diff options
Diffstat (limited to 'Physical_Chemsitry')
34 files changed, 0 insertions, 10756 deletions
diff --git a/Physical_Chemsitry/Chapter1.ipynb b/Physical_Chemsitry/Chapter1.ipynb deleted file mode 100755 index f8eddd0e..00000000 --- a/Physical_Chemsitry/Chapter1.ipynb +++ /dev/null @@ -1,208 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:2d2c58ce752ea004ea6f79edf5332f8f357b96a3e81270455c238eb5e0794fa8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1 - matter and its atomic nature"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 3"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Angle required\n",
- "#Initialization of variables\n",
- "import math\n",
- "l=0.71 *10**-8 #cm\n",
- "n=200. #lines/cm\n",
- "v=0.00145 #radian\n",
- "#calculations\n",
- "d=1/n\n",
- "phi2=2*l/d +v**2\n",
- "phi=math.sqrt(phi2)\n",
- "#results\n",
- "print '%s %.2e %s' %('Angle required =',phi,'radian')\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle required = 2.22e-03 radian\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 6"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Interplanar distance\n",
- "#Initialization of variables\n",
- "import math\n",
- "angle=37.25 #degrees\n",
- "l=1.539 #A\n",
- "n=1. #order\n",
- "#calculations\n",
- "d=n*l/(2*math.sin(angle/180.*math.pi))\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Interplanar distance =\",d,\"A\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Interplanar distance = 1.271 A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the ratio of radii\n",
- "#Initialization of variables\n",
- "import math\n",
- "r1=math.sqrt(3.)\n",
- "r2=1\n",
- "#calculations\n",
- "ratio=r1-r2\n",
- "#results\n",
- "print '%s %.3f' %('Ratio of radii =',ratio)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ratio of radii = 0.732\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 21"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Avagadro number\n",
- "#Initialization of variables\n",
- "d=2.64 #g/cc\n",
- "l=4.016*10**-8 #cm\n",
- "n=4\n",
- "M=25.94 #g/mol\n",
- "#calculations\n",
- "m=d*l**3 /n\n",
- "N0=M/m\n",
- "#results\n",
- "print '%s %.3e %s' %(\"Avagadro number =\",N0,\" molecule/mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Avagadro number = 6.068e+23 molecule/mol\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - pg 28"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the angle required\n",
- "#Initialization of variables\n",
- "import math\n",
- "import numpy\n",
- "A=numpy.array([-1, -1, -1 ])\n",
- "B=numpy.array([1, 1, -1])\n",
- "#calculations\n",
- "Ad=math.sqrt(1+1+1)\n",
- "Bd=math.sqrt(1+1+1)\n",
- "dot=numpy.dot(A,B) /(Ad*Bd) \n",
- "theta=math.acos(dot) *180./math.pi\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Angle =\",theta,\" degrees\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Angle = 109.47 degrees\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter10.ipynb b/Physical_Chemsitry/Chapter10.ipynb deleted file mode 100755 index 8ecfc863..00000000 --- a/Physical_Chemsitry/Chapter10.ipynb +++ /dev/null @@ -1,202 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:5f5d6cc341d2ab999dbe1e5a4234f374f3f6728cc09cef7967ab9dd07f94519e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10 - Second law of thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 249"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Change ine entropy\n",
- "#Initialization of variables\n",
- "import math\n",
- "T2=100+273.2 #K\n",
- "T1=50+273.2 #K\n",
- "n=1 #mol\n",
- "R=1.987 #cal/deg mol\n",
- "#calculations\n",
- "dS=5/2 *n*R*2.303*math.log10(T2/T1)\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Change in entropy =\",dS,\" eu\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in entropy = 0.572 eu\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Entropy change\n",
- "#Initialization of variables\n",
- "H=380 #cal\n",
- "T=273.2+32.1 #K\n",
- "#calculations\n",
- "dS=H/T\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Entropy change =\",dS,\" eu\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy change = 1.24 eu\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 257"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Work done\n",
- "#Initialization of variables\n",
- "Ha=0\n",
- "Hb=0\n",
- "#calculations\n",
- "H=Ha+Hb\n",
- "q=H\n",
- "U=0\n",
- "w=q-H\n",
- "#results\n",
- "print '%s %d %s' %(\"Work done =\",w,\"J\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Work done = 0 J\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 260"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in entropy\n",
- "#Initialization of variables\n",
- "import math\n",
- "prob=0.001\n",
- "R=1.\n",
- "N=6.023*10**23\n",
- "#calculations\n",
- "dS=1.987*2.303*math.log10(prob) /N\n",
- "#results\n",
- "print '%s %.1e %s' %(\"change in entropy =\",dS,\" eu\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "change in entropy = -2.3e-23 eu\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 263"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Change in A \n",
- "#Initialization of variables\n",
- "T=373.2 #K\n",
- "c=1.987 #cal/deg\n",
- "#calculations\n",
- "w=c*T\n",
- "A=-w\n",
- "#results\n",
- "print '%s %d %s' %(\"Change in A =\",A,\"cal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in A = -741 cal\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter11.ipynb b/Physical_Chemsitry/Chapter11.ipynb deleted file mode 100755 index b98e97b5..00000000 --- a/Physical_Chemsitry/Chapter11.ipynb +++ /dev/null @@ -1,95 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:00f3ac32d98b00732d6a51b0562de7dab713dab2a42986a603b1c1eafc01eefd"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11 - Condensed phases"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 271"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Vapor Pressure\n",
- "#Initialization of variables\n",
- "A=7.6546\n",
- "B=1686.8\n",
- "T=60+273.2\n",
- "#calculations\n",
- "logP=A-B/T\n",
- "P=10**logP\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Vapor Pressure =\",P,\"mm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Vapor Pressure = 391.0 mm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 277"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in C values\n",
- "#Initialization of variables\n",
- "alpha=4.92*10**-5 #deg**-1\n",
- "beta=7.85*10**-7 #atm**-1\n",
- "d=8.93 #g/cm**3\n",
- "T=298.15 #K\n",
- "#calculations\n",
- "dC=63.54*T*alpha**2 *1.987/(d*beta*82.06)\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Change in c values = \",dC,\" cal/deg mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in c values = 0.158 cal/deg mol\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter12.ipynb b/Physical_Chemsitry/Chapter12.ipynb deleted file mode 100755 index 92f1a058..00000000 --- a/Physical_Chemsitry/Chapter12.ipynb +++ /dev/null @@ -1,348 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:65a756a76ade1ec4b9387e0db33894444f541be4a142adaa7f77f2e89d036580"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 12 - Physical Equilibria"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 295"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the no of degrees of freedom\n",
- "#Initialization of variables\n",
- "p=3\n",
- "c=2\n",
- "#calculations\n",
- "f=2-p+c\n",
- "#results\n",
- "print '%s %d' %(\"no. of degrees of freedom =\",f)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "no. of degrees of freedom = 1\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 301"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in temperature\n",
- "#Initialization of variables\n",
- "T=273.2 #K\n",
- "vw=1.0001 #cm^3 /g\n",
- "vi=1.0907 #cm^3 /g\n",
- "hf=79.7 #cal/g\n",
- "P1=76 #cm\n",
- "P2=4.6 #cm\n",
- "#calculations\n",
- "dT=T*(vw-vi)*(P2-P1)*13.6*980.7/(hf*4.184*10**7)\n",
- "#results\n",
