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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 06: Chemical Equilibrium"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.1, Page Number 117"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "dHcCH4 = -891.0        #Std. heat of combustion for CH4, kJ/mol\n",
      "dHcC8H18 = -5471.0     #Std. heat of combustion for C8H18, kJ/mol\n",
      "\n",
      "T = 298.15\n",
      "SmCO2, SmCH4, SmH2O, SmO2, SmC8H18 = 213.8,186.3,70.0,205.2, 316.1\n",
      "dnCH4 = -2.\n",
      "dnC8H18 = 4.5\n",
      "R = 8.314\n",
      "#Calculations\n",
      "dACH4 = dHcCH4*1e3 - dnCH4*R*T - T*(SmCO2 + 2*SmH2O - SmCH4 - 2*SmO2)\n",
      "dAC8H18 = dHcC8H18*1e3 - dnC8H18*R*T - T*(8*SmCO2 + 9*SmH2O - SmC8H18 - 25.*SmO2/2) \n",
      "#Results \n",
      "print 'Maximum Available work through combustion of CH4 %4.1f kJ/mol'%(dACH4/1000)\n",
      "print 'Maximum Available work through combustion of C8H18 %4.1f kJ/mol'%(dAC8H18/1000)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum Available work through combustion of CH4 -813.6 kJ/mol\n",
        "Maximum Available work through combustion of C8H18 -5320.9 kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.2, Page Number 118"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "dHcCH4 = -891.0        #Std. heat of combustion for CH4, kJ/mol\n",
      "dHcC8H18 = -5471.0     #Std. heat of combustion for C8H18, kJ/mol\n",
      "\n",
      "T = 298.15\n",
      "SmCO2, SmCH4, SmH2O, SmO2, SmC8H18 = 213.8,186.3,70.0,205.2, 316.1\n",
      "dnCH4 = -2.\n",
      "dnC8H18 = 4.5\n",
      "R = 8.314\n",
      "#Calculations\n",
      "dGCH4 = dHcCH4*1e3  - T*(SmCO2 + 2*SmH2O - SmCH4 - 2*SmO2)\n",
      "dGC8H18 = dHcC8H18*1e3 - T*(8*SmCO2 + 9*SmH2O - SmC8H18 - 25.*SmO2/2) \n",
      "#Results \n",
      "print 'Maximum nonexapnasion work through combustion of CH4 %4.1f kJ/mol'%(dGCH4/1000)\n",
      "print 'Maximum nonexapnasion work through combustion of C8H18 %4.1f kJ/mol'%(dGC8H18/1000)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum nonexapnasion work through combustion of CH4 -818.6 kJ/mol\n",
        "Maximum nonexapnasion work through combustion of C8H18 -5309.8 kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.4, Page Number 123"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "dGf298 = 370.7     #Std. free energy of formation for Fe (g), kJ/mol\n",
      "dHf298 = 416.3     #Std. Enthalpy of formation for Fe (g), kJ/mol\n",
      "T0 = 298.15        #Temperature in K\n",
      "T = 400.           #Temperature in K\n",
      "R = 8.314\n",
      "\n",
      "#Calculations\n",
      "\n",
      "dGf = T*(dGf298*1e3/T0 + dHf298*1e3*(1./T - 1./T0))\n",
      "\n",
      "#Results \n",
      "print 'Std. free energy of formation for Fe(g) at 400 K is %4.1f kJ/mol'%(dGf/1000)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Std. free energy of formation for Fe(g) at 400 K is 355.1 kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.5, Page Number 127"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "\n",
      "#Variable Declaration\n",
      "nHe = 1.0          #Number of moles of He\n",
      "nNe = 3.0          #Number of moles of Ne\n",
      "nAr = 2.0          #Number of moles of Ar\n",
      "nXe = 2.5          #Number of moles of Xe\n",
      "T = 298.15         #Temperature in K\n",
      "P = 1.0            #Pressure, bar\n",
      "R = 8.314\n",
      "\n",
      "#Calculations\n",
      "n = nHe + nNe + nAr + nXe\n",
      "dGmix = n*R*T*((nHe/n)*log(nHe/n) + (nNe/n)*log(nNe/n) +(nAr/n)*log(nAr/n) + (nXe/n)*log(nXe/n))\n",
      "dSmix = n*R*((nHe/n)*log(nHe/n) + (nNe/n)*log(nNe/n) +(nAr/n)*log(nAr/n) + (nXe/n)*log(nXe/n))\n",
