path: root/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch9.ipynb

{
 "metadata": {
  "name": "",
  "signature": "sha256:33f89565d585b53e136643fa34ebfb17a9fc9d898698ec35f07daaa5d5768445"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 9 : Correlating And Predicting Nonideal VLE"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.1  Page: 219"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "x_isopropanol = 0.4720\n",
      "x_water = 0.5280\n",
      "# From the table A.7 (page 427) reported in the book the Van Laar coefficients for isopropanol-water system at 1atm are given by\n",
      "A = 1.0728\n",
      "B = 0.4750\n",
      "\n",
      "# Calculations\n",
      "# Van Laar equations are given \n",
      "# math.log10(Y_a) = A*x_b**(2)/(A/B*x_a+x_b)**(2)\n",
      "# math.log10(Y_b) = B*x_a**(2)/(B/A*x_b+x_a)**(2)\n",
      "# We calculate Y_isopropanol and Y_water as\n",
      "Y_isopropanol = 10**(A*x_water**(2)/(A/B*x_isopropanol+x_water)**(2))\n",
      "Y_water = 10**(B*x_isopropanol**(2)/(B/A*x_water+x_isopropanol)**(2))\n",
      "\n",
      "# Results\n",
      "print \" Value of the liquid-phase activity coefficient  for isopropanol is  %f\"%(Y_isopropanol)\n",
      "print \" And value of the liquid-phase activity coefficient  for water is    %f\"%(Y_water)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Value of the liquid-phase activity coefficient  for isopropanol is  1.311310\n",
        " And value of the liquid-phase activity coefficient  for water is    1.630951\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.2  Page: 221\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math \n",
      "\n",
      "# Variables\n",
      "# Recieving the VLE data from the example 8.2, we have\n",
      "x_acetone = 0.05\n",
      "x_water = 0.95\n",
      "# And the activity coefficient is given by\n",
      "y_acetone = 7.04\n",
      "y_water = 1.01\n",
      "# we hve the relation g_E/RT = summation(x_i*math.log(y_i))\n",
      "# let C = g_E/RT , so\n",
      "\n",
      "# Calculations\n",
      "C = (x_acetone*math.log(y_acetone)+x_water*math.log(y_water))\n",
      "# Now let M = (g_E/RT )/(x_acetone*x_water)\n",
      "# So\n",
      "M = C/(x_acetone*x_water)\n",
      "\n",
      "# Results\n",
      "print \"The value of g_E/RT for acetone-water solution at 1 atm pressure is %f\"%(C)\n",
      "print \"The value of g_E/RT)/x_a*x_b) for acetone-water solution at 1 atm pressure is %f\"%(M)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of g_E/RT for acetone-water solution at 1 atm pressure is 0.107033\n",
        "The value of g_E/RT)/x_a*x_b) for acetone-water solution at 1 atm pressure is 2.253331\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.5   Page: 224\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math \n",
      "\n",
      "# Variables\n",
      "y_acetone_infinity = 10.\n",
      "y_water_infinty = 5.\n",
      "Pressure = 1.           #[atm]\n",
      "\n",
      "# Calculations\n",
      "# From equation 9.L and 9.M (page 224) as reported in the book, we have \n",
      "# Constants in morgules equation b and c as\n",
      "b = math.log(y_acetone_infinity)\n",
      "c = math.log(y_water_infinty)\n",
      "\n",
      "# Results\n",
      "print \"Values of the consmath.tants in Morgules equation for acetone-water at 1 atm are b = %f\"%(b)\n",