- "print '%s %.4f %s' %(\"change in temperature =\",dT,\"deg\")\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "change in temperature = 0.0071 deg\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 302"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of dPbydT\n",
- "#Initialization of variables\n",
- "V=6.84 #cm^3 /g\n",
- "#calculations\n",
- "dPbydT=-1.7*4.184*10**7 /(2.19*V*0.06*1.01*10**6)\n",
- "#results\n",
- "print '%s %d %s' %(\"dPbydT =\",dPbydT,\" atm/deg\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dPbydT = -78 atm/deg\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 303"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Pressure\n",
- "#Initialization of variables\n",
- "P=6 #atm\n",
- "T=273.2+25 #K\n",
- "P=23.8 #mm\n",
- "V=0.018 #lt/mol\n",
- "R=0.08206 #lt am/deg mol\n",
- "#calculations\n",
- "dPa=V*P*4536/(R*T*760)\n",
- "Pa=dPa+P\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Pressure =\",Pa,\" mm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Pressure = 23.9 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the mole fraction of methanol in vapor\n",
- "#Initialization of variables\n",
- "x=0.25\n",
- "Ps1=96 #mm\n",
- "Ps2=43.9 #mm\n",
- "#calculations\n",
- "P1=x*Ps1\n",
- "P2=(1-x)*Ps2\n",
- "P=P1+P2\n",
- "Xdash=P1/P\n",
- "#results\n",
- "print '%s %.3f' %(\"mole fraction of methanol in vapor =\",Xdash)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "mole fraction of methanol in vapor = 0.422\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Molal elevation constant\n",
- "#Initialization of variables\n",
- "Hv=539.6 #cal/g\n",
- "T=273.2+100 #K\n",
- "#calculations\n",
- "Kb=1.987*T**2 /(1000*Hv)\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Molal elevation constant =\",Kb,\" deg /mole /kg\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molal elevation constant = 0.513 deg /mole /kg\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Molecular weight of solute\n",
- "#Initialization of variables\n",
- "ms=0.5 #mol/kg\n",
- "m=5 #g\n",
- "mw=100 #g\n",
- "Ws=1000 #g/kg\n",
- "#calculations\n",
- "Ma=m*Ws/(ms*mw)\n",
- "#results\n",
- "print '%s %d %s' %(\"Molecular weight of solute =\",Ma,\"g/mol \")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molecular weight of solute = 100 g/mol \n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - pg 311"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the molality of the solution\n",
- "#Initialization of variables\n",
- "dT=0.23 #C\n",
- "Kb=1.86 #deg/mol/kg\n",
- "#calculations\n",
- "m=dT/Kb\n",
- "#results\n",
- "print '%s %.2f %s' %(\"molality of solution =\",m,\"m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "molality of solution = 0.12 m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - pg 313"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Osmotic Pressure\n",
- "#Initialization of variables\n",
- "p=0.1 #m\n",
- "T=30+273.2 #K\n",
- "R=0.08206 #lt atm /deg/mol\n",
- "P1=1 #atm\n",
- "#calculations\n",
- "w=1000/p\n",
- "V=w/1000.\n",
- "dP=R*T/V\n",
- "P=dP+P1\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Osmotic Pressure =\",P,\" atm \")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Osmotic Pressure = 3.49 atm \n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter13.ipynb b/Physical_Chemsitry/Chapter13.ipynb deleted file mode 100755 index 79973f5f..00000000 --- a/Physical_Chemsitry/Chapter13.ipynb +++ /dev/null @@ -1,401 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:b3b8e3a1df1ec221a5596f2fd615d0e7df4ba35a4012aeb734d5820e6983cd64"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 13 - Thermodynamic changes accompanying chemical reaction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 320"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the dHp value\n",
- "#Initialization of variables\n",
- "n1=10 #mol\n",
- "n2=12 #mol\n",
- "#calculations\n",
- "dn=n1-n2\n",
- "#results\n",
- "print '%s %d %s' %(\"dHp = dEv-\",dn,\"*RT\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "dHp = dEv- -2 *RT\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 322"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Enthalpy\n",
- "#Initialization of variables\n",
- "Ht1=-22063 #cal\n",
- "T=298.15 #K\n",
- "#calculations\n",
- "H=Ht1 +0.5293*T + 0.3398*10**-3 *T**2 - 2.039*10**-7 *T**3\n",
- "#results\n",
- "print '%s %d %s' %(\"Enthalpy =\",H,\"cal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Enthalpy = -21880 cal\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 326"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Entropy\n",
- "#Initialization of variables\n",
- "Cp=0.797 #cal/deg/mol\n",
- "#calculations\n",
- "S=Cp/3.\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Entropy =\",S,\" eu/mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Entropy = 0.266 eu/mol\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 328"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Change in entropy\n",
- "#Initialization of variables\n",
- "T1=77.32 #K\n",
- "P=1 #atm\n",
- "T2=126 #K\n",
- "Pc=33.5 #atm\n",
- "#calculations\n",
- "dS=27/32. *1.987*P/Pc *(T2/T1)**3\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Change in entropy =\",dS,\"eu/mol\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in entropy = 0.22 eu/mol\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 330"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Change in entropy, enthalpy and free energy\n",
- "#Initialization of variables\n",
- "S1=57.47\n",
- "S2=50.34\n",
- "S3=49\n",
- "H1=8.09\n",
- "H2=21.06\n",
- "H3=0\n",
- "F1=12.39\n",
- "F2=20.72\n",
- "F3=0\n",
- "#calculations\n",
- "dS=S1-S2-0.5*S3\n",
- "dH=H1-H2-0.5*H3\n",
- "dF=F1-F2-0.5*F3\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Change in entropy =\",dS,\" eu\")\n",
- "print '%s %.2f %s' %(\"\\n Change in enthalpy =\",dH,\" kcal\")\n",
- "print '%s %.2f %s' %(\"\\n Change in free energy =\",dF,\"kcal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Change in entropy = -17.37 eu\n",
- "\n",
- " Change in enthalpy = -12.97 kcal\n",
- "\n",
- " Change in free energy = -8.33 kcal\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 334"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the change in free energy\n",
- "#Initialization of variables\n",
- "import math\n",
- "P1=0.01\n",
- "P2=0.1\n",
- "P3=0.01\n",
- "dF0=-54640 #cal\n",
- "T=298.15 #K\n",
- "R=1.987 #cal/deg\n",
- "#calculations\n",
- "Qp=P1/(P2*P3**0.5)\n",
- "dF=dF0+R*T*math.log(Qp)\n",
- "#results\n",
- "print '%s %d %s' %(\"change in free energy =\",dF,\"cal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "change in free energy = -54640 cal\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 335"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Equilibrium constant\n",
- "#Initialization of variables\n",
- "print \"From table 13.4 \"\n",
- "logKfwater=40.04724\n",
- "logKfH2=0\n",
- "logKfO2=0\n",
- "#calculations\n",
- "logK=logKfwater-logKfH2-0.5*logKfO2\n",
- "K=10**logK\n",
- "#results\n",
- "print '%s %.4e' %(\"Equilibrium constant = \",K)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 13.4 \n",
- "Equilibrium constant = 1.1149e+40\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - pg 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of Kc\n",
- "#Initialization of variables\n",
- "Kp=1.1*10**40 #atm**-0.5\n",
- "dn=-0.5\n",
- "R=0.08206 #lt atm/deg mol\n",
- "T=298.15 #K\n",
- "#calculations\n",
- "Kc=Kp*(R*T)**(-dn)\n",
- "#results\n",
- "print '%s %.1e %s' %(\"Kc =\",Kc,\" (mol/lt)^-0.5\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kc = 5.4e+40 (mol/lt)^-0.5\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - pg 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Density of equilibrium mixture\n",
- "#Initialization of variables\n",
- "import numpy\n",
- "Kp=0.141 #atm\n",
- "P=1 #atm\n",
- "nu=2\n",
- "R=0.08206 #lt atm/deg mol\n",
- "T=298.15 #K\n",
- "M=92.02 #g/mol\n",
- "#calculations\n",
- "p=([Kp+ 4*P,0, -Kp])\n",
- "z=numpy.roots(p)\n",
- "alpha=z[0]\n",
