From 435840cef00c596d9e608f9eb2d96f522ea8505a Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 5 May 2015 14:21:39 +0530 Subject: add books --- Physical_Chemsitry/Chapter_7.ipynb | 1029 ------------------------------------ 1 file changed, 1029 deletions(-) delete mode 100755 Physical_Chemsitry/Chapter_7.ipynb (limited to 'Physical_Chemsitry/Chapter_7.ipynb') 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 -- cgit