path: root/Introduction_To_Chemical_Engineering_Thermodynamics_by_G_Halder/9-Solution_Thermodynamics_Properties.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 9: Solution Thermodynamics Properties"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.10: Fugacity_of_Component_in_Mixture.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.10\n",
"// Page: 354\n",
"\n",
"printf('Example: 9.10 - Page: 354\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"x1 = 0.3;// [mole fraction of component 1 in the mixture]\n",
"x2 = 0.7;// [mole fraction of component 2 in the mixture]\n",
"phi1 = 0.7;// [fugacity coeffecient of component 1 in the mixture]\n",
"phi2 = 0.85;// [fugacity coeffecient of component 2 in the mixture]\n",
"P = 50;// [bar]\n",
"T = 273 + 100;// [K]\n",
"//*************//\n",
"\n",
"phi = exp(x1*log(phi1) + x2*log(phi2));// [fugacity coeffecient of the mixture]\n",
"f = phi*P;// [bar]\n",
"printf('Fugacity of the gaseous mixture is %.3f bar',f);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.11: Fugacity_of_Component_in_Mixture.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.11\n",
"// Page: 354\n",
"\n",
"printf('Example: 9.11 - Page: 354\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"x1 = 0.3;// [mole fraction of hydrogen in the mixture]\n",
"x2 = 0.25;// [mole fraction of nitrogen in the mixture]\n",
"x3 = 0.45;// [mole fraction of oxygen in the mixture]\n",
"phi1 = 0.7;// [fugacity coeffecient of oxygen in the mixture]\n",
"phi2 = 0.85;// [fugacity coeffecient of nitrogen in the mixture]\n",
"phi3 = 0.75;// [fugacity coeffecient of oxygen in the mixture]\n",
"P = 60;// [bar]\n",
"T = 273 + 150;// [K]\n",
"//***********//\n",
"\n",
"phi = exp(x1*log(phi1) + x2*log(phi2) + x3*log(phi3));// [fugacity coeffecient of the mixture]\n",
"f = phi*P;// [bar]\n",
"printf('Fugacity of the gaseous mixture is %.3f bar',f);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.12: Fugacity_of_Solids_and_Liquids.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.12\n",
"// Page: 356\n",
"\n",
"printf('Example: 9.12 - Page: 356\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"T = 372.12;// [K]\n",
"Psat = 100;// [kPa]\n",
"P = 300; //[kPa]\n",
"Vspecific = 1.043*10^(-3);//[cubic m/kg]\n",
"M = 18*10^(-3);// [molecular weight of water, kg/mol]\n",
"R = 8.314;// [J/mol K]\n",
"//***************//\n",
"\n",
"Psat = Psat/100;// [bar]\n",
"P = P/100;// [bar]\n",
"Vl = Vspecific*M;// [cubic m/mol]\n",
"// Vapour is assumed to be like an ideal gas.\n",
"phi = 1;\n",
"fsat = Psat*phi;// [bar]\n",
"fl = fsat*exp(Vl*(P - Psat)*10^5/(R*T));// [bar]\n",
"printf('Fugacity of liquid water is %.4f bar',fl);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.13: Fugacity_of_Solids_and_Liquids.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.13\n",
"// Page: 357\n",
"\n",
"printf('Example: 9.13 - Page: 357\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"Vl = 90.45*10^(-6);// [molar volume of liquid butadiene, cubic m/mol]\n",
"fsat = 4.12;// [bar]\n",
"P = 10;// [bar]\n",
"Psat = 4.12;// [bar]\n",
"T = 313;// [K]\n",
"R = 8.314;// [J/mol K]\n",
"//************//\n",
"\n",
"fl = fsat*exp(Vl*(P - Psat)*10^5/(R*T));// [bar]\n",
"printf('The fugacity of the liquid water is %.4f bar',fl);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.14: Fugacity_of_Solids_and_Liquids.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.14\n",
"// Page: 357\n",
"\n",
"printf('Example: 9.14 - Page: 357\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"b = 0.0391;// [cubic dm/mol]\n",
"P1 = 1000;// [atm]\n",
"T = 1000 + 273;// [K]\n",
"R = 0.0892;// [L bar/K mol]\n",