- "wbyV=P*M/(R*T*(1+(nu-1)*alpha))\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Density of the equilibrium mixture =\",wbyV,\" g/lt\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Density of the equilibrium mixture = 3.18 g/lt\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - pg 340"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Final pressure after equilibrium\n",
- "#Initialization of variables\n",
- "x=0.5\n",
- "P=0.468 #atm\n",
- "#calculations\n",
- "P1=x*P\n",
- "P2=x*P\n",
- "Kp=P1*P2\n",
- "#results\n",
- "print '%s %.4f %s' %(\"Final pressure after equilibrium =\",Kp,\" atm^2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Final pressure after equilibrium = 0.0548 atm^2\n"
- ]
- }
- ],
- "prompt_number": 12
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter14.ipynb b/Physical_Chemsitry/Chapter14.ipynb deleted file mode 100755 index b1948488..00000000 --- a/Physical_Chemsitry/Chapter14.ipynb +++ /dev/null @@ -1,603 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f51aa40f01063be99829dfd06c53ed90029beb6946cceb4b92469b9c50e91012"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14 - Development and use of activity concepts"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 350"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the moles of Iodine present\n",
- "#Initialization of variables\n",
- "x1=0.0200\n",
- "Kx=812.\n",
- "#calculations\n",
- "print \"Neglecting 2x in comparision with x1,\"\n",
- "x=x1/Kx\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Moles of Iodine present =\",x,\" mole\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Neglecting 2x in comparision with x1,\n",
- "Moles of Iodine present = 2.46e-05 mole\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 350 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Concentration of H+ ions\n",
- "#Initialization of variables\n",
- "Kc=1.749*10**-5 #M\n",
- "n1=0.1 #mole\n",
- "n2=0.01 #mole\n",
- "#calculations\n",
- "c=n1/n2 *Kc\n",
- "#results\n",
- "print '%s %.1e %s' %(\"Concentration of Hplus ions =\",c,\" M\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Concentration of Hplus ions = 1.7e-04 M\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 351"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Concentraton of Hplus ions\n",
- "#Initialization of variables\n",
- "import math\n",
- "c=0.01 #M\n",
- "kc=1.749*10**-5 #M\n",
- "#calculations\n",
- "x2=c*kc\n",
- "x=math.sqrt(x2)\n",
- "#results\n",
- "print '%s %.1e %s' %(\"Concentration of Hplus ions =\",x,\"M\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Concentration of Hplus ions = 4.2e-04 M\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 351"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Concentration of OH- ions\n",
- "#Initialization of variables\n",
- "import math\n",
- "K2=1.0008*10**-14 #m^2\n",
- "K1=1.754*10**-5 #m\n",
- "c=0.1\n",
- "#calculations\n",
- "print \"Neglecting x w.r.t c,\"\n",
- "x2=c*K2/K1\n",
- "x=math.sqrt(x2)\n",
- "#results\n",
- "print '%s %.1e %s' %(\"Concentration of OH minus ions =\",x,\" m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Neglecting x w.r.t c,\n",
- "Concentration of OH minus ions = 7.6e-06 m\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Concentration of H plus ions\n",
- "#Initialization of variables\n",
- "import math\n",
- "print \"from table 14.1,\"\n",
- "r1=7.47*10**-5 #m\n",
- "r2=4.57*10**-3 #m\n",
- "mp=1.008*10**-14 #m**2\n",
- "#calculations\n",
- "r3=r2/r1\n",
- "mH2=r3*mp\n",
- "mH=math.sqrt(mH2)\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Concentration of Hplus ions = \",mH,\" M\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from table 14.1,\n",
- "Concentration of Hplus ions = 7.85e-07 M\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 354"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Concentraton of H+ ions\n",
- "#Initialization of variables\n",
- "print \"from table 14.1,\"\n",
- "import math\n",
- "r1=1.75*10**-5 #m\n",
- "r2=1.772*10**-4 #m\n",
- "mp=1.008*10**-14 #m**2\n",
- "#calculations\n",
- "r3=r2/r1\n",
- "mH2=r3*mp\n",
- "mH=math.sqrt(mH2)\n",
- "#results\n",
- "print '%s %.1e %s' %(\"Concentration of Hplus ions =\",mH,\" M\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "from table 14.1,\n",
- "Concentration of Hplus ions = 3.2e-07 M\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Concentration of H+ ions\n",
- "#Initialization of variables\n",
- "import math\n",
- "c=1*10**-6 #m\n",
- "K=1.754*10**-5 #m\n",
- "Kp=1.008*10**-14 #m**2\n",
- "#calculations\n",
- "mH=c\n",
- "#Iteration 1\n",
- "mOH=Kp/mH\n",
- "mA=mH-mOH\n",
- "mHA=mH*mA/K\n",
- "mH2=mH-mHA+mOH\n",
- "#Iteration 2\n",
- "mOH2=Kp/mH2\n",
- "mA2=mH2-mOH2\n",
- "mHA2=mH2*mA2/K\n",
- "mH3=mH2-mHA2+mOH2\n",
- "#From x2\n",
- "x2=math.sqrt(Kp)\n",
- "x1=c\n",
- "mOH3=Kp/x2\n",
- "y2=x1\n",
- "#From x1\n",
- "mOH4=Kp/c\n",
- "mA4=mH-mOH4\n",
- "mHA4=mH*mA4/K\n",
- "y1=c-mHA4-mA4\n",
- "#upon further iterations, we get\n",
- "mHplus=mH3\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Concentration of H plus ions =\",mHplus,\"m\")\n",
- "print 'The answer is a bit different due to rounding off error.'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Concentration of H plus ions = 9.13e-07 m\n",
- "The answer is a bit different due to rounding off error.\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - pg 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the values of dS0, dH0, Krm\n",
- "#Initialization of variableH\n",
- "print \"From table 14-3,\"\n",
- "HH=0\n",
- "HHcoo=-98.\n",
- "HHcooh=-98.\n",
- "SH=0\n",
- "SHcoo=21.9\n",
- "SHcooh=39.1\n",
- "KH=0\n",
- "KHcoo=58.64\n",
- "KHcooh=62.38\n",
- "#calculationH\n",
- "dH=HH+HHcoo-HHcooh\n",
- "dS=SH+SHcoo-SHcooh\n",
- "dK=KH+KHcoo-KHcooh\n",
- "K=10**dK\n",
- "#results\n",
- "print '%s %.1f %s' %(\" dS0 =\",dS,\"eu\")\n",
- "print '%s %.1f %s' %(\"\\n dH0 =\",dH,\"kcal\")\n",
- "print '%s %.2f' %(\"\\n log Krm =\",dK)\n",
- "print '%s %.1e %s' %(\"\\n Krm =\",K,\"m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 14-3,\n",
- " dS0 = -17.2 eu\n",
- "\n",
- " dH0 = 0.0 kcal\n",
- "\n",
- " log Krm = -3.74\n",
- "\n",
- " Krm = 1.8e-04 m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - pg 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Activity of cl and ca\n",
- "#Initialization of variables\n",
- "mca=0.01 #m\n",
- "mcl=0.02 #m\n",
- "#calculations\n",
- "Mu=0.5*(mca*4 + mcl*1)\n",
- "print \"From table 14-5,\"\n",
- "aca=6 #A\n",
- "acl=3 #A\n",
- "print \"From table 14-6,\"\n",
- "gaca=0.555 \n",
- "gacl=0.843\n",
- "Aca=gaca*mca\n",
- "Acl=gacl*mcl\n",
- "#results\n",
- "print '%s %.4f' %(\"Activity of cl = \",Acl)\n",
- "print '%s %.4f' %(\"\\n Activity of ca = \",Aca)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 14-5,\n",
- "From table 14-6,\n",
- "Activity of cl = 0.0169\n",
- "\n",
- " Activity of ca = 0.0056\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - pg 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Concentration of H+ ions\n",
- "#Initialization of variables\n",
- "import math\n",
- "m1=0.1 #m\n",
- "m2=0.1 #m\n",
- "K=1.754*10**-5 #m\n",
- "#calculations\n",
- "mu=0.5*(m1*1**2 + m2*1**2)\n",
- "print(\"From table 14.5,\")\n",
- "aH=9 #A\n",
- "aA=4.5 #A\n",
- "print(\"From table 14.6\")\n",
- "gH=0.825\n",
- "gA=0.775\n",
- "gHA=1\n",
- "x1=gHA*K/(gH*gA)\n",
- "print(\"Assuming x to be small w.r.t m1,\")\n",
- "x=math.sqrt(x1*m1)\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Concentration of H plus ions =\",x,\" m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 14.5,\n",
- "From table 14.6\n",
- "Assuming x to be small w.r.t m1,\n",
- "Concentration of H plus ions = 1.66e-03 m\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 11 - pg 372"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the concentration of H+ ions\n",
- "#Initialization of variables\n",
- "import math\n",
- "import numpy\n",
- "K=1.754*10**-5 #m\n",
- "c=0.1\n",
- "#calculations\n",
- "print(\"Neglecting x w.r.t c,\")\n",
