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{
"metadata": {
"name": "ch11_1",
"signature": "sha256:ea5fa2ce45062851dc4892fcc5b621c33d72fa0443d476b1ef4133dade9e39b0"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 11 : Properties of a Component in a Mixture"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.1 Page Number : 385"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"import math\n",
"\n",
"# Variables\n",
"Vol_total = 3;\t\t\t#[m**(3)] - Total volume of solution\n",
"x_ethanol = 0.6;\t\t\t#Mole fraction of ethanol\n",
"x_water = 0.4;\t\t\t#Mole fraction of water\n",
"\n",
"# Calculations\n",
"#The partial molar volumes of the components in the mixture are\n",
"V_ethanol_bar = 57.5*10**(-6);\t\t\t#[m**(3)/mol]\n",
"V_water_bar = 16*10**(-6);\t\t\t#[m**(3)/mol]\n",
"\n",
"#The molar volumes of the pure components are\n",
"V_ethanol = 57.9*10**(-6);\t\t\t#[m**(3)/mol]\n",
"V_water = 18*10**(-6);\t\t\t#[m**(3)/mol]\n",
"\n",
"#The molar volume of the solution is\n",
"V_sol = x_ethanol*V_ethanol_bar + x_water*V_water_bar;\t\t\t#[m**(3)/mol]\n",
"#Total number of moles can be calculated as \n",
"n_total = Vol_total/V_sol;\t\t\t#[mol]\n",
"\n",
"#Moles of the components are\n",
"n_ethanol = n_total*x_ethanol;\t\t\t#[mol]\n",
"n_water = n_total*x_water;\t\t\t#[mol]\n",
"\n",
"#Finally the volume of the pure components required can be calculated as\n",
"Vol_ethanol = V_ethanol*n_ethanol;\n",
"Vol_water = V_water*n_water;\n",
"\n",
"# Results\n",
"print \"Required volume of ethanol is %f cubic metre\"%(Vol_ethanol);\n",
"print \"Required volume of water is %f cubic metre\"%(Vol_water);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Required volume of ethanol is 2.548166 cubic metre\n",
"Required volume of water is 0.528117 cubic metre\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.2 Page Number : 385"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"# Variables\n",
"T = 25+273.15;\t\t\t#[K] - Temperature\n",
"P = 1;\t\t\t#[atm]\n",
"#Component 1 = water\n",
"#component 2 = methanol\n",
"a = -3.2;\t\t\t#[cm**(3)/mol] - A constant\n",
"V2 = 40.7;\t\t\t#[cm**(3)/mol] - Molar volume of pure component 2 (methanol)\n",
"#V1_bar = 18.1 + a*x_2**(2)\n",
"\n",
"# Calculations and Results\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",
"print \"The expression for the partial molar volume of methanol2 isV2_bar = V2 + a*x_1**2 [cm**3/mol]\";\n",
"\n",
"#At infinite dilution, x_2 approach 0 and thus x_1 approach 1, therefore\n",
"x_1 = 1;\t\t\t# Mole fraction of component 1(water) at infinite dilution\n",
"V2_bar_infinite = V2 + a*(x_1**(2));\t\t\t#[cm**(3)/mol]\n",
"\n",
"print \"The partial molar volume of methanol at infinite dilution is %f cm**3/mol\"%(V2_bar_infinite);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The expression for the partial molar volume of methanol2 isV2_bar = V2 + a*x_1**2 [cm**3/mol]\n",
"The partial molar volume of methanol at infinite dilution is 37.500000 cm**3/mol\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.4 Page Number : 387"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" \n",
"# Variables\n",
"#H = a*x_1 + b*x_2 +c*x_1*x_2\n",
"\n",
"#The values of the constants are\n",
"a = 15000;\t\t\t#[J/mol]\n",
"b = 20000;\t\t\t#[J/mol]\n",
"c = -2000;\t\t\t#[J/mol]\n",
"\n",
"# Calculations and Results\n",
"#(1)\n",
"#Enthalpy of pure component 1 = H1 is obtained at x_2 = 0, thus \n",
"x_2 = 0;\n",
"x_1 = 1;\n",
"H1 = a*x_1 + b*x_2 +c*x_1*x_2;\t\t\t#[J/mol]\n",
"print \"a).The enthalpy of pure component 1 is %f J/mol\"%(H1);\n",
"\n",
"#Similarly for component 2,\n",
"#Enthalpy of pure component 2 = H2 is obtained at x_1 = 0, thus \n",
"x_1_prime = 0;\n",
"x_2_prime = 1;\n",
"H2 = a*x_1_prime + b*x_2_prime +c*x_1_prime*x_2_prime;\t\t\t#[J/mol]\n",
"print \" The enthalpy of pure component 2 is %f J/mol\"%(H2);\n",
"\n",
"\n",
"#(c)\n",