      "\n",
      "#Results \n",
      "print 'Std. free energy Change on mixing is %3.1e J'%(dGmix)\n",
      "print 'Std. entropy Change on mixing is %4.1f J'%(dSmix)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Std. free energy Change on mixing is -2.8e+04 J\n",
        "Std. entropy Change on mixing is -93.3 J\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.6, Page Number 128"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "dGfFe  = 0.0       #Std. Gibbs energy of formation for Fe (S), kJ/mol\n",
      "dGfH2O = -237.1     #Std. Gibbs energy of formation for Water (g), kJ/mol\n",
      "dGfFe2O3 = -1015.4  #Std. Gibbs energy of formation for Fe2O3 (s), kJ/mol\n",
      "dGfH2 = 0.0        #Std. Gibbs energy of formation for Hydrogen (g), kJ/mol\n",
      "T0 = 298.15        #Temperature in K\n",
      "R = 8.314\n",
      "nFe, nH2, nFe2O3, nH2O = 3,-4,-1,4\n",
      "\n",
      "#Calculations\n",
      "dGR = nFe*dGfFe + nH2O*dGfH2O + nFe2O3*dGfFe2O3 + nH2*dGfH2  \n",
      "\n",
      "#Results \n",
      "print 'Std. Gibbs energy change for reaction is %4.2f kJ/mol'%(dGR)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Std. Gibbs energy change for reaction is 67.00 kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.7, Page Number 128"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "dGR  = 67.0        #Std. Gibbs energy of formation for reaction, kJ, from previous problem\n",
      "dHfFe  = 0.0       #Enthalpy of formation for Fe (S), kJ/mol\n",
      "dHfH2O = -285.8    #Enthalpy of formation for Water (g), kJ/mol\n",
      "dHfFe2O3 = -1118.4 #Enthalpy of formation for Fe2O3 (s), kJ/mol\n",
      "dHfH2 = 0.0        #Enthalpy of formation for Hydrogen (g), kJ/mol\n",
      "T0 = 298.15        #Temperature in K\n",
      "T = 525.           #Temperature in K\n",
      "R = 8.314\n",
      "nFe, nH2, nFe2O3, nH2O = 3,-4,-1,4\n",
      "\n",
      "#Calculations\n",
      "dHR = nFe*dHfFe + nH2O*dHfH2O + nFe2O3*dHfFe2O3 + nH2*dHfH2  \n",
      "dGR2 = T*(dGR*1e3/T0 + dHR*1e3*(1./T - 1./T0))\n",
      "\n",
      "#Results \n",
      "print 'Std. Enthalpy change for reactionat %4.1f is %4.2f kJ/mol'%(T, dHR)\n",
      "print 'Std. Gibbs energy change for reactionat %4.1f is %4.0f kJ/mol'%(T, dGR2/1e3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Std. Enthalpy change for reactionat 525.0 is -24.80 kJ/mol\n",
        "Std. Gibbs energy change for reactionat 525.0 is  137 kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.8, Page Number 130"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "\n",
      "#Variable Declaration\n",
      "dGfNO2  = 51.3     #Std. Gibbs energy of formation for NO2 (g), kJ/mol\n",
      "dGfN2O4 = 99.8     #Std. Gibbs energy of formation for N2O4 (g), kJ/mol\n",
      "T0 = 298.15        #Temperature in K\n",
      "pNO2 = 0.350       #Partial pressure of NO2, bar\n",
      "pN2O4 = 0.650      #Partial pressure of N2O4, bar\n",
      "R = 8.314\n",
      "nNO2, nN2O4 = -2, 1 #Stoichiomentric coeff of NO2 and N2O4 respectively in reaction\n",
      "\n",
      "#Calculations\n",
      "dGR = nN2O4*dGfN2O4*1e3 + nNO2*dGfNO2*1e3 + R*T0*log(pN2O4/(pNO2)**2)\n",
      "\n",
      "#Results \n",
      "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR/1e3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Std. Gibbs energy change for reaction is 1.337 kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.9, Page Number 131"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp\n",
      "\n",
      "#Variable Declaration\n",
      "dGfCO2  = -394.4   #Std. Gibbs energy of formation for CO2 (g), kJ/mol\n",