- "x2=K\n",
- "x=math.sqrt(K)\n",
- "mu=x\n",
- "print(\"From tables 14-5 and 14-6,\")\n",
- "gH=0.963\n",
- "gA=0.960\n",
- "x22=K/(gH*gA)\n",
- "p=([1,x22, -c*x22])\n",
- "z=numpy.roots(p)\n",
- "alpha=z[1]\n",
- "#results\n",
- "print '%s %.2e %s' %(\"concentration of H plus ions =\",alpha,\" m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Neglecting x w.r.t c,\n",
- "From tables 14-5 and 14-6,\n",
- "concentration of H plus ions = 1.37e-03 m\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 12 - pg 373"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Solubility of Agcl\n",
- "#Initialization of variables\n",
- "print(\"From table 14.3\")\n",
- "import math\n",
- "K1=-13.5089\n",
- "K2=-22.9792\n",
- "K3=19.2218\n",
- "c=0.1 #m\n",
- "#calculations\n",
- "logK=K1-K2-K3\n",
- "K=10**logK\n",
- "mu=0.5*(c*1**2 + c*1**2)\n",
- "print(\"From tables 14-5 and 14-6,\")\n",
- "gAg=0.745\n",
- "gCl=0.755\n",
- "x2=K/(gAg*gCl)\n",
- "x=math.sqrt(x2)\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Solubility of Agcl =\",x,\"m\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From table 14.3\n",
- "From tables 14-5 and 14-6,\n",
- "Solubility of Agcl = 1.78e-05 m\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 13 - pg 376"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Concentration of Na and Cl in both cases\n",
- "#Initialization of variables\n",
- "import numpy\n",
- "Cna=0.11\n",
- "Ccl=0.1\n",
- "#calculations\n",
- "p=([99, - 2.1, Cna*Ccl])\n",
- "z=numpy.roots(p)\n",
- "alpha=z[1]\n",
- "Na1=Cna-10*alpha\n",
- "Cl1=Ccl-10*alpha\n",
- "#results\n",
- "print '%s %.4f %s' %(\" Concentration of Na in 1 =\",Na1,\"M\")\n",
- "print '%s %.4f %s' %(\"\\n Concentration of Cl in 1 =\",Cl1,\" M\")\n",
- "print '%s %.4f %s' %(\"\\n Concentration of Na in 2 =\",alpha,\"M\")\n",
- "print '%s %.4f %s' %(\"\\n Concentration of Cl in 2 =\",alpha,\"M\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Concentration of Na in 1 = 0.0157 M\n",
- "\n",
- " Concentration of Cl in 1 = 0.0057 M\n",
- "\n",
- " Concentration of Na in 2 = 0.0094 M\n",
- "\n",
- " Concentration of Cl in 2 = 0.0094 M\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter15.ipynb b/Physical_Chemsitry/Chapter15.ipynb deleted file mode 100755 index 8ce44ac5..00000000 --- a/Physical_Chemsitry/Chapter15.ipynb +++ /dev/null @@ -1,322 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:633d2351722f70bcb5591bec1c39746dd272ca892962c335e9bfb247bc80518a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15 - Electrochemistry"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 384"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Weight of copper leaving\n",
- "#Initialization of variables\n",
- "I=0.5 #amp\n",
- "t=55 #min\n",
- "we=31.77\n",
- "#calculations\n",
- "Q=I*t*60\n",
- "n=Q/96496.\n",
- "w=n*we\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Weight of copper leaving =\",w,\" g\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Weight of copper leaving = 0.543 g\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 386"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the values of tplus and tminus\n",
- "#Initialization of variables\n",
- "w1=0.7532 #g\n",
- "w2=0.9972 #g\n",
- "wdep=0.4 #g\n",
- "we=31.77 #g\n",
- "#calculations\n",
- "dn=w2/we - w1/we\n",
- "t=dn/(wdep/we)\n",
- "dne=wdep/we\n",
- "dnmig=dn-dne\n",
- "tplus=-dnmig/dne\n",
- "tminus=1-tplus\n",
- "#results\n",
- "print '%s %.3f' %(\"tplus =\",tplus)\n",
- "print '%s %.3f' %(\"\\n tminus=\",tminus)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "tplus = 0.390\n",
- "\n",
- " tminus= 0.610\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 393"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Equivalent conductance\n",
- "#Initialization of variables\n",
- "R1=312 #ohms\n",
- "R2=1043 #ohms\n",
- "c=0.01 #N\n",
- "kdash=0.002768 #ohm^-1cm^-1\n",
- "#calculations\n",
- "k=kdash*R1\n",
- "kdash2=k/R2\n",
- "ambda=kdash2/(c/1000.)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Equivalent conductance =\",ambda,\"ohm^-1 cm^2 equiv^-1\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equivalent conductance = 82.8 ohm^-1 cm^2 equiv^-1\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 393"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Conductance for acetic acid\n",
- "#Initialization of variables\n",
- "l1=349.8 \n",
- "l2=40.9\n",
- "#calculations\n",
- "l=l1+l2\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Conductance for acetic acid =\",l,\" ohm^-1 cm^2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Conductance for acetic acid = 390.7 ohm^-1 cm^2\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 395"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Specific conductanc\n",
- "#Initialization of variables\n",
- "l1=63.6\n",
- "l2=79.8\n",
- "n=1 #mg/lt\n",
- "we=116.7 #g/equiv\n",
- "#calculations\n",
- "l=l1+l2\n",
- "c=n*10**-3 /we\n",
- "k=c*l/1000.\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Specific conductance =\",k,\" ohm^-1 cm^-1\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Specific conductance = 1.23e-06 ohm^-1 cm^-1\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the SEP of the cell\n",
- "#Initialization of variables\n",
- "e1=0.763 #volt\n",
- "e2=0.337 #volt\n",
- "#calculations\n",
- "e0=e1+e2\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Standard electrode potential of the cell =\",e0,\"volts\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Standard electrode potential of the cell = 1.100 volts\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 403"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Emf of the cell\n",
- "#Initialization of variables\n",
- "import math\n",
- "aZn=0.1\n",
- "aCu=0.01\n",
- "e1=0.763 #volt\n",
- "e2=0.337 #volt\n",
- "#calculations\n",
- "e0=e1+e2\n",
- "Q=aZn/aCu\n",
- "E=e0- 0.05915*math.log10(Q) /2\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Emf of the cell =\",E,\" volts\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Emf of the cell = 1.070 volts\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - pg 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Decomposition potential\n",
- "#Initialization of variables\n",
- "e1=1.2 #volts\n",
- "e2=0.15 #volts\n",
- "e3=0.45 #volts\n",
- "#calculations\n",
- "E=e1+e2+e3\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Decomposition potential =\",E,\" volt\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Decomposition potential = 1.8 volt\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter16.ipynb b/Physical_Chemsitry/Chapter16.ipynb deleted file mode 100755 index edd91fdd..00000000 --- a/Physical_Chemsitry/Chapter16.ipynb +++ /dev/null @@ -1,129 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:732bbb076f615c901ca952cd233aa9a732520f864fa8a5b9d8200618c4707cce"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16 - Typical mechanisms and rate laws"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 422"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Rate constant\n",
- "#Initialization of variables\n",
- "import math\n",
- "P1=69.2 #mm\n",
- "P2=39.8#mm\n",
- "t=20 #min\n",
- "#calculations\n",
- "k=2.303*math.log10(P1/P2) /(t*60.)\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Rate constant =\",k,\"sec^-1\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rate constant = 4.61e-04 sec^-1\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 422"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Rate constant\n",
- "#Initialization of variables\n",
- "import math\n",
- "t=10. #min\n",
- "x=90.\n",
- "#calculations\n",
- "k=2.303*math.log10(100/(100-x)) /t\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Rate constant =\",k,\"min^-1\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rate constant = 0.230 min^-1\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 427"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the rate constant\n",