"#From part (b), we have the relation\n",
"#H1_bar = a + c*(x_2**(2))\n",
"#H2_bar = b + c*(x_1**(2))\n",
"\n",
"#For enthalpy of component 1 at infinite dilution, x_1 approach 0 and thus x_2 approach 1, therefore\n",
"x_1_c = 0;\n",
"x_2_c = 1;\n",
"H1_infinite = a + c*(x_2_c**(2));\t\t\t#[cm**(3)/mol]\n",
"print \"C).The enthalpy of componenet 1 at infinite dilution at x_1 = 0) is %f J/mol\"%(H1_infinite);\n",
"\n",
"#At x_1 = 0.2\n",
"x_1_c1 = 0.2;\n",
"x_2_c1 = 0.8;\n",
"H1_bar_c1 = a + c*(x_2_c1**(2));\t\t\t#[J/mol]\n",
"print \" The enthalpy of componenet 1 at at x_1 = 0.2) is %f J/mol\"%(H1_bar_c1);\n",
"\n",
"#At x_1 = 0.8\n",
"x_1_c2 = 0.8;\n",
"x_2_c2 = 0.2;\n",
"H1_bar_c2 = a + c*(x_2_c2**(2));\t\t\t#[J/mol]\n",
"print \" The enthalpy of componenet 1 at at x_1 = 0.8) is %f J/mol\"%(H1_bar_c2);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a).The enthalpy of pure component 1 is 15000.000000 J/mol\n",
" The enthalpy of pure component 2 is 20000.000000 J/mol\n",
"C).The enthalpy of componenet 1 at infinite dilution at x_1 = 0) is 13000.000000 J/mol\n",
" The enthalpy of componenet 1 at at x_1 = 0.2) is 13720.000000 J/mol\n",
" The enthalpy of componenet 1 at at x_1 = 0.8) is 14920.000000 J/mol\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.9 Page Number : 395"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"# Variables\n",
"n = 1*10**(3);\t\t\t#[mol] - No of moles\n",
"P = 0.1;\t\t\t#[MPa] - Pressure of the surrounding\n",
"T = 300;\t\t\t#[K] - Temperature of the surrounding\n",
"x_1 = 0.79;\t\t\t#Mole fraction of N2 in the air\n",
"x_2 = 0.21;\t\t\t#Mole fraction of O2 in the air\n",
"R=8.314;\t\t\t#[J/mol*K]\n",
"\n",
"# Calculations\n",
"#Change in availability when x_1 moles of component 1 goes from pure state to that in the mixture is\n",
"#x_1*(si_1 - si_2) = x_1*[H1 - H1_bar - T_0*(S1 - S1_bar)]\n",
"#Similarly change in availability of x_2 moles of component 2 is\n",
"#x_2*(si_1 - si_2) = x_2*[H2 - H2_bar - T_0*(S2 - S2_bar)]\n",
"\n",
"#and thus total availability change when 1 mol of mixture is formed from x_1 mol of component 1 and x_2 mol of component 2 is equal to reversible work\n",
"#W_rev = x_1*[H1 - H1_bar - T_0*(S1 - S1_bar)] + x_2*[H2 - H2_bar - T_0*(S2 - S2_bar)]\n",
"#W_rev = -[delta_H_mix] +T_0*[delta_S_mix]\n",
"\n",
"#If T = T_0 that is,temperature of mixing is same as that of surroundings, W_rev = -delta_G_mix.\n",
"#W_rev = -delta_G_mix = R*T*(x_1*math.log(x_1) + x_2*math.log(x_2))\n",
"W_rev = R*T*(x_1*math.log(x_1) + x_2*math.log(x_2));\t\t\t#[J/mol]\n",
"\n",
"#Therefore total work transfer is given by\n",
"W_min = (n*W_rev)/1000;\t\t\t#[kJ]\n",
"\n",
"# Results\n",
"print \"The minimum work required is %f kJ\"%(W_min);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimum work required is -1281.910728 kJ\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 11.10 Page Number : 400"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"# Variables\n",
"x_A = 0.20;\t\t\t# Mole fraction of A\n",
"x_B = 0.35;\t\t\t# Mole fraction of B\n",
"x_C = 0.45;\t\t\t# Mole fraction of C\n",
"\n",
"phi_A = 0.7;\t\t\t# Fugacity coefficient of A\n",
"phi_B = 0.6;\t\t\t# Fugacity coefficient of B\n",
"phi_C = 0.9;\t\t\t# Fugacity coefficient of C\n",
"\n",
"P = 6.08;\t\t\t#[MPa] - Pressure\n",
"T = 384;\t\t\t#[K] - Temperature\n",
"\n",
"# Calculations\n",
"#We know that\n",
"#math.log(phi) = x_1*math.log(phi_) + x_2*math.log(phi_2) + x_3*math.log(phi_3)\n",
"math.log_phi = x_A*math.log(phi_A) + x_B*math.log(phi_B) + x_C*math.log(phi_C);\t\t\t# Fugacity coefficient\n",
"phi = math.exp(math.log_phi);\n",
"\n",
"#Thus fugacity is given by,\n",
"f_mixture = phi*P;\t\t\t#[MPa]\n",
"\n",
"# Results\n",
"print \"The fugacity of the mixture is %f MPa\"%(f_mixture);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fugacity of the mixture is 4.515286 MPa\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