      "dGfH2 = 0.0        #Std. Gibbs energy of formation for H2 (g), kJ/mol\n",
      "dGfCO = 237.1      #Std. Gibbs energy of formation for CO (g), kJ/mol\n",
      "dGfH2O = 137.2     #Std. Gibbs energy of formation for H24 (l), kJ/mol\n",
      "T0 = 298.15        #Temperature in K\n",
      "R = 8.314\n",
      "nCO2, nH2, nCO, nH2O = 1,1,1,1 #Stoichiomentric coeff of CO2,H2,CO,H2O respectively in reaction\n",
      "\n",
      "#Calculations\n",
      "dGR = nCO2*dGfCO2 + nH2*dGfH2 + nCO*dGfCO + nH2O*dGfH2O\n",
      "Kp = exp(-dGR*1e3/(R*T0))\n",
      "\n",
      "#Results \n",
      "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR/1e3)\n",
      "print 'Equilibrium constant for reaction is %5.3f '%(Kp)\n",
      "if Kp > 1: print 'Kp >> 1. hence, mixture will consists of product CO2 and H2'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Std. Gibbs energy change for reaction is -0.020 kJ/mol\n",
        "Equilibrium constant for reaction is 3323.254 \n",
        "Kp >> 1. hence, mixture will consists of product CO2 and H2\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.11, Page Number 133"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp, sqrt\n",
      "\n",
      "#Variable Declaration\n",
      "dGfCl2  = 0.0     #Std. Gibbs energy of formation for CO2 (g), kJ/mol\n",
      "dGfCl = 105.7     #Std. Gibbs energy of formation for H2 (g), kJ/mol\n",
      "dHfCl2 = 0.0      #Std. Gibbs energy of formation for CO (g), kJ/mol\n",
      "dHfCl = 121.3     #Std. Gibbs energy of formation for H24 (l), kJ/mol\n",
      "T0 = 298.15       #Temperature in K\n",
      "R = 8.314\n",
      "nCl2, nCl= -1,2 #Stoichiomentric coeff of Cl2,Cl respectively in reaction\n",
      "PbyP0 = 0.01\n",
      "#Calculations\n",
      "dGR = nCl*dGfCl + nCl2*dGfCl2 \n",
      "dHR = nCl*dHfCl + nCl2*dHfCl2 \n",
      "func = lambda T: exp(-dGR*1e3/(R*T0) - dHR*1e3*(1./T - 1./T0)/R)\n",
      "Kp8 = func(800)\n",
      "Kp15 = func(1500)\n",
      "Kp20 = func(2000)\n",
      "DDiss = lambda K: sqrt(K/(K+4*PbyP0))\n",
      "alp8 = DDiss(Kp8)\n",
      "alp15 = DDiss(Kp15)\n",
      "alp20 = DDiss(Kp20)\n",
      "\n",
      "#Results \n",
      "print 'Part A'\n",
      "print 'Std. Gibbs energy change for reaction is %5.3f kJ/mol'%(dGR)\n",
      "print 'Std. Enthalpy change for reaction is %5.3f kJ/mol'%(dHR)\n",
      "print 'Equilibrium constants at 800, 1500, and 2000 K are %4.3e, %4.3e, and %4.3e'%(Kp8,Kp15,Kp20)\n",
      "\n",
      "print 'Part B'\n",
      "print 'Degree of dissociation at 800, 1500, and 2000 K are %4.3e, %4.3e, and %4.3e'%(alp8,alp15,alp20)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Part A\n",
        "Std. Gibbs energy change for reaction is 211.400 kJ/mol\n",
        "Std. Enthalpy change for reaction is 242.600 kJ/mol\n",
        "Equilibrium constants at 800, 1500, and 2000 K are 4.223e-11, 1.042e-03, and 1.349e-01\n",
        "Part B\n",
        "Degree of dissociation at 800, 1500, and 2000 K are 3.249e-05, 1.593e-01, and 8.782e-01\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.12, Page Number 134"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp\n",
      "\n",
      "#Variable Declaration\n",
      "dGfCaCO3  = -1128.8     #Std. Gibbs energy of formation for CaCO3 (s), kJ/mol\n",
      "dGfCaO = -603.3         #Std. Gibbs energy of formation for CaO (s), kJ/mol\n",
      "dGfCO2 = -394.4         #Std. Gibbs energy of formation for O2 (g), kJ/mol\n",
      "dHfCaCO3 = -1206.9      #Std. Enthalpy Change of formation for CaCO3 (s), kJ/mol\n",
      "dHfCaO = -634.9         #Std. Enthalpy Change of formation for CaO (s), kJ/mol\n",
      "dHfCO2 = -393.5         #Std. Enthalpy Change of formation for O2 (g), kJ/mol\n",
