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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 4 - Second and Third laws of thermodynamics"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1 - pg 192"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the maximum efficiency in all cases\n",
      "#initialisation of variables\n",
      "T = 100. #C\n",
      "T1= 25. #C\n",
      "T2= 150. #C\n",
      "T3= 357. #C\n",
      "T4= 500. #C\n",
      "T5= 2000. #C\n",
      "T6= 5*10**6\n",
      "T7= 1000. #C\n",
      "#CALCULATIONS\n",
      "e= (T-T1)/(T+273)\n",
      "e1= (T2-T1)/(273+T2)\n",
      "e2= (T3-T)/(273+T3)\n",
      "e3= (T5-T4)/(T5+273)\n",
      "e4= (T6-T7)/T6\n",
      "#RESULTS\n",
      "print '%s %.2f' % (' maximum efficiency in case 1= ',e)\n",
      "print '%s %.2f' % (' \\n maximum efficiency in case 2 =  ',e1)\n",
      "print '%s %.2f' % (' \\n maximum efficiency in case 3 =  ',e2)\n",
      "print '%s %.2f' % (' \\n maximum efficiency in case 4 =  ',e3)\n",
      "print '%s %.2f' % (' \\n maximum efficiency in case 5 =  ',e4)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " maximum efficiency in case 1=  0.20\n",
        " \n",
        " maximum efficiency in case 2 =   0.30\n",
        " \n",
        " maximum efficiency in case 3 =   0.41\n",
        " \n",
        " maximum efficiency in case 4 =   0.66\n",
        " \n",
        " maximum efficiency in case 5 =   1.00\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2 - pg 194"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the maximum efficiency and minimum work\n",
      "#initialisation of variables\n",
      "T= 20. #C\n",
      "T1= -10. #C\n",
      "q= 1000. #cal\n",
      "#CALCULATIONS\n",
      "e= (273+T1)/(T-T1)\n",
      "w= (T-T1)*q/(273+T1)\n",
      "#RESULTS\n",
      "print '%s %.1f' % (' maximum efficiency = ',e)\n",
      "print '%s %d %s' % (' \\n minimum work =',w,'cal')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " maximum efficiency =  8.8\n",
        " \n",
        " minimum work = 114 cal\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3 - pg 197"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the net work done on and by the gas\n",
      "#initialisation of variables\n",
      "T= 1000 #K\n",
      "T1= 400 #/K\n",
      "w= 1000 #cal\n",
      "E= 0 #cal\n",
      "gam=7/5.\n",
      "#CALCULATIONS\n",
      "q= w-E\n",
      "W= q*(T-T1)/T\n",
      "q1= W-q\n",
      "W1= -q1\n",
      "dE=5/2.*(T1-T)\n",
      "dH=7/2. *(T1-T)\n",
      "w2=dE-E\n",
      "w3=(T-T1)/(gam-1)\n",
      "#RESULTS\n",
      "print '%s %.1f %s' % ('net work done by gas=',W,' cal')\n",
      "print '%s %.1f %s' % ('net work done on gas =',W1,'cal')\n",
      "print '%s %.1f %s' %('Change in Internal energy = ',dE,'R cal')\n",
      "print '%s %.1f %s' %('Change in Enthalpy = ',dH,'R cal')\n",
      "print '%s %.1f %s' %('Work for adiabatic compression =',w3,'R cal')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "net work done by gas= 600.0  cal\n",
        "net work done on gas = 400.0 cal\n",
        "Change in Internal energy =  -1500.0 R cal\n",
        "Change in Enthalpy =  -2100.0 R cal\n",
        "Work for adiabatic compression = 1500.0 R cal\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4 - pg 199"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the entropy of substances\n",
      "#initialisation of variables\n",
      "import numpy as np\n",
      "Hv= np.array([1960.,1560.,4880.,37000.,5500.,27400.,60700.,9720.,30900.]) #cal mole^-1\n",
      "Tb= np.array([112.,87.3,239.,1806.,259.,1180.,2466.,373.,1029.]) #K\n",
      "#CALCULATIONS\n",
      "Sv= np.round(Hv/Tb,1)\n",
      "#RESULTS\n",
