lattice energy =',U,'kcal mole**-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " lattice energy = 181.9 kcal mole**-1\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - pg 507" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the least energy required for transfer\n", + "#initialisation of variables\n", + "import math\n", + "k= 13\n", + "e= 4.8*10**-10 #ev\n", + "h= 6.624*10**-27 #ergs sec\n", + "N= 6.023*10**23 #molecules\n", + "l= 1836 #A\n", + "#CALCULATIONS\n", + "I= e**4*0.080/(l*N*1.28*10**-13*2*k**2*(h/(2*math.pi))**2)\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' least energy required for transfer=',I,' ev')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " least energy required for transfer= 0.08 ev\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12 - pg 509" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the difference between potentials\n", + "#initialisation of variables\n", + "i= 54.4 #ev\n", + "i1= 24.6 #ev\n", + "k= 2.5 \n", + "#CALCULATIONS\n", + "I= i/(4*k**2)\n", + "I1= i1/(4*k**2)\n", + "d= I-I1\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' difference between first and second potential=',d,'ev')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " difference between first and second potential= 1.2 ev\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/Chapter_9.ipynb b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_9.ipynb new file mode 100755 index 00000000..1a10b96e --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/Chapter_9.ipynb @@ -0,0 +1,101 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:39eab768daff11aebdd87a93356bef21d5d2b1bdfeb10c1efcd7ba6d3163e0fd" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9 - Statistical Mechanics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 525" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Absolute Entropy\n", + "#initialisation of variables\n", + "import math\n", + "T= 298.16 #K\n", + "M= 4.003 #gm\n", + "S= 2.3151 #cal mol^-1 deg^-1\n", + "R= 1.987 #cal/molK\n", + "#CALCULATIONS\n", + "S1= 2.5*R*math.log(T)+1.5*R*math.log(M)-S\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Absolute Entropy=',S1,'cal mol^-1 deg^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Absolute Entropy= 30.122 cal mol^-1 deg^-1\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 528" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the heat required\n", + "#initialisation of variables\n", + "h= 6.624*10**-27#erg/sec\n", + "N= 6.023*10**23\n", + "c= 3*10**10 #m/sec\n", + "w= 2359.6 #cm**-1\n", + "T= 2000 #K\n", + "K= 1.380*10**-16\n", + "R= 1.987 #cal mol**-1 k**-1\n", + "#CALCULATIONS\n", + "x= h*c*w/(K*T)\n", + "y= 2.71**x\n", + "H= 3.5*R+(N*h*c*w/(T*4.184*10**7*(y-1)))\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Heat=',H,'cal mol**-1 deg**-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Heat= 7.715 cal mol**-1 deg**-1\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/README.txt b/Physical_Chemsitry_by_William_F_Sheehan/README.txt new file mode 100755 index 00000000..c00237f7 --- /dev/null +++ b/Physical_Chemsitry_by_William_F_Sheehan/README.txt @@ -0,0 +1,10 @@ +Contributed By: Jaya Pratyusha Kothuri +Course: btech +College/Institute/Organization: Sri Mittapalli College of Engineering +Department/Designation: Computer Science and Engineering +Book Title: Physical Chemsitry +Author: William F Sheehan +Publisher: Allyn And Bacon, U. S. A. +Year of publication: 1961 +Isbn: 0070153019 +Edition: 1 \ No newline at end of file diff --git a/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap3.png b/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap3.png new file mode 100755 index 00000000..832091f2 Binary files /dev/null and b/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap3.png differ diff --git a/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap4.png b/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap4.png new file mode 100755 index 00000000..e2646978 Binary files /dev/null and b/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap4.png differ diff --git a/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap6.png b/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap6.png new file mode 100755 index 00000000..1b82351b Binary files /dev/null and b/Physical_Chemsitry_by_William_F_Sheehan/screenshots/chap6.png differ -- cgit