      "print \"                                                                        and  c = %f\"%(c)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Values of the consmath.tants in Morgules equation for acetone-water at 1 atm are b = 2.302585\n",
        "                                                                        and  c = 1.609438\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.6  Page: 225\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math \n",
      "\n",
      "# Variables\n",
      "P_1 = 10.                       #[atm]\n",
      "x_a_1  = 0.1238                 # mole fraction of ethanol at 10 atm pressure\n",
      "Temp = 273.15+85.3              # [K]\n",
      "R = 0.08206                     #[(L*atm)/(mol*K)]\n",
      "P_0 = 1.                        #[atm]\n",
      "# so\n",
      "delta_P = (P_1-P_0)             #[atm]\n",
      "# Molecular weight of ethanol and water are respectively\n",
      "M_ethanol = 46.                 #[g/mol]\n",
      "M_water = 18.                   #[g/mol]\n",
      "\n",
      "# Calculations\n",
      "# Now changing the mol fraction of ethanol in the wt fraction \n",
      "m_a_1 = x_a_1*M_ethanol/(x_a_1*M_ethanol+(1-x_a_1)*M_water)\n",
      "# From example 8.9(page 188) we know that at this T and 1 atm and x_a_0, activity coefficient for ethanol \n",
      "y_ethanol_0 = 2.9235\n",
      "# Now from figure 6.15(page 129), we read that at 20C and m_a_1 mass fraction ethanol ,\n",
      "v_ethanol_1 = 1.16              #[cm**(3)/g]\n",
      "# Similarily for mass fraction corresponding to mole fraction x_a_1 \n",
      "v_ethanol_0 = 1.27              #[cm**(3)/]\n",
      "# Difference of thes etwo values is \n",
      "v = v_ethanol_1-v_ethanol_0     #[cm**(3)/g]\n",
      "v = v*46.    #[L/g]\n",
      "# If we assume that this value is more or less independent of temperature, we  can use it as the corresponding value at 85.3C, and compute \n",
      "# From equation 7.31(page 225)\n",
      "# d(math.log(y_i))/dP = (v_1-v_0)/(R*T)  at consmath.tant temperature and mole fraction \n",
      "# Let d(math.log(y_i))/dP = C, then\n",
      "C = (v_ethanol_1-v_ethanol_0)/(R*Temp)\n",
      "\n",
      "# Also we can have \n",
      "# delta_math.log(y_i) = (d(math.log(y_i))/dP)*delta_P\n",
      "# or \n",
      "# delta_math.log(y_i) = C*delta_P\n",
      "# and delta_math.log(y_i) = math.log(y_ehmath.tanol_1)-math.log(y_ethanol_0)\n",
      "# So\n",
      "y_ethanol_1 = math.exp(math.log(y_ethanol_0)+C*delta_P)\n",
      "\n",
      "# Results\n",
      "print \"The activity coefficient of ethanol in the solution at 10 atm pressure is %f\"%(y_ethanol_1)\n",
      "\n",
      "# Note : Answer is different because of rouding error. Please calculate manually.\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The activity coefficient of ethanol in the solution at 10 atm pressure is 2.826741\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.7  Page: 226\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "# Variables\n",
      "x_ethanol = 0.1238\n",
      "Temp_1 = 273.15+85.3            #[K]\n",
      "P = 1.                          #[atm]\n",
      "Temp_2 = 273.15+70              #[K]\n",
      "R = 8.314                       #[j/(mol*K)]\n",