- "#Initialization of variables\n",
- "t=242. #sec\n",
- "P=229. #mm\n",
- "P0=363. #mm\n",
- "#calculations\n",
- "k=(1./P -1./P0)/t\n",
- "#results\n",
- "print '%s %.2e %s' %(\"rate constant=\",k,\"sec^-1 mm^-1\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "rate constant= 6.66e-06 sec^-1 mm^-1\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter17.ipynb b/Physical_Chemsitry/Chapter17.ipynb deleted file mode 100755 index 6e295125..00000000 --- a/Physical_Chemsitry/Chapter17.ipynb +++ /dev/null @@ -1,183 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:4d58417f5757dcfb290ed1ee0e2191d724d6d044069d2c21c72e96a05505c726"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17 - Resolving Kinetic data"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 446"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Order of the reaction\n",
- "#Initialization of variablesx1=5\n",
- "import math\n",
- "from math import log\n",
- "x2=20.\n",
- "x1=5.\n",
- "n1=7.49\n",
- "n2=5.14\n",
- "#calculations\n",
- "n=(log(n1)-log(n2))/(log(100-x1) - log(100-x2))\n",
- "#results\n",
- "print '%s %.2f' %(\"Order of the reaction = \",n)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Order of the reaction = 2.19\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 448"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Order of the reaction\n",
- "#Initialization of variablesx1=5\n",
- "import math\n",
- "from math import log\n",
- "p2=169.\n",
- "p1=363.\n",
- "t1=410.\n",
- "t2=880.\n",
- "#calculations\n",
- "ndash=(log(t2) - log(t1))/(log(p1) - log(p2))\n",
- "n=ndash+1\n",
- "#results\n",
- "print '%s %.2f' %(\"Order of the reaction = \",n)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Order of the reaction = 2.00\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 454"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Activation energy and Z\n",
- "#Initialization of variables\n",
- "import math\n",
- "R=1.987 #cal/deg/mol\n",
- "k1=4.45*10**-5\n",
- "k2=2.52*10**-6\n",
- "T1=283+273.2 #K\n",
- "T2=356+273.2 #K\n",
- "#calculations\n",
- "Ea=2.303*R*1.7530 /(1/T1 - 1/T2)\n",
- "logZ= math.log10(k1) +Ea/(2.303*R*T1)\n",
- "Z=10**logZ\n",
- "#results\n",
- "print '%s %d %s' %(\"Activation energy =\",Ea,\"cal/mol\")\n",
- "print '%s %.1e %s' %(\"\\n Z =\",Z,\"lt /mol sec\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Activation energy = 38456 cal/mol\n",
- "\n",
- " Z = 5.7e+10 lt /mol sec\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 456"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Equlibrium constant\n",
- "#Initialization of variables\n",
- "g1=0.661\n",
- "g2=0.899\n",
- "g3=0.405\n",
- "g4=0.803\n",
- "g5=0.946\n",
- "g6=0.614\n",
- "k=1.33\n",
- "#calculations\n",
- "k0=k*g3/(g1*g2)\n",
- "k2=k0*g4*g5/g6\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Equlibrium constant =\",k2,\"lt/mol min\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equlibrium constant = 1.12 lt/mol min\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter18.ipynb b/Physical_Chemsitry/Chapter18.ipynb deleted file mode 100755 index d460229d..00000000 --- a/Physical_Chemsitry/Chapter18.ipynb +++ /dev/null @@ -1,102 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:63ceed8607dc113635851a8fc16e00b5a679fa077f828a47388aa54622bce4ae"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18 - Catalysis"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 472"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Volume \n",
- "#Initialization of variables\n",
- "V1=0.284 #cm^3 /g\n",
- "V2=1.43 #cm^3 /g\n",
- "P1=142.4 #mm\n",
- "P2=760. #mm\n",
- "#calculations\n",
- "z=(1/V1 - 1/V2)/(1/P1 - 1/P2)\n",
- "invVm=1/V2 - z/P2\n",
- "Vm=1/invVm\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Volume =\",Vm,\"cm^3/g\")\n",
- "print 'The answer in the textbook is a bit different due to rounding off error.'\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Volume = 20.5 cm^3/g\n",
- "The answer in the textbook is a bit different due to rounding off error.\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 477"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the total area\n",
- "#Initialization of variables\n",
- "Vm=2.86 #cc/g\n",
- "P=1 #atm\n",
- "R=82.06 #cm^3 atm/deg mol\n",
- "T=273.2 #deg\n",
- "N=6.023*10**23\n",
- "sigma=16.2*10**-16 #cm^2 /molecule\n",
- "#calculations\n",
- "n=P*Vm/(R*T)\n",
- "A=N*n*sigma\n",
- "#results\n",
- "print '%s %.2e %s' %(\"total area =\",A,\" cm^2 (g catalyst)^-1\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "total area = 1.24e+05 cm^2 (g catalyst)^-1\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter19.ipynb b/Physical_Chemsitry/Chapter19.ipynb deleted file mode 100755 index 5b2d735e..00000000 --- a/Physical_Chemsitry/Chapter19.ipynb +++ /dev/null @@ -1,356 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f00e66b23ba11c1502ba60271f6e9b549183dbf77dfd61ae21a7394bb7c1f4ce"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19 - Photochemistry"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 488"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Overall transmittance\n",
- "#Initialization of variables\n",
- "r1=0.727\n",
- "r2=0.407\n",
- "#calculations\n",
- "r3=r1*r2\n",
- "#results\n",
- "print '%s %.3f' %(\"Overall transmittance = \",r3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Overall transmittance = 0.296\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 488"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Extinction coefficient\n",
- "import math\n",
- "#Initialization of variables\n",
- "r=0.450\n",
- "c=0.02 #M\n",
- "l=4 #cm\n",
- "#calculations\n",
- "e=-math.log10(r) /(c*l)\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Extinction coefficient =\",e,\"litres mole^-1 cm^-1\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Extinction coefficient = 4.33 litres mole^-1 cm^-1\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 488"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Transmittance of the solution\n",
- "#Initialization of variables\n",
- "import math\n",
- "from math import log10\n",
- "r1=0.850\n",
- "r2=0.50\n",
- "#calculations\n",
- "Da=-log10(r1)\n",
- "Db=-log10(r2)\n",
- "D=Da+Db\n",
- "r3=10**(-D)\n",
- "#results\n",
- "print '%s %.3f' %(\"Transmittance of solution =\",r3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Transmittance of solution = 0.425\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 491"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Extinction coefficient\n",
- "#Initialization of variables\n",
- "c=0.000025 #M\n",
- "l=2 #cm\n",
- "D=0.417\n",
- "#calculations\n",
- "e=D/(c*l)\n",
- "#result\n",
- "print '%s %d %s' %(\"Extinction coefficient =\",e,\" liters mole^-1 cm^-1\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Extinction coefficient = 8340 liters mole^-1 cm^-1\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 491"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Kc for dissociation\n",
- "#Initialization of variables\n",
- "c=0.5 #M\n",
- "c1=0.000025 #M\n",
- "D2=0.280\n",
- "D1=0.417\n",
- "#calculations\n",
- "c2=D2*c1/(D1)\n",
- "dC=c1-c2\n",
- "SCN=c- 6*c2 -4*dC\n",
- "K=dC*SCN**2 /c2\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Kc for dissociation =\",K,\" M^2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Kc for dissociation = 0.12 M^2\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 492"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of m\n",
- "#Initialization of variables\n",
- "import math\n",
- "from math import log\n",
- "D2=0.249\n",
- "D1=0.172\n",
- "a2=0.00752\n",
- "a1=0.00527\n",
- "#calculations\n",
- "m=(log(D2) -log(D1))/(log(a2) - log(a1))\n",
- "#results\n",
- "print '%s %.2f' %(\"m = \",m)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "m = 1.04\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 495"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Increase in optical density\n",
- "#Initialization of variables\n",