      "T0 = 298.15             #Temperature in K\n",
      "R = 8.314\n",
      "nCaCO3, nCaO, nO2 = -1,1,1 #Stoichiomentric coeff of CaCO3, CaO, O2 respectively in reaction\n",
      "\n",
      "#Calculations\n",
      "dGR = nCaO*dGfCaO + nO2*dGfCO2 + nCaCO3*dGfCaCO3\n",
      "dHR = nCaO*dHfCaO + nO2*dHfCO2 + nCaCO3*dHfCaCO3\n",
      "\n",
      "func = lambda T: exp(-dGR*1e3/(R*T0) - dHR*1e3*(1./T - 1./T0)/R)\n",
      "\n",
      "Kp10 = func(1000)\n",
      "Kp11 = func(1100)\n",
      "Kp12 = func(1200)\n",
      "\n",
      "#Results \n",
      "print 'Std. Gibbs energy change for reaction is %4.1f kJ/mol'%(dGR)\n",
      "print 'Std. Enthalpy change for reaction is %4.1f kJ/mol'%(dHR)\n",
      "print 'Equilibrium constants at 1000, 1100, and 1200 K are %4.4f, %4.3fe, and %4.3f'%(Kp10,Kp11,Kp12)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Std. Gibbs energy change for reaction is 131.1 kJ/mol\n",
        "Std. Enthalpy change for reaction is 178.5 kJ/mol\n",
        "Equilibrium constants at 1000, 1100, and 1200 K are 0.0956, 0.673e, and 3.423\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.13, Page Number 135"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp\n",
      "\n",
      "#Variable Declaration\n",
      "dGfCG  = 0.0            #Std. Gibbs energy of formation for CaCO3 (s), kJ/mol\n",
      "dGfCD = 2.90            #Std. Gibbs energy of formation for CaO (s), kJ/mol\n",
      "rhoG = 2.25e3           #Density of Graphite, kg/m3\n",
      "rhoD = 3.52e3           #Density of dimond, kg/m3\n",
      "T0 = 298.15             #Std. Temperature, K\n",
      "R = 8.314               #Ideal gas constant, J/(mol.K) \n",
      "P0 = 1.0                #Pressure, bar\n",
      "M = 12.01               #Molceular wt of Carbon\n",
      "#Calculations\n",
      "P = P0*1e5 + dGfCD*1e3/((1./rhoG-1./rhoD)*M*1e-3)\n",
      "\n",
      "#Results \n",
      "print 'Pressure at which graphite and dimond will be in equilibrium is %4.2e bar'%(P/1e5)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Pressure at which graphite and dimond will be in equilibrium is 1.51e+04 bar\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.14, Page Number 143"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp\n",
      "\n",
      "#Variable Declaration\n",
      "beta = 2.04e-4          #Thermal exapansion coefficient, /K\n",
      "kapa = 45.9e-6          #Isothermal compressibility, /bar\n",
      "T = 298.15              #Std. Temperature, K\n",
      "R = 8.206e-2            #Ideal gas constant, atm.L/(mol.K) \n",
      "T1 = 320.0              #Temperature, K\n",
      "Pi = 1.0                #Initial Pressure, bar\n",
      "V = 1.00                #Volume, m3\n",
      "a = 1.35                #van der Waals constant a for nitrogen, atm.L2/mol2\n",
      "\n",
      "#Calculations\n",
      "dUbydV = Pf = (beta*T1-kapa*P0)/kapa\n",
      "dVT = V*kapa*(Pf-Pi)\n",
      "dVbyV = dVT*100/V\n",
      "Vm = Pi/(R*T1)\n",
      "dUbydVm = a/(Vm**2)\n",
      "\n",
      "#Results \n",
      "print 'dUbydV = %4.2e bar'%(dUbydV)\n",
      "print 'dVbyV = %4.3f percent'%(dVbyV)\n",
      "print 'dUbydVm = %4.0e atm'%(dUbydVm)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "dUbydV = 1.42e+03 bar\n",
        "dVbyV = 6.519 percent\n",
        "dUbydVm = 9e+02 atm\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.15, Page Number 144"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import exp, log\n",
      "\n",
      "#Variable Declaration\n",
      "m = 1000.0                #mass of mercury, g\n",
      "Pi, Ti  = 1.00, 300.0     #Intial pressure and temperature, bar, K\n",
      "Pf, Tf  = 300., 600.0     #Final pressure and temperature, bar, K\n",
