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 "metadata": {
  "name": "ch14_1",
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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 14 : Activity Coefficients Models for Liquid Mixtures"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.4  Page Number : 461"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math\n",
      "\n",
      "# Variables,\n",
      "T = 300;\t\t\t#[K] - Temperature\n",
      "b = 100;\t\t\t#[cal/mol]\n",
      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
      "# R*T*math.log(Y_1) = b*x_2**(2)\n",
      "# R*T*math.log(Y_2) = b*x_1**(2)\n",
      "\n",
      "#For equimolar mixture\n",
      "x_1 = 0.5;\t\t\t#Mole fraction of component 1\n",
      "x_2 = 0.5;\t\t\t#Mole fraction of component 2\n",
      "\n",
      "# Calculations and Results\n",
      "#The excess Gibbs free energy is given by\n",
      "# G_excess = R*T*(x_1*math.log(Y_1) + x_2*math.log(Y_2)) = b*x_1*x_2**(2) + b*x_2*x_1**(2) = b*x_1*(x_1 + x_2) = b*x_1*x_2\n",
      "G_excess = b*x_1*x_2;\t\t\t#[cal/mol]\n",
      "\n",
      "#The ideal Gibbs free energy change of mixing is given by,\n",
      "delta_G_id_mix = R*T*(x_1*math.log(x_1)+x_2*math.log(x_2));\t\t\t#[cal/mol]\n",
      "\n",
      "#The Gibbs free energy of mixing is given by\n",
      "delta_G_mix = delta_G_id_mix + G_excess;\t\t\t#[cal/mol]\n",
      "\n",
      "\n",
      "# delta_S_mix = delta_S_id_mix = - R*sum(x_i*math.log(x_i))\n",
      "\n",
      "#delta_G_mix = delta_H_mix - T*delta_S_mix = delta_H_mix + R*T*(x_1*math.log(x_1)+x_2*math.log(x_2))\n",
      "delta_H_mix = b*x_1*x_2;\t\t\t#[cal/mol]\n",
      "\n",
      "print \"The value of Gibbs free energy change for equimolar mixture formation is %f cal/mol\"%(delta_G_mix);\n",
      "print \"The value of enthalpy change for equimolar mixture formation is %f cal/mol\"%(delta_H_mix);\n",
      "\n",
      "#Work required for separation of mixture into pure components is\n",
      "W = delta_G_mix;\n",
      "print \"The least amount of work required for separation at 300 K is %f cal/mol\"%(W);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of Gibbs free energy change for equimolar mixture formation is -388.185034 cal/mol\n",
        "The value of enthalpy change for equimolar mixture formation is 25.000000 cal/mol\n",
        "The least amount of work required for separation at 300 K is -388.185034 cal/mol\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.10  Page Number : 466"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "import math \n",
      "from numpy import zeros\n",
      "from scipy.stats import linregress\n",
      "\n",
      "# Variables,\n",
      "T = 25 +173.15;\t\t\t#[K] - Temperature\n",
      "x_1=[0.0115,0.0160,0.0250,0.0300,0.0575,0.1125,0.1775,0.2330,0.4235,0.5760,0.6605,0.7390,0.8605,0.9250,0.9625];\n",
      "y_1=[8.0640,7.6260,7.2780,7.2370,5.9770,4.5434,3.4019,2.8023,1.7694,1.3780,1.2302,1.1372,1.0478,1.0145,1.0070];\n",
      "y_2=[1.0037,1.0099,1.0102,1.0047,1.0203,1.0399,1.1051,1.1695,1.4462,1.8520,2.2334,2.6886,3.7489,4.8960,5.6040];\n",
      "\n",
      "x_2 = zeros(15);\t\t\t# x_2 = (1 - x_1)\n",
      "G_RT = zeros(15);\t\t\t# G_RT = G_excess/(R*T)\n",