"deff('[Vreal] = f1(P)','Vreal = R*T/P + b');\n",
"deff('[Videal] = f2(P)','Videal = R*T/P');\n",
"//**************//\n",
"\n",
"// We know that:\n",
"// RTlog(f/P) = integral('Vreal - Videal',P,0,P)\n",
"f = P1*exp((1/(R*T))*integrate('f1(P) - f2(P)','P',0,P1));// [atm]\n",
"phi = f/P1;\n",
"printf('The fugacity of the gas is %d atm \n',f);\n",
"printf('The fugacity coeffecient of the gas is %.3f atm',phi);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.15: Fugacity_of_Solids_and_Liquids.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.15\n",
"// Page: 359\n",
"\n",
"printf('Example: 9.15 - Page: 359\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"Vl = 73*10^(-6);// [cubic m/mol]\n",
"P = 275;// [bar]\n",
"Psat = 4.360;// [bar]\n",
"T = 110 + 273;// [K]\n",
"R = 8.314;// [J/mol K]\n",
"//**************//\n",
"\n",
"// Acetone vapour is assumed to behave like ideal gas.\n",
"phi = 1;\n",
"fsat = Psat;// [bar]\n",
"fl = fsat*exp(Vl*(P - Psat)*10^5/(R*T));// [bar]\n",
"printf('Fugacity of liquid butadiene at 313 K & 10 bar is %.3f bar',fl); "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.16: Gibbs_Theorem.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.16\n",
"// Page: 362\n",
"\n",
"printf('Example: 9.16 - Page: 362\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"V1 = 2.8;// [Volume of Oxygen, L]\n",
"V2 = 19.6;// [Volume of hydrogen, L]\n",
"R = 1.987;// [cal/K mol]\n",
"//**************//\n",
"\n",
"n1 = V1/22.4;// [moles of Oxygen]\n",
"n2 = V2/22.4;// [moles of Hydrogen]\n",
"n = n1 + n2;// [total number of moles]\n",
"x1 = n1/n;// [mole fraction of Oxygen]\n",
"x2 = n2/n;// [mole fraction of Hydrogen]\n",
"// From Eqn. 9.88:\n",
"deltaS_mix = - (R*(x1*log(x1) + x2*log(x2)));// [cal/K mol]\n",
"printf('The entropy change of mixiong is %.3f cal/K mol',deltaS_mix)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.17: Gibbs_Theorem.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.17\n",
"// Page: 363\n",
"\n",
"printf('Example: 9.17 - Page: 363\n\n');\n",
"\n",
"// Mathematics is involved in proving but just that no numerical computations are involved.\n",
"// For prove refer to this example 9.17 on page number 363 of the book.\n",
"\n",
"printf(' Mathematics is involved in proving but just that no numerical computations are involved.\n\n');\n",
"printf(' For prove refer to this example 9.17 on page 363 of the book.');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.18: Gibbs_Theorem.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.18\n",
"// Page: 364\n",
"\n",
"printf('Example: 9.18 - Page: 364\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"n1 = 0.7;// [moles of helium]\n",
"n2 = 0.3;// [moles of argon]\n",
"R = 8.314;// [J/mol K]\n",
"T = 273 + 25;// [K]\n",
"//******************//\n",
"\n",
"n = n1 + n2;// [total moles]\n",
"x1 = n1/n;// [mole fraction of helium]\n",
"x2 = n2/n;// [mole fraction of argon]\n",
"deltaG_mix = n*R*T*(x1*log(x1) + x2*log(x2));// [J]\n",
"printf('The free energy change of mixing is %.2f J\n',deltaG_mix);\n",
"\n",
"// Since the gases are ideal:\n",
"deltaH_mix = 0;// [J]\n",
"printf('The enthalpy change of mixing is %d J\n',deltaH_mix);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.19: Gibbs_Theorem.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.19\n",
"// Page: 364\n",
"\n",
"printf('Example: 9.19 - Page: 364\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"V = 20;// [Volume of vessel, L]\n",
"V1 = 12;// [Volume of Hydrogen, L]\n",
"V2 = 10;// [Volume of Nitrogen, L]\n",
"P = 1;// [atm]\n",
"T = 298;// [K]\n",
"P1 = 1;// [atm]\n",
"P2 = 1;// [atm]\n",
"R = 0.082;// [L atm/K mol]\n",
"//************//\n",
"\n",