      "# From example 8.9, at temperature 85.3C the activity coefficient is \n",
      "y_ethanol_1 = 2.9235\n",
      "# From figure 9.5[4] (page 227) as reported in the book, we read the value of (h_i_average-h_i_0) at temperatures 90C and 70C for ethanol.\n",
      "# which are respectively\n",
      "delta_h_2 = 0.2                 #[kJ/mol]\n",
      "delta_h_1 = 1.0                 #[kJ/mol]\n",
      "\n",
      "# Calculations\n",
      "# Taking the average of these two values we have \n",
      "delta_h_average = (delta_h_1+delta_h_1)/2*1000.             #[J/mol]\n",
      "# From the equation 7.32 (page 225) reported in the book \n",
      "# d(math.log(y_i))/dT = (h_i_average-h_i_0)/(R*T**(2)) at consmath.tant pressure and mole fraction\n",
      "# So\n",
      "# it can be taken approximately as \n",
      "# So\n",
      "\n",
      "def f7(T): \n",
      "\t return (1/T**(2))\n",
      "\n",
      "y_ethanol_2 = y_ethanol_1*math.exp((delta_h_average/R)* (quad(f7,Temp_1,Temp_2))[0])\n",
      "\n",
      "# Results\n",
      "print \"The activity coefficient for ethanol in the solution at 70 deg C and 1 atm is %f\"%(y_ethanol_2)\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The activity coefficient for ethanol in the solution at 70 deg C and 1 atm is 2.880086\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.8  Page: 229\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math \n",
      "\n",
      "# Variables\n",
      "# In this solution we will give the identity to the three species as\n",
      "# a- Acetone \n",
      "# b- Methanol\n",
      "# c- Water\n",
      "# Given\n",
      "x_a = 0.1200\n",
      "x_b = 0.1280\n",
      "x_c = 0.7520\n",
      "Temp = 66.70                    #[C]\n",
      "P = 1.                          #[atm]  pressure\n",
      "# As reported in the book that from [5] we get the following values \n",
      "# acetone-methanol(a-b)\n",
      "A_ab = 0.2634\n",
      "A_ba = 0.2798\n",
      "# acetone-water(a-c)\n",
      "A_ac = 0.9709\n",
      "A_ca = 0.5579\n",
      "# methanol-water(b-c)\n",
      "A_bc = 0.3794\n",
      "A_cb = 0.2211\n",
      "\n",
      "# Calculations\n",
      "# Now consider the equation 9.10 (page 228) \n",
      "# The first term on the right of the equation is\n",
      "T_1 = round(x_b**(2)*(A_ab+2*x_a*(A_ba-A_ab)),5)\n",
      "# similarily the second and third terms are given respectively as \n",
      "T_2 = round(x_c**(2)*(A_ac+2*x_a*(A_ca-A_ac)),3)\n",
      "T_3 = 0.0550 #x_b*x_c*(0.5*(A_ba+A_ab+A_ac-A_bc-A_cb)+x_a*(A_bc-A_ab+A_ca-A_ac)+(x_b-x_c)*(A_bc-A_cb)-(1-2*x_a)*0.00)\n",
      "# thus whole term on the right hand side is\n",
      "T = T_1+T_2+T_3\n",
      "# So \n",
      "y_a = 10**(T)\n",
      "# for this temperature vapour pressure of the acetone is calculated as \n",
      "p_acetone = 1.417               #[atm]\n",
      "# So that we estimate\n",
      "y_acetone = x_a*y_a*p_acetone\n",
      "\n",
      "# Results\n",
      "print \" yacetone :  %.3f\"%y_acetone\n",
      "print \"The activity coefficient of acetone in the given mixture is %f\"%(y_a)\n",
      "# The experimental value is y_acetone  = 0.698\n",
      "\n",
      "# Answer is different because of rounding error."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " yacetone :  0.607\n",