      "x1x2_G_RT = zeros(15);\t\t\t# x1x2_G_RT = (x_1*x_2/(G_excess/(R*T)))\n",
      "G_RTx1x2 = zeros(15);\t\t\t# G_RTx1x1 = G_excess/(R*T*x_1*x_2)\n",
      "\n",
      "# Calculations and Results\n",
      "for i in range(15):\n",
      "    x_2[i]=(1-x_1[i]);\n",
      "    G_RT[i]=(x_1[i]*math.log(y_1[i]))+(x_2[i]*math.log(y_2[i]));\n",
      "    x1x2_G_RT[i]=(x_1[i]*x_2[i])/G_RT[i];\n",
      "    G_RTx1x2[i]=1/x1x2_G_RT[i];\n",
      "\n",
      "\n",
      "slop,intr,sig,d,e=linregress(x_1,x1x2_G_RT);\n",
      "\n",
      "A = 1/intr;\n",
      "B = 1/(slop+(1/A));\n",
      "print \" The value of van Laar parameters are A = %f and B = %f\"%(A,B);\n",
      "\n",
      "# Now from Margules equation\n",
      "# G_RTx1x2 = G_excess/(R*T*x_1*x_2) = B1*x_1 + A1*x_1 = A1 + (B1 - A1)*x_1\n",
      "#slope = (B1 - A1) and intercept = A1\n",
      "\n",
      "# Again employing the concept of linear regression of the data ( x_1 , G_RTx1x2 ) to find the value of intercept and slope of the above equation\n",
      "#Let slope = slop1 and intercept = intr1\n",
      "\n",
      "slop1,intr1,sig1,d,e=linregress(x_1,G_RTx1x2);\n",
      "\n",
      "A1 = intr1;\n",
      "B1 = slop1 + A1;\n",
      "print \" The value of Margules parameters are A = %f and B = %f\"%(A1,B1);\n",
      "\n",
      "print \" Because of the higher value of correlation factor for Van Laar model it fits the data better.\"\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The value of van Laar parameters are A = 2.271326 and B = 1.781420\n",
        " The value of Margules parameters are A = 2.293274 and B = 1.746000\n",
        " Because of the higher value of correlation factor for Van Laar model it fits the data better.\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.11  Page Number : 470"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "# Variables,\n",
      "T = 60 + 273.15;\t\t\t#[K] - Temperature\n",
      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
      "#component 1 = acetone\n",
      "#component 2 = water\n",
      "x_1 = 0.3;\t\t\t# Mole fraction of component 1\n",
      "x_2 = 1 - x_1;\t\t\t#Mole fraction of component 2\n",
      "V_mol_1 = 74.05;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 1\n",
      "V_mol_2 = 18.07;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2\n",
      "\n",
      "#for Wilson equation\n",
      "a_12 = 291.27;\t\t\t#[cal/mol]\n",
      "a_21 = 1448.01;\t\t\t#[cal/mol]\n",
      "\n",
      "#For NRTL\n",
      "b_12 = 631.05;\t\t\t#[cal/mol]\n",
      "b_21 = 1197.41;\t\t\t#[cal/mol]\n",
      "alpha = 0.5343;\n",
      "\n",
      "# Calculations and Results\n",
      "#Froom wilson equation\n",
      "A_12=(V_mol_2/V_mol_1)*(math.exp(-a_12/(R*T)));\n",
      "A_21 = (V_mol_1/V_mol_2)*(math.exp(-a_21/(R*T)));\n",
      "Beta = A_12/(x_1+x_2*A_12) - A_21/(x_2+x_1*A_21);\n",
      "#math.log(Y1) = -math.log(x_1 + x_2*A_12) + x_2*Beta; \n",
      "Y1 = math.exp(-math.log(x_1+x_2*A_12)+x_2*Beta);\n",
      "#similarly for Y2\n",
      "Y2 = math.exp(-math.log(x_2+x_1*A_21)-x_1*Beta);\n",
      "print \"The value of activity coefficients for Wilson equation are Y1 = %f \\t and \\t Y2 = %f\"%(Y1,Y2);\n",
      "\n",
      "#From NRTL equation,\n",
      "t_12 = b_12/(R*T);\n",
      "t_21 = b_21/(R*T);\n",