      "print '%s' % (' Entropy (cal mole deg^-1)')\n",
      "print Sv\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Entropy (cal mole deg^-1)\n",
        "[ 17.5  17.9  20.4  20.5  21.2  23.2  24.6  26.1  30. ]\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5 - pg 201"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the entropy at constant pressure and volume\n",
      "#initialisation of variables\n",
      "import math\n",
      "T= 300. #K\n",
      "T1= 400. #K\n",
      "k= 6.0954\n",
      "k1= 3.2533*10**-3\n",
      "k2= -1.0171*10**-6\n",
      "R= 1.98719 #cal/mol K\n",
      "#CALCULATIONS\n",
      "S= 2*(k*math.log(T1/T)+k1*(T1-T)+k2*(T1**2-T**2)/2)\n",
      "S1= S-2*R*math.log(T1/T)\n",
      "#RESULTS\n",
      "print '%s %.4f %s' % (' Entropy=',S,' cal  deg^-1')\n",
      "print '%s %.4f %s' % (' \\n Entropy =',S1,'cal  deg^-1')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Entropy= 4.0865  cal  deg^-1\n",
        " \n",
        " Entropy = 2.9432 cal  deg^-1\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8 - pg 216"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the final temperature \n",
      "#initialisation of variables\n",
      "T1= 273.16 #K\n",
      "R= 1.987 #cal /mol K\n",
      "R1= 0.08205 #J /mol K\n",
      "n= 10 #moles\n",
      "V1= 22.4 #lit\n",
      "a= 1.36\n",
      "Cv= 4.9\n",
      "#CALCULATIONS\n",
      "T2= T1-(R*a*(n-1)/(R1*n*Cv*V1))\n",
      "#RESULTS\n",
      "print '%s %.2f %s' % (' temperature=',T2,' K')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " temperature= 272.89  K\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9 - pg 218"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the inversion Temperature\n",
      "#initialisation of variables\n",
      "a= 1.360 #l^2 atm mole^-1\n",
      "b= 0.0317 #l mole^-1\n",
      "R= 0.08205 #J/mol K\n",
      "#CALCULATIONS\n",
      "T= 2*a/(b*R)\n",
      "#RESULTS\n",
      "print '%s %.1f %s' % ('Inversion Temperature=',T,'K')\n",
      "print 'The answer is a bit different due to rounding off error in textbook'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Inversion Temperature= 1045.8 K\n",
        "The answer is a bit different due to rounding off error in textbook\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 10 - pg 218"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the joule thomson coefficient\n",
      "#initialisation of variables\n",
      "a= 1.360 #l^2 atm mole^-1\n",
      "b= 0.0317 #l mole^-1\n",
      "R= 0.08205 #J/mol K\n",
      "R1= 1.987 #cal/mole K\n",
      "Cp= 6.9 #cal mole^-1 deg^-1\n",
      "T= 273.2 #K\n",
      "#CALCULATIONS\n",
      "u= ((2*a/(R*T))-b)/(Cp*(R/R1))\n",
      "#RESULTS\n",
      "print '%s %.3f %s' % (' Joule thomson coefficient=',u,' atm^-1')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Joule thomson coefficient= 0.315  atm^-1\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 12 - pg 221"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate the increase in entropy\n",
      "#initialisation of variables\n",
      "import math\n",
      "p= 4/3. #atm\n",
      "p1= 1 #atm\n",
      "R= 1.9872 #cal /mole K\n",
      "#CALCULATIONS\n",
      "S= 2*R*math.log(p/p1)\n",
      "#RESULTS\n",
      "print '%s %.4f %s' %(' increase in entropy=',S,'cal deg^-1')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " increase in entropy= 1.1434 cal deg^-1\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13 - pg 222"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate the increase in entropy\n",
      "#initialisation of variables\n",
      "import math\n",
      "p1= 1 #atm\n",
      "R= 1.9872 #cal /mole K\n",
      "#CALCULATIONS\n",