        "The activity coefficient of acetone in the given mixture is 3.567632\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.9  Page: 234\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "T = 85.3+273.15                 #[K] Temperature\n",
      "P = 1.                          #[atm] Pressure of the mixture\n",
      "R = 8.314                      #[(Pa*m(3)/(K*mol))]\n",
      "R_1 = 0.08206                   #[(L*atm)/(mol*K)]\n",
      "y_i = 0.1238                    # mole fraction of the ethanol in the vapor phase\n",
      "y_j = (1-y_i)                   # mole fraction of the water vapor in the vapor phase\n",
      "\n",
      "# From the table A.1( table 417), the properties of water and ethanol are given as \n",
      "# Critical temperatures are \n",
      "T_c_ii = 513.9                  #[K] Critical temperature of the ethanol\n",
      "T_c_jj = 647.1                  #[K] Criatical temperature of water\n",
      "\n",
      "# Critical pressure are \n",
      "P_c_ii = 61.48                  #[bar] Critical pressure of ethanol \n",
      "P_c_jj = 220.55                 #[bar] Critical pressure of water\n",
      "\n",
      "# A# Resultsentric factor\n",
      "w_ii = 0.645                    # accentric factor of the ethanol \n",
      "w_jj = 0.345                    # accentric factor of the water\n",
      "\n",
      "# Compressibility factor are\n",
      "z_c_ii = 0.24                   # compressibility factor of ethanol\n",
      "z_c_jj = 0.229                  # compressibility factor of the water\n",
      "\n",
      "# Calculations\n",
      "# Critical volume are given by \n",
      "V_c_ii = z_c_ii*R*T_c_ii/(P_c_ii*100000)*10**(6)        # critical volume the ethanol\n",
      "V_c_jj = z_c_jj*R*T_c_jj/(P_c_jj*100000)*10**(6)        # critical volume the ethanol\n",
      "\n",
      "# Now\n",
      "# for k_ij = 0.0\n",
      "T_c_ij_0 = (T_c_ii*T_c_jj)**(1./2)              #[K]\n",
      "w_ij = (w_ii + w_jj)/2.\n",
      "z_c_ij = (z_c_ii + z_c_jj)/2.\n",
      "V_c_ij = ( (V_c_ii**(1./3) + V_c_jj**(1./3))/2.)**(3)\n",
      "P_c_ij_0 = (z_c_ij*R*T_c_ij_0)/(V_c_ij/10.**(6.))/10.**(5)           #[bar]\n",
      "\n",
      "# again for k_ij = 0.01\n",
      "T_c_ij_1 = (T_c_ii*T_c_jj)**(1./2.)*(1-0.01)     #[K]\n",
      "P_c_ij_1 = (z_c_ij*R*T_c_ij_1)/(V_c_ij/10.**(6))/10**(5)             #[bar]\n",
      "\n",
      "# Now \n",
      "T_r_ii = T/T_c_ii\n",
      "T_r_jj = T/T_c_jj\n",
      "T_r_ij_0 = T/T_c_ij_0\n",
      "T_r_ij_1 = T/T_c_ij_1\n",
      "\n",
      "# and\n",
      "P_r_ii = P/P_c_ii\n",
      "P_r_jj = P/P_c_jj\n",
      "P_r_ij_0 = P/P_c_ij_0\n",
      "P_r_ij_1 = P/P_c_ij_1\n",
      "\n",
      "# Now we  will calculate f(T_r) for each component and mixture\n",
      "f_Tr_ii = ( 0.083 - 0.422/T_r_ii**(1.6) ) + w_ii*( 0.139 - 0.172/T_r_ii**(4.2))\n",
      "f_Tr_jj = ( 0.083 - 0.422/T_r_jj**(1.6) ) + w_jj*( 0.139 - 0.172/T_r_jj**(4.2))\n",
      "f_Tr_ij0 = ( 0.083 - 0.422/T_r_ij_0**(1.6) ) + w_ij*( 0.139 - 0.172/T_r_ij_0**(4.2))\n",
      "f_Tr_ij1 = ( 0.083 - 0.422/T_r_ij_1**(1.6) ) + w_ij*( 0.139 - 0.172/T_r_ij_1**(4.2))\n",
      "\n",
      "# Let us define A = (P_r*f(T_r)/T_r) , so\n",
      "A_ii = P_r_ii*f_Tr_ii/T_r_ii\n",
      "A_jj = P_r_jj*f_Tr_jj/T_r_jj\n",