"n1 = P1*V1/(R*T);// [number of moles of Hydrogen]\n",
"n2 = P2*V2/(R*T);// [number of moles of Nitrogen]\n",
"n = n1 + n2;// [total number of moles]\n",
"Pfinal = n*R*T/V;// [atm]\n",
"p1 = Pfinal*n1;// [partial pressure of Hydrogen, atm]\n",
"p2 = Pfinal*n2;// [partial pressure of Nitrogen, atm]\n",
"deltaG_mix = R*T*(n1*log(p1/P1) + n2*log(p2/P2));// [J]\n",
"printf('Free Energy change of mixing is %.2f J\n',deltaG_mix);\n",
"\n",
"// Since mixing is ideal:\n",
"deltaH_mix = 0;// [J]\n",
"printf('Enthalpy change in mixing is %.2f J\n',deltaH_mix);\n",
"\n",
"deltaS_mix = - (deltaG_mix/T);// [J/K]\n",
"printf('Entropy Change in mixing is %.3f J/K\n',deltaS_mix);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.1: Partial_Molar_Properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.1\n",
"// Page: 338\n",
"\n",
"printf('Example: 9.1 - Page: 338\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"V1_bar = 52.37*10^(-6);// [partial molar volume of ethanol, cubic m/mol]\n",
"y1 = 0.5;// [mole fraction of ethanol]\n",
"Density = 800.21;// [kg/cubic m]\n",
"M1 = 46*10^(-3);// //[Molecular wt. of ethanol,kg/mol]\n",
"M2 = 18*10^(-3);// [Molecular wt. of water,kg/cmol]\n",
"//*************//\n",
"\n",
"y2 = 1 - y1;// [mole fraction of water]\n",
"M = y1*M1 + y2*M2;// [Molecular wt. of mixture, kg/mol]\n",
"V = M/Density;// [Volume of mixture, cubic m/mol]\n",
"// From Eqn. 9.9:\n",
"V2_bar = (V - y1*V1_bar)/y2;// [partial molar volume of water, cubic m/mol]\n",
"printf('Partial molar volume of water is %.2e cubic m/mol\n',V2_bar);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.20: Excess_Property_Of_Mixture.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.20\n",
"// Page: 367\n",
"\n",
"printf('Example: 9.20 - Page: 367\n\n');\n",
"\n",
"// Mathematics is involved in proving but just that no numerical computations are involved.\n",
"// For prove refer to this example 9.20 on page number 367 of the book.\n",
"\n",
"printf(' Mathematics is involved in proving but just that no numerical computations are involved.\n\n');\n",
"printf(' For prove refer to this example 9.20 on page 367 of the book.');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.21: Henry_Law.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.21\n",
"// Page: 373\n",
"\n",
"printf('Example: 9.21 - Page: 373\n\n');\n",
"\n",
"// Mathematics is involved in proving but just that no numerical computations are involved.\n",
"// For prove refer to this example 9.21 on page number 373 of the book.\n",
"\n",
"printf(' Mathematics is involved in proving but just that no numerical computations are involved.\n\n');\n",
"printf(' For prove refer to this example 9.21 on page 373 of the book.');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.2: Partial_Molar_Properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.2\n",
"// Page: 338\n",
"\n",
"printf('Example: 9.2 - Page: 338\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"Vol = 2;// [Volume of the mixture, cubic m/mol]\n",
"y1 = 0.4;// [mole fraction of alcohol, cubic m/mol]\n",
"V1_bar = 38.3*10^(-6);// [partial molar volume of alcohol, cubic m/mol]\n",
"V2_bar = 17.2*10^(-6);// [partial molar volume of water, cubic m/mol]\n",
"V1 = 39.21*10^(-6);// [molar volume of alcohol, cubic m/mol]\n",
"V2 = 18*10^(-6);// [molar volume of water, cubic m/mol]\n",
"//*************//\n",
"\n",
"// From Eqn. 9.9:\n",
"V = y1*V1_bar + (1 - y1)*V2_bar;// [molar volume of the solution]\n",
"n = Vol/V;// [number of moles of solution]\n",
"n1 = y1*n;// [number of moles of alcohol required]\n",
"n2 = (1 - y1)*n;// [number of moles of water required]\n",
"V_alcohol = V1*n1;// [Volume of alcohol required, cubic m]\n",