      "G_12 = math.exp(-alpha*t_12);\n",
      "G_21 = math.exp(-alpha*t_21);\n",
      "\n",
      "#math.log(Y1) = x_1**(2)*[t_12*(G_12/(x_1+x_2*G_12))**(2) + (G_12*t_12)/((G_12/(x_1+x_2*G_12))**(2))]\n",
      "Y1_prime = math.exp(x_2**(2)*(t_21*(G_21/(x_1+x_2*G_21))**(2)+(G_12*t_12)/(((x_2+x_1*G_12))**(2))));\n",
      "#Similarly for Y2\n",
      "Y2_prime = math.exp(x_1**(2)*(t_12*(G_12/(x_2+x_1*G_12))**(2)+(G_21*t_21)/(((x_1+x_2*G_21))**(2))));\n",
      "\n",
      "print \"The value of activity coefficients for NRTL equation are Y1 = %f \\t and \\t Y2 = %f\"%(Y1_prime,Y2_prime);\n",
      "\n",
      "\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of activity coefficients for Wilson equation are Y1 = 2.172258 \t and \t Y2 = 1.254121\n",
        "The value of activity coefficients for NRTL equation are Y1 = 2.143030 \t and \t Y2 = 1.262506\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.12  Page Number : 474"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "# Variables \n",
      "T = 307;\t\t\t#[K]\n",
      "x_1 = 0.047;\n",
      "x_2 = 1 - x_1;\n",
      "\n",
      "# The subgroups in the  two components are\n",
      "# Acetone (1) : 1 CH3, 1 CH3CO\n",
      "# n-pentane (2) : 2 CH3, 3 CH2\n",
      "\n",
      "#Group volume (Rk) and surface area (Qk) parameters of the subgroup are\n",
      "k_CH3 = 1;\n",
      "k_CH2 = 2;\n",
      "k_CH3CO = 19;\n",
      "Rk_CH3 = 0.9011;\n",
      "Rk_CH2 = 0.6744;\n",
      "Rk_CH3CO = 1.6724;\n",
      "Qk_CH3 = 0.848;\n",
      "Qk_CH2 = 0.540;\n",
      "Qk_CH3CO = 1.4880;\n",
      "\n",
      "# Interaction parameters of the subgroups in kelvin (K) are\n",
      "a_1_1 = 0;\n",
      "a_1_2 = 0;\n",
      "a_1_19 = 476.40;\n",
      "a_2_1 = 0;\n",
      "a_2_2 = 0;\n",
      "a_2_19 = 476.40;\n",
      "a_19_1 = 26.76;\n",
      "a_19_2 = 26.76;\n",
      "a_19_19 = 0;\n",
      "\n",
      "# Calculations\n",
      "r_1 = 1*Rk_CH3 + 1*Rk_CH3CO;\n",
      "r_2 = 2*Rk_CH3 + 3*Rk_CH2;\n",
      "q_1 = 1*Qk_CH3 + 1*Qk_CH3CO;\n",
      "q_2 = 2*Qk_CH3 + 3*Qk_CH2;\n",
      "\n",
      "J_1 = r_1/(r_1*x_1+r_2*x_2);\n",
      "J_2 = r_2/(r_1*x_1+r_2*x_2);\n",
      "L_1 = q_1/(q_1*x_1+q_2*x_2);\n",
      "L_2 = q_2/(q_1*x_1+q_2*x_2);\n",
      "t_1_1 = math.exp(-a_1_1/T);\n",
      "t_1_2 = math.exp(-a_1_2/T);\n",
      "t_1_19 = math.exp(-a_1_19/T);\n",
      "t_2_1 = math.exp(-a_2_1/T);\n",
      "t_2_2 = math.exp(-a_2_2/T);\n",
      "t_2_19 = math.exp(-a_2_19/T);\n",
      "t_19_1 = math.exp(-a_19_1/T);\n",
      "t_19_2 = math.exp(-a_19_2/T);\n",
      "t_19_19 = math.exp(-a_19_19/T);\n",
      "\n",
      "e_1_1 = 1*Qk_CH3/q_1;\n",
      "e_2_1 = 0;\n",
      "e_19_1 = (1*Qk_CH3CO/q_1);\n",
      "e_1_2 = 2*Qk_CH3/q_2;\n",
      "e_2_2 = 3*Qk_CH2/q_2;\n",
      "e_19_2 = 0;\n",
      "\n",
      "B_1_1 = e_1_1*t_1_1 + e_2_1*t_2_1 + e_19_1*t_19_1;\n",
      "B_1_2 = e_1_1*t_1_2 + e_2_1*t_2_2 + e_19_1*t_19_2;\n",
      "B_1_19 = e_1_1*t_1_19 + e_2_1*t_2_19 + e_19_1*t_19_19;\n",
      "B_2_1 = e_1_2*t_1_1 + e_2_2*t_2_1 + e_19_2*t_19_1;\n",
      "B_2_2 = e_1_2*t_1_2 + e_2_2*t_2_2 + e_19_2*t_19_2;\n",
      "B_2_19 = e_1_2*t_1_19 + e_2_2*t_2_19 + e_19_2*t_19_19;\n",
      "\n",
      "theta_1 = (x_1*q_1*e_1_1 + x_2*q_2*e_1_2)/(x_1*q_1 + x_2*q_2);\n",