- "c=0.1 #M\n",
- "V=100 #ml\n",
- "v1=25 #ml\n",
- "D=0.980\n",
- "d1=0.090\n",
- "d2=0.150\n",
- "#calculations\n",
- "a=v1*c/V\n",
- "b=(V-v1)*c/V\n",
- "Da=a*d1/c\n",
- "Db=b*d2/c\n",
- "Ddash=Da+Db\n",
- "dD=D-Ddash\n",
- "#results\n",
- "print '%s %.3f' %(\"Increase in optical density =\",dD)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Increase in optical density = 0.845\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - pg 496"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the value of lambda for the reaction to occur\n",
- "#Initialization of variables\n",
- "E=50000. #cal/mol\n",
- "#calculations\n",
- "lam=2.8593/E\n",
- "#results\n",
- "print '%s %d %s' %(\"For the reaction to occur lambda <\",lam*10**8,\"A\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For the reaction to occur lambda < 5718 A\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - pg 497"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Amount of reactant disappeared\n",
- "#Initialization of variables\n",
- "lam=3000*10**-8 #cm\n",
- "ield=0.420\n",
- "Et=70000 #cal\n",
- "#calculations\n",
- "E=2.8593/lam\n",
- "n=ield*Et/E\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Amount of reactant disappeared =\",n,\" mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Amount of reactant disappeared = 0.308 mol\n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter2.ipynb b/Physical_Chemsitry/Chapter2.ipynb deleted file mode 100755 index b7de8073..00000000 --- a/Physical_Chemsitry/Chapter2.ipynb +++ /dev/null @@ -1,172 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:2335de0a1036d6583f9ddb709f27717d7dd705b5f69b2600b20c842c855e802b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2 - Particles Atomic and subatomic"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 40"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Increase in kinetic energy\n",
- "#Initialization of variables\n",
- "m1=1.008142\n",
- "m2=1.008982\n",
- "#calculations\n",
- "dm=m1-m2\n",
- "dt=abs(dm) *931\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Increase in kinetic energy =\",dt,\"Mev\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Increase in kinetic energy = 0.782 Mev\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 44"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Thickness\n",
- "#Initialization of variables\n",
- "import math\n",
- "d=8.642 #g/cc\n",
- "M=112.41 #g/mol\n",
- "ratio=0.01/100\n",
- "nb=2400\n",
- "#calculations\n",
- "n=d*6.02*10**23 /M\n",
- "sigma=nb*10**-24\n",
- "x=-2.303*math.log10(ratio) /(sigma*n)\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Thickness =\",x,\"cm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thickness = 0.083 cm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 49"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Threshold\n",
- "#Initialization of variables\n",
- "M1=4\n",
- "M2=14\n",
- "E=-1.2 #Mev\n",
- "#calculations\n",
- "R1=1.5*10**-13 *(M1)**(1/3.)\n",
- "R2=1.5*10**-13 *(M2)**(1/3.)\n",
- "V1=2*7*(4.8*10**-10)**2 /(R1+R2)\n",
- "V2=V1/(1.6*10**-6)\n",
- "x=(M1+M2)*V2/M2\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Threshold =\",x,\" Mev\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Threshold = 4.3 Mev\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Time taken\n",
- "#Initialization of variables\n",
- "import math\n",
- "t=1622. #years\n",
- "per=1. #percent\n",
- "#calculations\n",
- "Nratio=1-per/100.\n",
- "x=t*math.log10(Nratio) / math.log10(0.5)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Time taken =\",x,\" years\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time taken = 23.5 years\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter3.ipynb b/Physical_Chemsitry/Chapter3.ipynb deleted file mode 100755 index 5b215119..00000000 --- a/Physical_Chemsitry/Chapter3.ipynb +++ /dev/null @@ -1,57 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:491d30e4e083aec11bc84fdc5735ea79764fcb01fd2223f482cd31ce71e9fd49"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3 - Waves and Quanta"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - pg 83"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Vibrational degrees of freedom\n",
- "#initialization of variables\n",
- "atoms=5\n",
- "#calculations\n",
- "f=3*atoms\n",
- "fvib=f-3-3\n",
- "#results\n",
- "print '%s %d' %(\"Vibrational degrees of freedom = \",fvib)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Vibrational degrees of freedom = 9\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter4.ipynb b/Physical_Chemsitry/Chapter4.ipynb deleted file mode 100755 index 82fe5388..00000000 --- a/Physical_Chemsitry/Chapter4.ipynb +++ /dev/null @@ -1,171 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:567a331a7783164e40dcbd30f4cdd5679a7a2dc34fb757d2d02e98a98daba061"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 4 - Molecular energy levels"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 104"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Moment of inertia\n",
- "#initialization of variables\n",
- "import math\n",
- "B=10.34 #cm**-1\n",
- "c=2.998*10**10 #cm/s\n",
- "h=6.625*10**-27 #erg sec\n",
- "#calculations\n",
- "I=h/(8*math.pi**2 *B*c)\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Moment of inertia =\",I,\" g cm^2\")"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Moment of inertia = 2.71e-40 g cm^2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 105"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Bond length\n",
- "#Initialization of variables\n",
- "import math\n",
- "ma=1.0080\n",
- "mb=35.457\n",
- "Na=6.0232*10**23\n",
- "I=2.707*10**-40 #g cm**2\n",
- "#calculations\n",
- "mu1=ma*mb/(ma+mb)\n",
- "mu2=mu1/Na\n",
- "r=math.sqrt(I/mu2)\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Bond length =\",r,\"cm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Bond length = 1.29e-08 cm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Example 3 - pg 110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Force constant\n",
- "#Initialization of variables\n",
- "import math\n",
- "c=2.998*10**10 #cm/s\n",
- "wave=2990 #cm**-1\n",
- "mu=1.627*10**-24 #g\n",
- "#calculations\n",
- "k=mu*(2*math.pi*c*wave)**2\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Force constant =\",k,\"dynes/cm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Force constant = 5.16e+05 dynes/cm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the anharmonicity constant\n",
- "#initialization of variables\n",
- "l1=2886. #cm^-1\n",
- "l2=5668. #cm^-1\n",
- "#calculations\n",
- "wave=2*l1-l2 \n",
- "wave2= wave+l1\n",
- "x=wave/(2*wave2)\n",
- "#results\n",
- "print '%s %.4f' %(\"anharmonicity constant = \",x)\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "anharmonicity constant = 0.0174\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter6.ipynb b/Physical_Chemsitry/Chapter6.ipynb deleted file mode 100755 index b154f805..00000000 --- a/Physical_Chemsitry/Chapter6.ipynb +++ /dev/null @@ -1,232 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:9bf5bbe9f730183d02f70c65bc17bfc01b5796cd6c1851929b94b0569d87691c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6 - valence electrons in molecules"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 150"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the wave function\n",
- "#initialization of variables\n",
- "import math\n",
- "from math import sqrt\n",
- "a2=1/8.\n",
- "#calculations\n",
- "b2=1-a2\n",
- "a1=sqrt(a2)\n",
- "b1=sqrt(b2)\n",
- "#results\n",
- "print '%s %.2f %s %.2f %s' %(\" Wave function is\",a1, \"phi1 +\",b1,\"phi2\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Wave function is 0.35 phi1 + 0.94 phi2\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 156"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Bond angle\n",
- "#initialization of variables\n",
- "import math\n",
- "sinu=2/math.sqrt(3.)\n",
- "cosu=math.sqrt(2/3.)\n",
- "#calculations\n",