      "rho = 13534.              #Density of mercury, kg/m3\n",
      "beta = 18.1e-4            #Thermal exapansion coefficient for Hg, /K \n",
      "kapa = 3.91e-6            #Isothermal compressibility for Hg, /Pa\n",
      "Cpm = 27.98               #Molar Specific heat at constant pressure, J/(mol.K) \n",
      "M = 200.59                #Molecular wt of Hg, g/mol\n",
      "\n",
      "#Calculations\n",
      "Vi = m*1e-3/rho\n",
      "Vf = Vi*exp(-kapa*(Pf-Pi))\n",
      "Ut = m*Cpm*(Tf-Ti)/M \n",
      "Up =  (beta*Ti/kapa-Pi)*1e5*(Vf-Vi) + (Vi-Vf+Vf*log(Vf/Vi))*1e5/kapa\n",
      "dU = Ut + Up\n",
      "Ht = m*Cpm*(Tf-Ti)/M\n",
      "Hp = ((1 + beta*(Tf-Ti))*Vi*exp(-kapa*Pi)/kapa)*(exp(-kapa*Pi)-exp(-kapa*Pf))\n",
      "dH =  Ht + Hp\n",
      "#Results\n",
      "print 'Internal energy change is %6.2e J/mol in which \\ncontribution of temeprature dependent term %6.4f percent'%(dU,Ut*100/dH)\n",
      "print 'Enthalpy change is %4.3e J/mol in which \\ncontribution of temeprature dependent term %4.1f percent'%(dH,Ht*100/dH)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Internal energy change is 4.06e+04 J/mol in which \n",
        "contribution of temeprature dependent term 99.9999 percent\n",
        "Enthalpy change is 4.185e+04 J/mol in which \n",
        "contribution of temeprature dependent term 100.0 percent\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.16, Page Number 145"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "T = 300.0                   #Temperature of Hg, K \n",
      "beta = 18.1e-4              #Thermal exapansion coefficient for Hg, /K \n",
      "kapa = 3.91e-6              #Isothermal compressibility for Hg, /Pa\n",
      "M = 0.20059                 #Molecular wt of Hg, kg/mol \n",
      "rho = 13534                 #Density of mercury, kg/m3\n",
      "Cpm = 27.98                 #Experimental Molar specif heat at const pressure for mercury, J/(mol.K)\n",
      "\n",
      "#Calculations\n",
      "Vm = M/rho\n",
      "DCpmCv = T*Vm*beta**2/kapa\n",
      "Cvm = Cpm - DCpmCv\n",
      "#Results\n",
      "print 'Difference in molar specific heats \\nat constant volume and constant pressure %4.2e J/(mol.K)'%DCpmCv\n",
      "print 'Molar Specific heat of Hg at const. volume is %4.2f J/(mol.K)'%Cvm"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Difference in molar specific heats \n",
        "at constant volume and constant pressure 3.73e-03 J/(mol.K)\n",
        "Molar Specific heat of Hg at const. volume is 27.98 J/(mol.K)\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 6.17, Page Number 147"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "T = 298.15                  #Std. Temperature, K \n",
      "P = 1.0                     #Initial Pressure, bar\n",
      "Hm0, Sm0 = 0.0,154.8        #Std. molar enthalpy and entropy of Ar(g), kJ, mol, K units\n",
      "Sm0H2, Sm0O2 = 130.7,205.2  #Std. molar entropy of O2 and H2 (g), kJ/(mol.K)\n",
      "dGfH2O = -237.1         #Gibbs energy of formation for H2O(l), kJ/mol  \n",
      "nH2, nO2 = 1, 1./2       #Stoichiomentric coefficients for H2 and O2 in water formation reaction \n",
      "\n",
      "#Calculations\n",
      "Gm0 = Hm0 - T*Sm0\n",
      "dGmH2O = dGfH2O*1000 - T*(nH2*Sm0H2 + nO2*Sm0O2)\n",
      "#Results\n",
      "print 'Molar Gibbs energy of Ar %4.3f kJ/mol'%(Gm0/1e3)\n",
      "print 'Molar Gibbs energy of Water %4.3f kJ/mol'%(dGmH2O/1e3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Molar Gibbs energy of Ar -46.154 kJ/mol\n",
        "Molar Gibbs energy of Water -306.658 kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 15
    }
   ],
   "metadata": {}
  }
 ]
}