      "theta_2 = (x_1*q_1*e_2_1 + x_2*q_2*e_2_2)/(x_1*q_1 + x_2*q_2);\n",
      "theta_19 = (x_1*q_1*e_19_1 + x_2*q_2*e_19_2)/(x_1*q_1 + x_2*q_2);\n",
      "\n",
      "s_1 = theta_1*t_1_1 + theta_2*t_2_1 + theta_19*t_19_1;\n",
      "s_2 = theta_1*t_1_2 + theta_2*t_2_2 + theta_19*t_19_2;\n",
      "s_19 = theta_1*t_1_19 + theta_2*t_2_19 + theta_19*t_19_19;\n",
      "\n",
      "# math.log(Y1_C) = 1 - J_1 + math.log(J_1) - 5*q_1*(1- (J_1/L_1) + math.log(J_1/L_1))\n",
      "# math.log(Y2_C) = 1 - J_2 + math.log(J_2) - 5*q_2*(1- (J_2/L_2) + math.log(J_2/L_2))\n",
      "Y1_C = math.exp(1 - J_1 + math.log(J_1) - 5*q_1*(1- (J_1/L_1) + math.log(J_1/L_1)));\n",
      "Y2_C = math.exp(1 - J_2 + math.log(J_2) - 5*q_2*(1- (J_2/L_2) + math.log(J_2/L_2)));\n",
      "\n",
      "# For species 1\n",
      "summation_theta_k_1 = theta_1*(B_1_1/s_1) + theta_2*(B_1_2/s_2) + theta_19*(B_1_19/s_19);\n",
      "summation_e_ki_1 = e_1_1*math.log(B_1_1/s_1) + e_2_1*math.log(B_1_2/s_2) + e_19_1*math.log(B_1_19/s_19);\n",
      "\n",
      "# For species 2\n",
      "summation_theta_k_2 = theta_1*(B_2_1/s_1) + theta_2*(B_2_2/s_2) + theta_19*(B_2_19/s_19);\n",
      "summation_e_ki_2 = e_1_2*math.log(B_2_1/s_1) + e_2_2*math.log(B_2_2/s_2) + e_19_2*math.log(B_2_19/s_19);\n",
      "\n",
      "# math.log(Y1_R) = q_1(1 - summation_theta_k_1 + summation_e_ki_1)\n",
      "# math.log(Y2_R) = q_2(1 - summation_theta_k_2 + summation_e_ki_2)\n",
      "Y1_R = math.exp(q_1*(1 - summation_theta_k_1 + summation_e_ki_1));\n",
      "Y2_R = math.exp(q_2*(1 - summation_theta_k_2 + summation_e_ki_2));\n",
      "\n",
      "# math.log(Y1) = math.log(Y1_C) + math.log(Y1_R)\n",
      "# math.log(Y2) = math.log(Y2_C) + math.log(Y2_R)\n",
      "Y1 = math.exp(math.log(Y1_C) + math.log(Y1_R));\n",
      "Y2 = math.exp(math.log(Y2_C) + math.log(Y2_R));\n",
      "\n",
      "# Results\n",
      "print \" The activity coefficients are Y1 = %f  and  Y2 = %f\"%(Y1,Y2);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The activity coefficients are Y1 = 4.992034  and  Y2 = 1.005260\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.13  Page Number : 481"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "# Variables,\n",
      "T = 25 + 273.15;\t\t\t#[K] - Temperature\n",
      "R = 1.987;\t\t\t#[cal/mol*K] - Universal gas constant\n",
      "#component 1 = chloroform\n",
      "#component 2 = carbon tetrachloride\n",
      "x_1 = 0.5;\t\t\t#Mole fraction of component 1 \t\t\t#Equimolar mixture\n",
      "x_2 = 0.5;\t\t\t#Mole fraction of component 2\n",
      "V_mol_1 = 81;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 1\n",
      "V_mol_2 = 97;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2\n",
      "del_1 = 9.2;\t\t\t#[(cal/cm**(3))**(1/2)] - Mole fraction of component 1\n",
      "del_2 = 8.6;\t\t\t#[(cal/cm**(3))**(1/2)] - Mole fraction of component 2\n",
      "\n",
      "# Calculations\n",
      "#Scatchard - Hilderbrand model\n",
      "phi_1 = (x_1*V_mol_1)/(x_1*V_mol_1+x_2*V_mol_2);\n",
      "phi_2 = (x_2*V_mol_2)/(x_1*V_mol_1+x_2*V_mol_2);\n",
      "\n",
      "#math.log(Y1) = (V_mol_1/(R*T))*phi_1**(2)*(del_1-del_2)**(2)\n",
      "Y1 = math.exp((V_mol_1/(R*T))*(phi_1**(2))*((del_1-del_2)**(2)));\n",
      "\n",