"V_water = V2*n2;// [Volume of water required, cubic m]\n",
"printf('Volume of alcohol required is %.3f cubic m while volume of water required is %.3f cubic m\n',V_alcohol,V_water);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.3: Partial_Molar_Properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.3\n",
"// Page: 339\n",
"\n",
"printf('Example: 9.3 - Page: 339\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"Vol = 2000;// [cubic cm]\n",
"y1_1 = 0.96;// [mass fraction of ethanol in laboratory alcohol]\n",
"y2_1 = 0.04;// [mass fraction of water in laboratory alcohol]\n",
"y1_2 = 0.56;// [mass fracion of ethanol in vodka]\n",
"y2_2 = 0.44;// [mass fraction of water in vodka]\n",
"Vbar_water1 = 0.816;// [cubic cm/g]\n",
"Vbar_ethanol1 = 1.273;// [cubic cm/g]\n",
"Vbar_water2 = 0.953;// [cubic cm/g]\n",
"Vbar_ethanol2 = 1.243;// [cubic cm/g]\n",
"Density_water = 0.997;// [cubic cm/g]\n",
"//***************//\n",
"\n",
"// Solution (i)\n",
"// From Eqn 9.9\n",
"Va = y1_1*Vbar_ethanol1 + y2_1*Vbar_water1;// [Volume of laboratory alcohol, cubic cm/g]\n",
"mass = Vol/Va;// [g]\n",
"// Let Mw be the mass of water added in laboratory alcohol.\n",
"// Material balance on ethanol:\n",
"Mw = mass*y1_1/y1_2 - mass;// [g]\n",
"Vw = Mw/Density_water;// [Volume of water added, cubic cm]\n",
"printf('Mass of water added is %d g\n',Mw);\n",
"\n",
"// Solution (ii)\n",
"Mv = mass + Mw;// [Mass of vodka, g]\n",
"Vv = y1_2*Vbar_ethanol2 + y2_2*Vbar_water2;// [Volume of ethanol, cubic cm/g]\n",
"V_vodka = Vv*Mv;// [Volume of vodka obtained after conversion, cubic cm]\n",
"printf('The volume of vodka obtained after conversion is %.d cubic cm\n',V_vodka);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.4: Partial_Molar_Properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.4\n",
"// Page: 339\n",
"\n",
"printf('Example: 9.4 - Page: 339\n\n');\n",
"\n",
"// Mathematics is involved in proving but just that no numerical computations are involved.\n",
"// For prove refer to this example 9.4 on page number 339 of the book.\n",
"\n",
"printf(' Mathematics is involved in proving but just that no numerical computations are involved.\n\n');\n",
"printf(' For prove refer to this example 9.4 on page 339 of the book.');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.5: Partial_Molar_Properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.5\n",
"// Page: 340\n",
"\n",
"printf('Example: 9.5 - Page: 340\n\n');\n",
"\n",
"// Mathematics is involved in proving but just that no numerical computations are involved.\n",
"// For prove refer to this example 9.5 on page number 340 of the book.\n",
"\n",
"printf(' Mathematics is involved in proving but just that no numerical computations are involved.\n\n');\n",
"printf(' For prove refer to this example 9.5 on page 340 of the book.');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.6: Partial_Molar_Properties.sce"
		   ]
		  },
  {
"cell_type": "code",
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"clear;\n",
"clc;\n",
"\n",
"// Example: 9.6\n",
"// Page: 341\n",
"printf('Example - 9.6 and Page number - 341\n\n');\n",
"\n",
"//Given\n",
"T = 25+273.15;// [K]\n",
"P = 1;// [atm]\n",
"// Component 1 = water\n",
"// Component 2 = methanol\n",
"a = -3.2;// [cubic cm/mol]  A constant\n",
"V2 = 40.7;// [cubic cm/mol]  Molar volume of pure component 2 (methanol)\n",
"// V1_bar = 18.1 + a*x_2^(2)\n",
"\n",
"// From Gibbs-Duhem equation at constant temperature and pressure we have\n",
"// x_1*dV1_bar + x_2*dV2_bar = 0\n",
"// dV2_bar = -(x_1/x_2)*dV1_bar = -(x_1/x_2)*a*2*x_2*dx_2 = -2*a*x_1*dx_2 = 2*a*x_1*dx_1\n",
"\n",