      "S= 0 #Initial and final states are alike\n",
      "#RESULTS\n",
      "print '%s %.4f %s' %(' increase in entropy=',S,'cal deg^-1')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " increase in entropy= 0.0000 cal deg^-1\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14 - pg 222"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate rhe increase in entropy in both cases\n",
      "#initialisation of variables\n",
      "import math\n",
      "T= 25. #C\n",
      "T1= 100. #C\n",
      "R= 1.9872 #cal /mole K\n",
      "p= 1 #atm\n",
      "p1= 10. #atm\n",
      "#CALCULATIONS\n",
      "S= 3.5*R*math.log((T1+273)/(T+273))\n",
      "S1= S+R*math.log(p/p1)\n",
      "#RESULTS\n",
      "print '%s %.2f %s' % (' increase in entropy in case 1 =',S,'cal deg^-1')\n",
      "print '%s %.2f %s' % (' \\n increase in entropy in case 2 =',S1,'cal deg^-1')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " increase in entropy in case 1 = 1.56 cal deg^-1\n",
        " \n",
        " increase in entropy in case 2 = -3.01 cal deg^-1\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 15 - pg 222"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the absolute entropy\n",
      "#initialisation of variables\n",
      "import math\n",
      "S= 45.77 #cal deg^-1\n",
      "T= 25. #C\n",
      "T1= 100. #C\n",
      "R= 1.9872 #cal /mole K\n",
      "#CALCULATIONS\n",
      "S0= S+ 3.5*R*math.log((T1+273)/(T+273))\n",
      "#RESULTS\n",
      "print '%s %.2f %s' % (' absolute entropy=',S0,'cal deg^-1')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " absolute entropy= 47.33 cal deg^-1\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16 - pg 226"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the change in entropy\n",
      "#initialisation of variables\n",
      "import math\n",
      "Cp= 18. #cal deg^-1\n",
      "T= 0. #C\n",
      "T1= -5. #C\n",
      "H2= -1440. #cal\n",
      "Cp1= 9. #cal deg^-1\n",
      "H= 0.\n",
      "#CALCULATIONS\n",
      "T2= (-Cp*(T-T1)-H2+Cp1*(273.16+T))/Cp1\n",
      "S= Cp*math.log((273.16+T)/(273.16+T1))-(Cp*(T-T1)/(T+273.16))\n",
      "#RESULTS\n",
      "print '%s %.4f %s' % (' Change in entropy=',S,'cal deg^-1')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Change in entropy= 0.0031 cal deg^-1\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 18 - pg 231"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Gibbs free energy\n",
      "#initialisation of variables\n",
      "H= -57.7979 #cal\n",
      "H1= -68.3174 #cal\n",
      "S= 45.106 #cal deg^-1\n",
      "S1= 16.716 #cal deg^-1\n",
      "T= 25 #C\n",
      "#CALCULATIONS\n",
      "H2= (H-H1)*1000\n",
      "S2= S-S1\n",
      "G= H2-(273.16+T)*S2\n",
      "#RESULTS\n",
      "print '%s %.1f %s' % (' Gibbs free energy=',G,'cal')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Gibbs free energy= 2054.7 cal\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 19 - pg 231"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Gibbs free energy\n",
      "#initialisation of variables\n",
      "H= -68317.4 #cal\n",
      "S= 16.716 #cal\n",
      "S1= 49.003 #cal\n",
      "S2= 31.211 #cal\n",
      "T= 25 #C\n",
      "#CALCULATIONS\n",
      "H1= 2*H\n",
      "S3= 2*S-(S1+2*S2)\n",
      "G= H1-(T+273.16)*S3\n",
      "#RESULTS\n",
      "print '%s %.1f %s' % (' Gibbs free energy=',G,'cal')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Gibbs free energy= -113380.4 cal\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 20 - pg 232"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the gibbs free energy change\n",
      "#initialisation of variables\n",
      "H= -57.7979 #kcal\n",
      "H1= -196.5 #kcal\n",
      "S1=45.106\n",