      "#Similarly, for Y2\n",
      "Y2 = math.exp((V_mol_2/(R*T))*(phi_2**(2))*((del_1-del_2)**(2)));\n",
      "\n",
      "# Results\n",
      "print \"The value of activity coefficients for Scatchard-Hilderbrand model are Y1 = %f \\t and \\t Y2 = %f\"%(Y1,Y2);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of activity coefficients for Scatchard-Hilderbrand model are Y1 = 1.010245 \t and \t Y2 = 1.017658\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.14  Page Number : 485"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "# Variables,\n",
      "T = 25 + 273.15;\t\t\t#[K] - Temperature\n",
      "mol_HCl = 0.001;\t\t\t#[mol/kg] - Molality of HCl\n",
      "A = 0.510;\t\t\t#[(kg/mol)**(1/2)]\n",
      "Z_positive = 1;\t\t\t#Stoichiometric coefficient of 'H' ion\n",
      "Z_negative = -1;\t\t\t#Stoichiometric coefficient of 'Cl' ion\n",
      "m_H_positive = mol_HCl;\t\t\t#\n",
      "m_Cl_negative = mol_HCl;\n",
      "\n",
      "# Calculations and Results\n",
      "# I = 1/2*[((Z_positive)**(2))*m_H_positive + ((Z_negative)**(2))*m_Cl_negative]\n",
      "I = 1./2*(((Z_positive)**(2))*m_H_positive + ((Z_negative)**(2))*m_Cl_negative);\n",
      "\n",
      "#Using Debye-Huckel limiting law wee get,\n",
      "# math.log(Y1) = -A*(abs(Z_positive*Z_negative))*(I**(1/2)))\n",
      "Y = 10**(-A*(abs(Z_positive*Z_negative))*(I**(1./2)));\n",
      "print \"The mean activity coefficient at 25 C using Debye-Huckel limiting law is Y = %f\"%(Y);\n",
      "\n",
      "#Using Debye-Huckel extended model we get\n",
      "#math.log(Y_prime) = (-A*(abs(Z_positive*Z_negative))*(I**(1/2)))/(1 + (I**(1/2)));\n",
      "Y_prime = 10**((-A*(abs(Z_positive*Z_negative))*(I**(1./2)))/(1 + (I**(1./2))));\n",
      "print \"The mean activity coefficient at 25 C using Debye-Huckel extended model is Y = %f\"%(Y_prime);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The mean activity coefficient at 25 C using Debye-Huckel limiting law is Y = 0.963546\n",
        "The mean activity coefficient at 25 C using Debye-Huckel extended model is Y = 0.964643\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.15  Page Number : 485"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "# Variables,\n",
      "T = 25 + 273.15;\t\t\t#[K] - Temperature\n",
      "mol_CaCl2 = 0.001;\t\t\t#[mol/kg] - Molality of HCl\n",
      "A = 0.510;\t\t\t#[(kg/mol)**(1/2)]\n",
      "Z_positive = 2;\t\t\t#Stoichiometric coefficient of 'Ca' ion\n",
      "Z_negative = -1;\t\t\t#Stoichiometric coefficient of 'Cl' ion\n",
      "m_Ca_positive = mol_CaCl2;\n",
      "m_Cl_negative = 2*mol_CaCl2;\n",
      "\n",
      "# Calculations\n",
      "# I = 1/2*[((Z_positive)**(2))*m_Ca_positive + ((Z_negative)**(2))*m_Cl_negative]\n",
      "I = 1./2*(((Z_positive)**(2))*m_Ca_positive + ((Z_negative)**(2))*m_Cl_negative);\n",
      "\n",
      "#Using Debye-Huckel limiting law wee get,\n",
      "# math.log(Y1) = -A*(abs(Z_positive*Z_negative))*(I**(1/2)))\n",
      "Y = 10**(-A*(abs(Z_positive*Z_negative))*(I**(1./2)));\n",
      "\n",
      "# Results\n",
      "print \"The mean activity coefficient at 25 C using Debye-Huckel limiting law is Y = %f\"%(Y);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The mean activity coefficient at 25 C using Debye-Huckel limiting law is Y = 0.879290\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.17  Page Number : 488"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "# Variables,\n",
