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| author | hardythe1 | 2015-05-05 14:21:39 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-05-05 14:21:39 +0530 |
| commit | 435840cef00c596d9e608f9eb2d96f522ea8505a (patch) | |
| tree | 4c783890c984c67022977ca98432e5e4bab30678 /Physical_Chemsitry/Chapter_7.ipynb | |
| parent | aa1863f344766ca7f7c20a395e58d0fb23c52130 (diff) | |
| download | Python-Textbook-Companions-435840cef00c596d9e608f9eb2d96f522ea8505a.tar.gz Python-Textbook-Companions-435840cef00c596d9e608f9eb2d96f522ea8505a.tar.bz2 Python-Textbook-Companions-435840cef00c596d9e608f9eb2d96f522ea8505a.zip | |
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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 |