- "tanu=sinu/cosu\n",
- "u=math.atan(sinu/cosu) *180/math.pi\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Bond angle =\",2*u,\"degrees\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Bond angle = 109.47 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Example 4 - pg 156"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Pauling strength\n",
- "#initialization of variables\n",
- "import math\n",
- "cosu=1/math.sqrt(3.)\n",
- "sinu=math.sqrt(2./3)\n",
- "#calculations\n",
- "f=1/2. + math.sqrt(3.) /2. *cosu + math.sqrt(3./2) *sinu\n",
- "#results\n",
- "print ' %s %d' %(\"Pauling strength = \",f)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Pauling strength = 2\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 157"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Wave function\n",
- "#initialization of variables\n",
- "import math\n",
- "alpha=60.*math.pi/180.\n",
- "#calculations\n",
- "cosa=math.cos(alpha)\n",
- "sina=math.sin(alpha)\n",
- "#results\n",
- "print '%s %.2f %s %.2f %s' %(\"Wave function =\",cosa,\"s +\",sina,\"pz\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Wave function = 0.50 s + 0.87 pz\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 9 - pg 169"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Bond energy\n",
- "#initialization of variables\n",
- "DHH=103. #kcal/mol\n",
- "#calculations\n",
- "DHHp=0.5*(DHH)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Bond energy =\",DHHp,\"kcal/mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Bond energy = 51.5 kcal/mol\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 10 - pg 174"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Exchange energy\n",
- "#initialization of variables\n",
- "DHH=42 #kcal/mol\n",
- "#calculations\n",
- "DHHp=0.5*(DHH)\n",
- "#results\n",
- "print '%s %.1f %s' %(\"Exchange energy =\",DHHp,\"kcal/mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Exchange energy = 21.0 kcal/mol\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter7.ipynb b/Physical_Chemsitry/Chapter7.ipynb deleted file mode 100755 index c4d52fe6..00000000 --- a/Physical_Chemsitry/Chapter7.ipynb +++ /dev/null @@ -1,301 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:163d017a43636b203253208922a2be0b72d9894b37f4d97c9da43a5e9b39c875"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7 - Gases and Introductory stastical thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Gas constant\n",
- "#Initialization of variables\n",
- "h=76. #cm\n",
- "d=13.5951 #g/cc\n",
- "g=980.655 #cm/s^2\n",
- "T=273.15 #K\n",
- "v=22414.6 #cm^3 /mol\n",
- "#calculations\n",
- "P=h*d*g\n",
- "R=P*v/(T)\n",
- "#results\n",
- "print '%s %.3e %s' %(\"Gas constant =\",R,\"ergs/deg. mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Gas constant = 8.315e+07 ergs/deg. mol\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the R value in calories\n",
- "#Initialization of variables\n",
- "cal=4.184*10**7 #ergs\n",
- "R=8.315*10**7 #ergs/deg/mol\n",
- "#calculations\n",
- "Rdash=R/cal\n",
- "#results\n",
- "print '%s %.3f %s' %(\"R in calories =\",Rdash,\" cal/ deg mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "R in calories = 1.987 cal/ deg mol\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the root mean square velocity\n",
- "#Initialization of variables\n",
- "import math\n",
- "from math import sqrt\n",
- "R=8.315*10**7 #ergs/deg/mol\n",
- "T=273.2 #deg\n",
- "M=4 #g/mol\n",
- "#calculations\n",
- "u2=3*T*R/M\n",
- "u=sqrt(u2)\n",
- "#results\n",
- "print '%s %.2e %s' %(\"root mean square velocity =\",u,\" cm/sec\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "root mean square velocity = 1.31e+05 cm/sec\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 194"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Partial pressure of N2, O2 and CO2\n",
- "#Initialization of variables\n",
- "n1=2.\n",
- "n2=10.\n",
- "n3=3.\n",
- "P=720. #mm of Hg\n",
- "#calculations\n",
- "n=n1+n2+n3\n",
- "x1=n1/n\n",
- "P1=x1*P\n",
- "x2=n2/n\n",
- "P2=x2*P\n",
- "x3=n3/n\n",
- "P3=x3*P\n",
- "#results\n",
- "print '%s %d %s' %(\"\\n Partial pressure of N2 =\",P1,\"mm\")\n",
- "print '%s %d %s' %(\"\\n Partial pressure of O2 =\",P2,\" mm\")\n",
- "print '%s %d %s' %(\"\\n Partial pressure of CO2 =\",P3,\"mm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " Partial pressure of N2 = 96 mm\n",
- "\n",
- " Partial pressure of O2 = 480 mm\n",
- "\n",
- " Partial pressure of CO2 = 144 mm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 6 - pg 197"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Total energy\n",
- "#Initialization of variables\n",
- "T=273.2+25 #K\n",
- "n=1 #mol\n",
- "R=1.987 #cal/deg mol\n",
- "#calculations\n",
- "Etr=1.5*n*R*T\n",
- "Erot=1.5*n*R*T\n",
- "Evib=0\n",
- "Eel=0\n",
- "Etot=Etr+Erot+Evib+Eel\n",
- "#results\n",
- "print '%s %d %s' %(\"Total energy =\",Etot,\"cal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total energy = 1777 cal\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 7 - pg 199"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Molecular diameter of He\n",
- "#Initialization of variables'\n",
- "import math\n",
- "b=24.1 #cm^2/mol\n",
- "N=6.023*10**23 #mole^-1\n",
- "#calculations\n",
- "d=(3*b/(2*math.pi*N))**(1./3)\n",
- "#results\n",
- "print '%s %.2e %s' %(\"Molecular diameter of He =\",d,\" cm\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Molecular diameter of He = 2.67e-08 cm\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 8 - pg 205"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Volume\n",
- "#Initialization of variables\n",
- "P=100. #atm\n",
- "T=200. #K\n",
- "n=1. #mole\n",
- "R=0.08206 #l atm/deg mol\n",
- "print \"From psychrometric charts,\"\n",
- "Tc=126.2 #K\n",
- "Pc=33.5 #K\n",
- "#calculations\n",
- "Pr=P/Pc\n",
- "Tr=T/Tc\n",
- "print \"From z charts,\"\n",
- "z=0.83\n",
- "V=z*n*R*T/P\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Volume =\",V,\" liter\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "From psychrometric charts,\n",
- "From z charts,\n",
- "Volume = 0.136 liter\n"
- ]
- }
- ],
- "prompt_number": 7
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter8.ipynb b/Physical_Chemsitry/Chapter8.ipynb deleted file mode 100755 index 7ee5bb63..00000000 --- a/Physical_Chemsitry/Chapter8.ipynb +++ /dev/null @@ -1,218 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:56268e206bf92e9d6c6eb22d6025708a41b67ffcb1bed3001f2340e19e39294f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8 - First law of thermodynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Joules required\n",
- "#Initialization of variables\n",
- "P=1.0132*10**6 #dynes/cm**2\n",
- "A=100 #cm**2\n",
- "z=10 #cm\n",
- "#calculations\n",
- "w=P*A*z*10**-7\n",
- "#results\n",
- "print '%s %.4e %s' %(\"Joules =\",w,\"J\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Joules = 1.0132e+02 J\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 211"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Calories required\n",
- "#Initialization of variables\n",
- "P=1.0132*10**6 #dynes/cm**2\n",
- "A=100 #cm**2\n",
- "z=10 #cm\n",
- "#calculations\n",
- "w=P*A*z*10**-7\n",
- "cal=w/4.184\n",
- "#results\n",
- "print '%s %.3f %s' %(\"Calories =\",cal,\"cal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Calories = 24.216 cal\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 3 - pg 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Heat of vaporization and the Change in energy\n",
- "#Initialization of variables\n",
- "T=373.2 #K\n",
- "n=1. #mol\n",
- "qp=9720. #cal/mol\n",
- "#calculations\n",
- "q=n*qp\n",
- "w=1.987*T\n",
- "dE=q-w\n",
- "#results\n",
- "print '%s %d %s' %(\"Heat of vaporization =\",q,\"cal\")\n",
- "print '%s %d %s' %(\"\\n Change in energy =\",dE,\"cal\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Heat of vaporization = 9720 cal\n",
- "\n",