      "T = 50 + 273.15;\t\t\t#[K] - Temperature\n",
      "R=8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
      "x_1 = 0.3;\t\t\t# Mole fraction of component 1\n",
      "x_2 = (1-x_1);\t\t\t# Mole fraction of component 2\n",
      "#Increment of 1% means Y2 = 1.01*Y1\n",
      "\n",
      "# Calculations\n",
      "#Excess volume of the mixture is given by,\n",
      "V_excess = 4*x_1*x_2;\t\t\t#[cm**(3)/mol]\n",
      "#Amd threfore\n",
      "V_1_excess = 4*x_2*x_2*10**(-6);\t\t\t#[m**(3)/mol] - Exces volume of component 1\n",
      "V_2_excess = 4*x_1*x_1*10**(-6);\t\t\t#[m**(3)/mol] - Exces volume of component 2\n",
      "\n",
      "#We have from equation 14.89 of the book,\n",
      "#V_i_excess/(R*T) = (del_math.log(Y_i)/del_P)_T,x\n",
      "\n",
      "#Rearranging above equation\n",
      "#d(math.log(Y1)) = (V1_excess/(R*T))dP\n",
      "#Integrating the above equation at constant 'T' and 'x' in the limit from 'Y1' to '1.01*Y1' in the LHS and from 'P' to 'P+delta_P' in the RHS\n",
      "#On simplification we get\n",
      "#math.log(1.01*Y1/Y1) = (V_1_exces/(R*T))*delta_P\n",
      "delta_P = math.log(1.01)/(V_1_excess/(R*T));\t\t\t#[N/m**(2)]\n",
      "delta_P = delta_P*10**(-6);\t\t\t#[MPa]\n",
      "\n",
      "# Results\n",
      "print \"The required pressure to increase the activity coefficient by 1%% is %f MPa\"%(delta_P);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The required pressure to increase the activity coefficient by 1% is 13.639411 MPa\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.21  Page Number : 490"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "# Variables\n",
      "T = 293.15;\t\t\t#[K] - Temperature\n",
      "R=8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
      "A = 1280;\t\t\t#[J/mol]\n",
      "\n",
      "#(dA/dT)_P,x = del_A (say)\n",
      "dA_dT = -7.0;\t\t\t#[J/mol-K]\n",
      "\n",
      "#For equilomar mixture,\n",
      "x_1 = 0.5;\t\t\t# Mole fraction of component 1\n",
      "x_2 = 0.5;\t\t\t# Mole fraction of component 2\n",
      "\n",
      "# Calculations\n",
      "#math.log(Y1) =  (A/(R*T))*x_2**(2)\n",
      "#math.log(Y2) =  (A/(R*T))*x_1**(2)\n",
      "Y1 = math.exp((A/(R*T))*x_2**(2));\n",
      "Y2 = math.exp((A/(R*T))*x_1**(2));\n",
      "\n",
      "#G_excess/(R*T*) = x_1*math.log(Y1) + x_2*math.log(Y2) = (A/(R*T))*x_1*x_2\n",
      "G_excess = A*x_1*x_2;\t\t\t#[J/mol] - Excess Gibbs free energy\n",
      "\n",
      "#H_excess/(R*T**(2)) = -[d/dT(G_excess/(R*T))]_P,x\n",
      "#H_excess/(R*T**(2)) = -((x_1*x_2)/R)*[d/dT(A/T)]_P,x\n",
      "#On simplification  and putting the values we get\n",
      "H_excess = A*x_1*x_2 - T*dA_dT*x_1*x_2;\t\t\t#[J/mol] - Excess enthalpy\n",
      "\n",
      "#Now excess entropy is given by\n",
      "S_excess = (H_excess - G_excess)/T;\t\t\t#[J/mol-K] - Excess entropy\n",
      "\n",
      "# Results\n",
      "print \"For equimolar mixture\";\n",
      "print \"Excess Gibbs free energy G_excess) is %f J/mol\"%(G_excess);\n",
      "print \"Excess enthalpy H_excess) is %f J/mol\"%(H_excess);\n",