"// At x_1 = 0: x_2 = 1 and thus V2_bar = V2\n",
"// Integrating the above equation from x_1 = 0 to x_1 in the RHS, and from V2_bar = V2 to V2 in the LHS, we get\n",
"// V2_bar = V2 + a*x_1^(2) -  Molar volume of component 2(methanol) in the mixture \n",
"\n",
"printf('The expression for the partial molar volume of methanol(2) is\nV2_bar = V2 + a*x_1^(2) [cubic cm/mol]\n\n');\n",
"\n",
"// At infinite dilution, x_2 approach 0 and thus x_1 approach 1, therefore\n",
"x_1 = 1;// Mole fraction of component 1(water) at infinite dilution\n",
"V2_bar_infinite = V2 + a*(x_1^(2));//[cubic cm/mol]\n",
"\n",
"printf('The partial molar volume of methanol at infinite dilution is %.1f cubic cm/mol',V2_bar_infinite);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.7: Partial_Molar_Properties.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.7\n",
"// Page: 342\n",
"\n",
"printf('Example: 9.7 - Page: 342\n\n');\n",
"\n",
"// Solution\n",
"\n",
"//*****Data******//\n",
"// Data = [X1 V*10^6(cubic m/mol)];\n",
"Data = [0 20;0.2 21.5;0.4 24.0;0.6 27.4;0.8 32.0;1 40];\n",
"//************//\n",
"\n",
"scf(1.1);\n",
"plot(Data(:,1),Data(:,2));\n",
"title('Example 9.7');\n",
"xlabel('Mole fraction');\n",
"ylabel('Molar Volume*10^(6)');\n",
"xgrid();\n",
"\n",
"// Solution (i)\n",
"printf('For X1 = 0.5\n');\n",
"// A tangent is drawn to the curve at X1 = 0.5.\n",
"// The intercept at X2 = 0 or X1 = 1, gives V1_bar.\n",
"V1_bar1 = 33.8*10^(-6);// [cubic m/mol]\n",
"// The intercept at X2 = 1 or X1 = 0, gives V2_bar.\n",
"V2_bar1 = 17*10^(-6);// [cubic m/mol]\n",
"printf('Partial molar volume of component 1 is %.2e cubic m/mol\n',V1_bar1);\n",
"printf('Partial molar volume of component 2 is %.2e cubic m/mol\n',V2_bar1);\n",
"printf('\n');\n",
"\n",
"// Solution (ii)\n",
"printf('For X2 = 0.75\n');\n",
"// A tangent is drawn to the curve at X1 = 0.75.\n",
"// The intercept at X2 = 0 or X1 = 1, gives V1_bar.\n",
"V1_bar2 = 36.6*10^(-6);// [cubic m/mol]\n",
"// The intercept at X2 = 1 or X1 = 0, gives V2_bar.\n",
"V2_bar2 = 12.4*10^(-6);// [cubic m/mol]\n",
"point1 = [0 V1_bar1; 1 V2_bar1];\n",
"point2 = [0 V1_bar2;1 V2_bar2];\n",
"scf(2);\n",
"plot(point1(:,1),point1(:,2),point2(:,1),point2(:,2));\n",
"legend('X1 = 0.5','X1 = 0.75');\n",
"xlabel('Mole fraction');\n",
"ylabel('Molar Volume');\n",
"printf('Partial molar volume of component 1 is %.2e cubic m/mol\n',V1_bar);\n",
"printf('Partial molar volume of component 2 is %.2e cubic m/mol\n',V2_bar);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.8: Gibbs_Duhem_Equatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.8\n",
"// Page: 352\n",
"\n",
"printf('Example: 9.8 - Page: 352\n\n');\n",
"\n",
"// Mathematics is involved in proving but just that no numerical computations are involved.\n",
"// For prove refer to this example 9.8 on page number 352 of the book.\n",
"\n",
"printf(' Mathematics is involved in proving but just that no numerical computations are involved.\n\n');\n",
"printf(' For prove refer to this example 9.8 on page 352 of the book.');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 9.9: Gibbs_Duhem_Equatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Example: 9.9\n",
"// Page: 352\n",
"\n",
"printf('Example: 9.9 - Page: 352\n\n');\n",
"\n",
"// Mathematics is involved in proving but just that no numerical computations are involved.\n",
"// For prove refer to this example 9.9 on page number 352 of the book.\n",
"\n",
"printf(' Mathematics is involved in proving but just that no numerical computations are involved.\n\n');\n",
"printf(' For prove refer to this example 9.9 on page 352 of the book.');"
   ]
   }
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