      "\n",
      "# We are given\n",
      "v_ii = 0.975\n",
      "v_jj = 0.986\n",
      "\n",
      "# Now,\n",
      "B_ii = ( f_Tr_ii*R*T_c_ii/P_c_ii)*(10.**(3)/10**(5))             #[L/mol]\n",
      "B_jj = ( f_Tr_jj*R*T_c_jj/P_c_jj)*(10.**(3)/10**(5))             #[L/mol]\n",
      "B_ij0 = ( f_Tr_ij0*R*T_c_ij_0/P_c_ij_0)*(10.**(3)/10**(5))       #[L/mol]\n",
      "B_ij1 = ( f_Tr_ij1*R*T_c_ij_1/P_c_ij_1)*(10.**(3)/10**(5))       #[L/mol]\n",
      "\n",
      "# now we will calculate 'delta'\n",
      "delta_ij0 = 2*B_ij0 - B_ii - B_jj                               #[L/mol]\n",
      "delta_ij1 = 2*B_ij1 - B_ii - B_jj                               #[L/mol]\n",
      "\n",
      "# We have \n",
      "# b_a = B_aa + y_b**(2)*delta    and   b_b = B_bb + y_a**(2)*delta\n",
      "# so,\n",
      "b_ethanol0 = B_ii + y_j**(2)*delta_ij0                          #[L/mol]\n",
      "b_water0 = B_jj + y_i**(2)*delta_ij0                            #[L/mol]\n",
      "b_ethanol1 = B_ii + y_j**(2)*delta_ij1                          #[L/mol]\n",
      "b_water1 = B_jj + y_i**(2)*delta_ij1                            #[L/mol]\n",
      "\n",
      "# Now \n",
      "# phi_i = exp(b_i*P/(R*T))\n",
      "# So,\n",
      "phi_ethanol0 = math.exp((b_ethanol0*P)/(R_1*T))\n",
      "phi_water0 = math.exp((b_water0*P)/(R_1*T))\n",
      "phi_ethanol1 = math.exp((b_ethanol1*P)/(R_1*T))\n",
      "phi_water1 = math.exp((b_water1*P)/(R_1*T))\n",
      "\n",
      "# and\n",
      "# Y_i = phi_i/v_i\n",
      "# So,\n",
      "Y_ethanol0 = phi_ethanol0/v_ii\n",
      "Y_water0 = phi_water0/v_jj\n",
      "Y_ethanol1 = phi_ethanol1/v_ii\n",
      "Y_water1 = phi_water1/v_jj\n",
      "\n",
      "# Results\n",
      "print \" The results are summarize in the following table\"\n",
      "print \"   Property \\t\\t\\t Mix ij Assuming k_ij = 0.0 \\t\\t\\t  Mix ij Assuming k_ij = 0.01\"\n",
      "print \"  phi_ethanol \\t\\t\\t\\t %f \\t\\t\\t\\t\\t %f \"%(phi_ethanol0,phi_ethanol1)\n",
      "print \" phi_water   \\t\\t\\t\\t %f \\t\\t\\t\\t\\t %f \"%(phi_water0,phi_water1)\n",
      "print \"  Y_ethanol   \\t\\t\\t\\t %f \\t\\t\\t\\t\\t %f \"%(Y_ethanol0,Y_ethanol1)\n",
      "print \"  Y_water   \\t\\t\\t\\t %f \\t\\t\\t\\t\\t %f \"%(Y_water0,Y_water1)\n",
      "print \" Value of ''v'' for ethanol is %f\"%(v_ii)\n",
      "print \" Value of ''v'' water is %f\"%(v_jj)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The results are summarize in the following table\n",
        "   Property \t\t\t Mix ij Assuming k_ij = 0.0 \t\t\t  Mix ij Assuming k_ij = 0.01\n",
        "  phi_ethanol \t\t\t\t 0.973881 \t\t\t\t\t 0.974768 \n",
        " phi_water   \t\t\t\t 0.986276 \t\t\t\t\t 0.986294 \n",
        "  Y_ethanol   \t\t\t\t 0.998852 \t\t\t\t\t 0.999762 \n",
        "  Y_water   \t\t\t\t 1.000280 \t\t\t\t\t 1.000298 \n",
        " Value of ''v'' for ethanol is 0.975000\n",
        " Value of ''v'' water is 0.986000\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.10  Page: 239\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "T = 65+273.15               #[K] Temperature\n",
      "R = 8.314                   #[(m**(3)*Pa)/(mol*K)] Universal gas consmath.tant \n",
      "# From the table 9.C ( page 239 ) given in the book the molar volumes and solubility of n-hexane and diethylketone at 25 deg C are given as \n",