      "print \"Excess entropy S_excess) is %f J/mol\"%(S_excess);\n",
      "print \"The value of activity coefficient Y1 is %f\"%(Y1);\n",
      "print \"The value of activity coefficient Y2 is %f\"%(Y2);\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "For equimolar mixture\n",
        "Excess Gibbs free energy G_excess) is 320.000000 J/mol\n",
        "Excess enthalpy H_excess) is 833.012500 J/mol\n",
        "Excess entropy S_excess) is 1.750000 J/mol\n",
        "The value of activity coefficient Y1 is 1.140305\n",
        "The value of activity coefficient Y2 is 1.140305\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 14.22  Page Number : 491"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      " \n",
      "# Given\n",
      "T = 60 + 273.15;\t\t\t#[K] - Temperature\n",
      "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
      "\n",
      "# math.log(Y1_inf) = math.log(Y2_inf) = 0.15 + 10/T\n",
      "\n",
      "\n",
      "# From van L# Variablesr equation \n",
      "# A = math.log(Y1_inf) and B = math.log(Y2_inf) and since it is given that math.log(Y1_inf) = math.log(Y2_inf), therefore A = B\n",
      "#(x_1*x_2)/(G_excess/R*T) = x_1/B + x_2/A = X_1/A + x_2/A = 1/A\n",
      "# G_excess/(R*T) = A*x_1*x_2\n",
      "\n",
      "# For equilomar mixture,\n",
      "x_1 = 0.5;\t\t\t# Mole fraction of component 1\n",
      "x_2 = 0.5;\t\t\t# Mole fraction of component 2\n",
      "\n",
      "# Calculations and Results\n",
      "# Expression for A can be written as\n",
      "# A = 0.15 + 10/T, where T is in C. Therefore\n",
      "A = 0.15 + 10/(T - 273.15);\n",
      "# Differentiating it with respect to temprature we get\n",
      "dA_dT = - 10/((T-273.15)**(2));\n",
      "\n",
      "# The excess Gibbs free energy can be calculated as\n",
      "G_excess = A*x_1*x_2*(R*T);\t\t\t#[J/mol]\n",
      "\n",
      "# The ideal Gibbs free energy change can  be calculated as\n",
      "delta_G_id_mix = R*T*(x_1*math.log(x_1) + x_2*math.log(x_2));\t\t\t#[J/mol]\n",
      "\n",
      "# Finally we have,\n",
      "delta_G_mix = G_excess + delta_G_id_mix;\t\t\t#[J/mol]\n",
      "\n",
      "print \"The Gibbs free energy change of mixing for equimolar mixture is %f J/mol\"%(delta_G_mix);\n",
      "\n",
      "\n",
      "# Now let us determine the excess enthalpy. We know that\n",
      "# H_excess/(R*T**(2)) = -[d/dT(G_excess/R*T)]_P,x = - x_1*x_2*[d/dT(A)]_P,x\n",
      "# Therefore at 'T' = 60 C the excess enthalpy is given by\n",
      "H_excess = -R*(T**(2))*x_1*x_2*dA_dT;\t\t\t#[J/mol]\n",
      "\n",
      "delta_H_id_mix = 0;\t\t\t#[J/mol] - Enthalpy change of mixing for ideal solution is zero.\n",
      "\n",
      "#Thus enthalpy change of mixing for an equimolar mixture at 333.15 K is given by\n",
      "delta_H_mix = delta_H_id_mix + H_excess;\t\t\t#[J/mol]\n",
      "\n",
      "\n",
      "print \"The enthalpy change of mixing for equimolar mixture is %f J/mol\"%(delta_H_mix);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Gibbs free energy change of mixing for equimolar mixture is -1700.608815 J/mol\n",
        "The enthalpy change of mixing for equimolar mixture is 640.806876 J/mol\n"
       ]
      }
     ],
     "prompt_number": 12
    }
   ],
   "metadata": {}
  }
 ]
}