      "S2=6.49\n",
      "S3=21.5\n",
      "S4=31.211\n",
      "T=298.16\n",
      "#calculations\n",
      "dH=3*H-H1\n",
      "dS=3*S1+2*S2-S3-3*S4\n",
      "dG=dH*1000-T*dS\n",
      "#results\n",
      "print '%s %d %s' %(\"Gibbs free energy change =\",dG,\"cal\")\n",
      "print 'The answer is a bit different due to rounding off error in textbook'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Gibbs free energy change = 13217 cal\n",
        "The answer is a bit different due to rounding off error in textbook\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 22 - pg 240"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Gibbs free energy and value of A in both cases\n",
      "#initialisation of variables\n",
      "import math\n",
      "p= 1. #atm\n",
      "p1= 3. #atm\n",
      "R= 1.987 #cal/mole K\n",
      "T= 27. #C\n",
      "b= 0.0428 #l mole^-1\n",
      "a= 3.61 #l^2 atm mole^-1\n",
      "#CALCULATIONS\n",
      "G= R*(273+T)*math.log(p/p1)\n",
      "A= R*(273+T)*math.log(p/p1)\n",
      "G1= R*(273+T)*math.log(p/p1)+(b-(a/(0.08205*(T+273))))*(p-p1)*(R/0.08205)\n",
      "#RESULTS\n",
      "print '%s %.1f %s' % (' Gibs free energy=',G,'cal')\n",
      "print '%s %.1f %s' % (' \\n Value of dA=',A,'cal')\n",
      "print '%s %.1f %s' % (' \\n Gibs free energy=',G1,'cal')\n",
      "print '%s %.1f %s' % (' \\n Value of dA=',A,'cal')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Gibs free energy= -654.9 cal\n",
        " \n",
        " Value of dA= -654.9 cal\n",
        " \n",
        " Gibs free energy= -649.9 cal\n",
        " \n",
        " Value of dA= -654.9 cal\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 24 - pg 244"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the fugacities of both components\n",
      "#initialisation of variables\n",
      "import math\n",
      "b= 0.0386 #l**2 atm mole**-1\n",
      "a= 1.348 #l mole**-1\n",
      "R= 0.08205 #cal /mole K\n",
      "T= 25 #C\n",
      "a1= 3.61 #l**2 atm mole**-1\n",
      "b1= 0.0428 #l mole**-1\n",
      "P= 50 #atm\n",
      "P1= 1 #atm\n",
      "#CALCULATIONS\n",
      "Bn= b-(a/(R*(273.2+T)))\n",
      "Bc= b1-(a1/(R*(273.2+T))) \n",
      "Fn= P1*math.e**(Bn*P1/(R*(273.2+T)))\n",
      "Fc= P1*math.e**(Bc*P1/(R*(273.2+T)))\n",
      "Fn1= P*math.e**(Bn*P/(R*(273.2+T)))\n",
      "Fc1= P*math.e**(Bc*P/(R*(273.2+T)))\n",
      "#RESULTS\n",
      "print '%s %.3f %.2f %s' % (' Fugacity of N2 at 1 and 50 atm are respectively =',Fn,Fn1,'atm')\n",
      "print '%s %.3f %.2f %s' % (' \\n Fugacity of CO2 at 1 and 50 atm are respectively =',Fc,Fc1,'atm')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Fugacity of N2 at 1 and 50 atm are respectively = 0.999 48.34 atm\n",
        " \n",
        " Fugacity of CO2 at 1 and 50 atm are respectively = 0.996 40.37 atm\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 25 - pg 245"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Increase in pressure in all cases\n",
      "#initialisation of variables\n",
      "import math\n",
      "P1= 23.756 #atm\n",
      "T= 25. #C\n",
      "P2= 1. #atm\n",
      "P3= 10. #atm\n",
      "P4= 100. #atm\n",
      "R= 82.02 #J/mole K\n",
      "v= 18.07 #ml\n",
      "#CALCULATIONS\n",
      "p1= P1/760.\n",
      "p2= 10**(math.log10(P1)+(v*(P2-p1)/(2.303*R*(273.2+T))))\n",
      "p3= 10**(math.log10(P1)+(v*(P3-p1)/(2.303*R*(273.2+T))))\n",
      "p4= 10**(math.log10(P1)+(v*(P4-p1)/(2.303*R*(273.2+T))))\n",
      "x= -(P1-p2)*100/P1\n",
      "x1= -(P1-p3)*100/P1\n",
      "x2= -(P1-p4)*100/P1\n",
      "#RESULTS\n",
      "print '%s %.2f %s' % ('Increase in pressure=',x,'percent')\n",
      "print '%s %.2f %s' % ('Increase in pressure=',x1,' percent')\n",