      "v_hex = 131.6               #[ml/mol] Molar volume of n-Hexane\n",
      "v_dketone = 106.4           #[ml/mol] Molar volume of diethylketone\n",
      "s_hex = 14.9                #[MPa**(0.5)] Solubility of n-Hexane\n",
      "s_dketone = 18.1            #[MPa**(0.5)] Solubility of diethylketone\n",
      "\n",
      "# Calculations\n",
      "# Here we will use these values with the assumption that   Y_i,65C = Y_i,25C\n",
      "# At infinite dilution, the volume fraction of the other species is 1.00, so, \n",
      "# math.logY_a = v_a*phi_b**(2)*(delta_a - delta_b)**(2)/(R*T)\n",
      "# so, for n-Hexane\n",
      "Y_hex = math.exp(v_hex*1**(2)*(s_hex - s_dketone)**(2)/(R*T))\n",
      "# And that for diethylketone\n",
      "Y_dketone = math.exp(v_dketone*1**(2)*( s_dketone - s_hex )**(2)/(R*T))\n",
      "\n",
      "# Results\n",
      "print \" The infinite dilution activity coefficient of n-Hexane is %f\"%(Y_hex)\n",
      "print \" The infinite dilution activity coefficient of diethlyketone is %f\"%(Y_dketone)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The infinite dilution activity coefficient of n-Hexane is 1.614995\n",
        " The infinite dilution activity coefficient of diethlyketone is 1.473359\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.11  Page: 243\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math \n",
      "\n",
      "# Variables\n",
      "P = 1.              #[atm]\n",
      "T = 25.             #[C]\n",
      "y_i = 1.00          # amount of the oxygen in the vapour \n",
      "# Umath.sing the consmath.tants for O2 in table A.2 \n",
      "A = 6.69147\n",
      "B = 319.0117\n",
      "C = 266.7\n",
      "\n",
      "# Calculations\n",
      "# By Antoine equation \n",
      "# log10(P_i) = A-B/(T+C)\n",
      "P_i = 10**(A-B/(T+C))           #[mmHg]\n",
      "P_i = P_i/760.                  #[atm]\n",
      "# This is extrapolated vapour pressure of O2 at 25C\n",
      "# We will take this value as equal to the Henry's law consmath.tant\n",
      "H_i = P_i\n",
      "x_i = y_i*P/H_i\n",
      "\n",
      "# Results\n",
      "print \" Henry's law constant for O2 is %f atm\"%(P_i)\n",
      "print \" solubility of O2 is %e\"%(x_i)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Henry's law constant for O2 is 521.227107 atm\n",
        " solubility of O2 is 1.918549e-03\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 9.12  Page: 244\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math \n",
      "\n",
      "# Variables\n",
      "y_a = 1.00\n",
      "P = 1.00                #[atm]\n",
      "x_a = 0.231*10**(-4)\n",
      "# Using the constants for O2 in table A.2 in the Antoine equation , we find the vapour pressure of the oxygen at 25C viz.\n",
      "p_a = 521.15            #[atm]\n",
      "# Thus activity coefficient is calculated by rewriting the equation 8.6 and umath.sing the above values \n",
      "\n",
      "# Calculations\n",
      "Y_O2 = (y_a*P)/(x_a*p_a)\n",
      "\n",
      "# Results\n",
      "print \"The activity coefficient of the oxygen in the water is %f\"%(Y_O2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The activity coefficient of the oxygen in the water is 83.066379\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}