- " Change in energy = 8978 cal\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 4 - pg 222"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the work done, Heat transferred and change in energy\n",
- "#Initialization of variables\n",
- "T1=25+273. #K\n",
- "T2=25+273. #K\n",
- "#calculations\n",
- "print \"Since, T2=T1, dE=0\"\n",
- "dE=0\n",
- "w=0\n",
- "q=dE+w\n",
- "#results\n",
- "print '%s %d %s' %(\"\\n Work done = \",w,\"J\")\n",
- "print '%s %d %s' %(\"\\n Heat transferred =\",q,\"J\")\n",
- "print '%s %d %s' %(\"\\n Change in energy = \",dE,\"J\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Since, T2=T1, dE=0\n",
- "\n",
- " Work done = 0 J\n",
- "\n",
- " Heat transferred = 0 J\n",
- "\n",
- " Change in energy = 0 J\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 5 - pg 224"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Vibrational energy\n",
- "#Initialization of variables\n",
- "R=1.987 #cal/deg mol\n",
- "#calculations\n",
- "Cvtr=1.5*R\n",
- "Cvrot=1.5*R\n",
- "Cvt=Cvtr+Cvrot\n",
- "print \"Observed Cv= 6.43\"\n",
- "Cvobs=6.43\n",
- "Cvvib=Cvobs-Cvt\n",
- "#results\n",
- "print '%s %.2f %s' %(\"Vibrational =\",Cvvib,\"cal/deg mol\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Observed Cv= 6.43\n",
- "Vibrational = 0.47 cal/deg mol\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter9.ipynb b/Physical_Chemsitry/Chapter9.ipynb deleted file mode 100755 index dad780a4..00000000 --- a/Physical_Chemsitry/Chapter9.ipynb +++ /dev/null @@ -1,103 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:d8a3e9ee0b493342a03a360ee0ca604f80a61547dd5ed22ef022608eecfbe8c4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9 - Boltzmann distribution law"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 1 - pg 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the no of particles\n",
- "#Initialization of variables\n",
- "import math\n",
- "r=2.1*10**-6 #cm\n",
- "n=889\n",
- "x=0.1 #cm\n",
- "T=298.2 #K\n",
- "#calculations\n",
- "V=4/3 *math.pi *r**3\n",
- "rho=19.3-1\n",
- "ffd=rho*V*980.7\n",
- "eps=ffd*x\n",
- "logN=-6.96*10**-14 /(2.303*1.38*10**-16 *T)\n",
- "N=10**logN *n\n",
- "#results\n",
- "print '%s %.1f' %(\"No. of particles =\",N)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "No. of particles = 163.9\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 2 - pg 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#calculate the Temperature\n",
- "#Initialization of variables\n",
- "import math\n",
- "x=1. #percent\n",
- "wave=1595 #cm**-1\n",
- "#calculations\n",
- "E=2.8593*wave\n",
- "Nratio=(100-x)/x\n",
- "logN=math.log10(Nratio)\n",
- "T=E/(2.303*1.987*logN)\n",
- "#results\n",
- "print '%s %d %s' %(\"Temperature =\",T,\"K\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Temperature = 499 K\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter_1.ipynb b/Physical_Chemsitry/Chapter_1.ipynb deleted file mode 100755 index ca5fb73c..00000000 --- a/Physical_Chemsitry/Chapter_1.ipynb +++ /dev/null @@ -1,533 +0,0 @@ -{
- "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/Chapter_10.ipynb b/Physical_Chemsitry/Chapter_10.ipynb deleted file mode 100755 index 4ee0e2f1..00000000 --- a/Physical_Chemsitry/Chapter_10.ipynb +++ /dev/null @@ -1,635 +0,0 @@ -{
- "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/Chapter_2.ipynb b/Physical_Chemsitry/Chapter_2.ipynb deleted file mode 100755 index f3c80669..00000000 --- a/Physical_Chemsitry/Chapter_2.ipynb +++ /dev/null @@ -1,437 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e0f60b1374b6cd8ea6c7a7a98303a8b74644396bbd6f803eaeda82697f2fd3e7"
- },
- "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": "markdown",
- "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/Chapter_3.ipynb b/Physical_Chemsitry/Chapter_3.ipynb deleted file mode 100755 index 09238726..00000000 --- a/Physical_Chemsitry/Chapter_3.ipynb +++ /dev/null @@ -1,878 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:e18fb187eca74984e6f3523731bde87a33ea02f1017bfbe68a1eb5fa826002ce"
- },
- "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",
- "n1= 10 #mol\n",
- "n2= 12 #mol\n",
- "#CALCULATIONS\n",
- "E= (x/y)*H\n",
- "H= (E+R*T*(n1-n2))/1000\n",
- "#RESULTS\n",
- "print '%s %.1f %s' %(' Enthalpy =',H,'kcal mole^-1 ')\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " Enthalpy = -1213.2 kcal mole^-1 \n"
- ]
- }
- ],
- "prompt_number": 21
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Physical_Chemsitry/Chapter_4.ipynb b/Physical_Chemsitry/Chapter_4.ipynb deleted file mode 100755 index 619fe75d..00000000 --- a/Physical_Chemsitry/Chapter_4.ipynb +++ /dev/null @@ -1,1022 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a82e4bec145fb0d4efd67c857ae44efe430c89bfa64d07388861ee610e670b10"
- },
- "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": 1,
- "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/Chapter_5.ipynb b/Physical_Chemsitry/Chapter_5.ipynb deleted file mode 100755 index 98aa3041..00000000 --- a/Physical_Chemsitry/Chapter_5.ipynb +++ /dev/null @@ -1,767 +0,0 @@ -{
- "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/Chapter_6.ipynb b/Physical_Chemsitry/Chapter_6.ipynb deleted file mode 100755 index 5c81a57c..00000000 --- a/Physical_Chemsitry/Chapter_6.ipynb +++ /dev/null @@ -1,711 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:b7db16137d946fe9cbf59acb0ec0858aab53c92eb7e66cd0a1bfe1d955619ce0"
- },
- "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",
- "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",
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LJbr3Xjj+eNhlF7jwQlh22awjEpmfTr+npyoqdTPbi1Ct/w04EpgITDWzvcxs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\nIiI5oaQuIiKSE0rqIiIiOfH/AVD524dASlmNAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x70f3750>"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The answers are a bit different due to rounding off error in textbook\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "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/Chapter_7.ipynb b/Physical_Chemsitry/Chapter_7.ipynb deleted file mode 100755 index d0564a44..00000000 --- a/Physical_Chemsitry/Chapter_7.ipynb +++ /dev/null @@ -1,1029 +0,0 @@ -{
- "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/Chapter_8.ipynb b/Physical_Chemsitry/Chapter_8.ipynb deleted file mode 100755 index c1054064..00000000 --- a/Physical_Chemsitry/Chapter_8.ipynb +++ /dev/null @@ -1,440 +0,0 @@ -{
- "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/Chapter_9.ipynb b/Physical_Chemsitry/Chapter_9.ipynb deleted file mode 100755 index 1a10b96e..00000000 --- a/Physical_Chemsitry/Chapter_9.ipynb +++ /dev/null @@ -1,101 +0,0 @@ -{
- "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/screenshots/chap11.png b/Physical_Chemsitry/screenshots/chap11.png Binary files differdeleted file mode 100755 index 6ed53889..00000000 --- a/Physical_Chemsitry/screenshots/chap11.png +++ /dev/null diff --git a/Physical_Chemsitry/screenshots/chap12.png b/Physical_Chemsitry/screenshots/chap12.png Binary files differdeleted file mode 100755 index a1675611..00000000 --- a/Physical_Chemsitry/screenshots/chap12.png +++ /dev/null diff --git a/Physical_Chemsitry/screenshots/chap13.png b/Physical_Chemsitry/screenshots/chap13.png Binary files differdeleted file mode 100755 index 42833795..00000000 --- a/Physical_Chemsitry/screenshots/chap13.png +++ /dev/null diff --git a/Physical_Chemsitry/screenshots/chap3.png b/Physical_Chemsitry/screenshots/chap3.png Binary files differdeleted file mode 100755 index 832091f2..00000000 --- a/Physical_Chemsitry/screenshots/chap3.png +++ /dev/null diff --git a/Physical_Chemsitry/screenshots/chap4.png b/Physical_Chemsitry/screenshots/chap4.png Binary files differdeleted file mode 100755 index e2646978..00000000 --- a/Physical_Chemsitry/screenshots/chap4.png +++ /dev/null diff --git a/Physical_Chemsitry/screenshots/chap5.png b/Physical_Chemsitry/screenshots/chap5.png Binary files differdeleted file mode 100755 index 5e065126..00000000 --- a/Physical_Chemsitry/screenshots/chap5.png +++ /dev/null |