      "print '%s %.1f %s' % ('Increase in pressure=',x2,' percent')\n",
      "print '%s %.3f %s' %('Vapor pressure at 1 atm',p2,'mm Hg')\n",
      "print '%s %.3f %s' %('Vapor pressure at 10 atm',p3,'mm Hg')\n",
      "print '%s %.3f %s' %('Vapor pressure at 100 atm',p4,'mm Hg')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Increase in pressure= 0.07 percent\n",
        "Increase in pressure= 0.74  percent\n",
        "Increase in pressure= 7.7  percent\n",
        "Vapor pressure at 1 atm 23.773 mm Hg\n",
        "Vapor pressure at 10 atm 23.932 mm Hg\n",
        "Vapor pressure at 100 atm 25.577 mm Hg\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 26 - pg 247"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the pressure required\n",
      "#initialisation of variables\n",
      "H= 1436.3 #cal mole^-1\n",
      "d= 0.9999 #g ml^-1\n",
      "d1= 0.9168 #g ml^-1\n",
      "P= 1. #atm\n",
      "m= 18.02 #gm\n",
      "R= 1.987 #cal/mole K\n",
      "T= 2 #C\n",
      "#CALCULATIONS\n",
      "V= (P/d)-(P/d1)\n",
      "H1= H*82.05/(m*R) \n",
      "P1= H1*(-T)/(273*V)\n",
      "#RESULTS\n",
      "print '%s %.1f %s' % (' pressure required to decrease=',P1,'atm')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " pressure required to decrease= 266.0 atm\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 27 - pg 249"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the heat of vapourisation\n",
      "#initialisation of variables\n",
      "H= 540. #cal gram ^-1\n",
      "T= 95. #C\n",
      "T1= 100. #C\n",
      "m= 18. #gms\n",
      "R= 1.987 #cal /mole K\n",
      "P= 760. #mm of Hg\n",
      "#CALCULATIONS\n",
      "H1= m*H\n",
      "P1= P/(10**(H1*(T1-T)/(2.303*R*(273+T)*(273+T1))))\n",
      "#RESULTS\n",
      "print '%s %.1f %s' % (' heat of vapourisation=',P1,'mm of Hg')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " heat of vapourisation= 636.0 mm of Hg\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 28 - pg 249"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the vapor pressure of water\n",
      "#initialisation of variables\n",
      "H= 9720 #cal mole^-1\n",
      "P= 1 #atm\n",
      "R= 1.987 #cal /mole K\n",
      "T= 100 #C\n",
      "T1= 95 #C\n",
      "#CALCULATIONS\n",
      "r= P*H/(R*(273+T)**2)\n",
      "dP= r*(T1-T)\n",
      "P1= (P+dP)*626/0.824\n",
      "#RESULTS\n",
      "print '%s %d %s' % (' vapour pressure=',P1,'mm Hg')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " vapour pressure= 626 mm Hg\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 29 - pg 250"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the vapour pressure\n",
      "#initialisation of variables\n",
      "G= 145 #cal\n",
      "R= 1.987 #cal/mole K\n",
      "T= 95 #C\n",
      "#CALCULATIONS\n",
      "P= 10**(-G/(2.303*R*(273+T)))*(624/0.820)\n",
      "#RESULTS\n",
      "print '%s %d %s' % (' vapour pressure=',P,'mm Hg')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " vapour pressure= 624 mm Hg\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 30 - pg 250"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the molar heat of vapourisation\n",
      "#initialisation of variables\n",
      "import math\n",
      "R= 1.987 #cal/mole K\n",
      "T1= 25 #C\n",
      "T2= 76.8 #C\n",
      "P2= 760. #mm\n",
      "P1= 115. #mm\n",
      "#CALCULATIONS\n",
      "H= 2.303*R*(273.2+T1)*(273.2+T2)*math.log10(P2/P1)/(T2-T1)\n",
      "#RESULTS\n",
      "print '%s %d %s' % (' molar heat of vapourisation=',H,'cal mole^-1')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " molar heat of vapourisation= 7561 cal mole^-1\n"
       ]
      }
     ],
     "prompt_number": 17
    }
   ],
   "metadata": {}
  }
 ]
}