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diff --git a/A_Textbook_Of_Chemical_Engineering_Thermodynamics/ch7.ipynb b/A_Textbook_Of_Chemical_Engineering_Thermodynamics/ch7.ipynb new file mode 100644 index 00000000..b0e53020 --- /dev/null +++ b/A_Textbook_Of_Chemical_Engineering_Thermodynamics/ch7.ipynb @@ -0,0 +1,888 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Properties of Solutions" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.2" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To find the volume of mixture\n", + "\n", + "# Variables\n", + "V = 0.1; \t\t\t#volume of mixture required (m**3)\n", + "Ve = 0.03; \t\t\t#volume of alcohol\n", + "Vw = 0.07; \t\t\t#volume of water\n", + "de = 789.; \t\t\t#density of ethanol (kg/m**3)\n", + "dw = 997.; \t\t\t#density of water (kg/m**3)\n", + "pe = 53.6*10**-6; \t\t\t#partial molar volume of ethanol (m**3/mol)\n", + "pw = 18.*10**-6; \t\t\t#partial molar volume of water (m**3/mol)\n", + "Me = 46.; \t\t\t#molecular wt of ethanol\n", + "Mw = 18.; \t\t\t#molecular wt of water\n", + "\n", + "# Calculations\n", + "#To find the volume of mixture\n", + "ne = (Ve*de*10**3)/Me; \t\t\t#number of moles of ethanol\n", + "nw = (Vw*dw*10**3)/Mw; \t\t\t#number of moles of water\n", + "xe = ne/(ne+nw) \t\t\t#mole fraction of ethanol\n", + "xw = 1-ne; \t\t\t#mole fraction of water\n", + "act_V = (ne*pe)+(nw*pw)\n", + "\n", + "# Results\n", + "if (V==act_V) :\n", + " print 'It is possible to prepare the required solution'\n", + "else:\n", + " Ve_act = (Ve/act_V)*V;\n", + " Vw_act = (Vw/act_V)*V;\n", + " print ' For the given volumes of ethanol and water, it is not possible to prepare 0.1 cubic m of mixture'\n", + " print ' Required volume of ethanol is %f cubic m'%Ve_act\n", + " print ' Required volume of water is %f cubic m'%Vw_act\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " For the given volumes of ethanol and water, it is not possible to prepare 0.1 cubic m of mixture\n", + " Required volume of ethanol is 0.030810 cubic m\n", + " Required volume of water is 0.071890 cubic m\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.3" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To find the required volume of methanol and water\n", + "\n", + "# Variables\n", + "V = 2.; \t\t\t#volume of desired solution (m**3)\n", + "x1 = 0.3; \t\t\t#moles fraction of methanol\n", + "x2 = 0.7; \t\t\t#moles fraction of water\n", + "V1 = 38.632*10**-6; \t\t\t#partial molar volume of methanol (m**3/mol)\n", + "V2 = 17.765*10**-6; \t\t\t#partial molar volume of water (m**3/mol)\n", + "mol_V1 = 40.727*10**-6; \t\t#molar volume of ethanol (m**3/mol)\n", + "mol_V2 = 18.068*10**-6; \t\t#molar volume of water (m**3/mol)\n", + "\n", + "# Calculations\n", + "#To find the required volume of methanol and water\n", + "V_mol = (x1*V1)+(x2*V2); \t\t\t#molar volume of desired solution\n", + "n = V/V_mol; \t\t\t #no. of moles in the desired solution\n", + "n1 = x1*n; \t\t\t#moles of methanol\n", + "n2 = x2*n; \t\t\t#moles of water\n", + "V_m = n1*mol_V1;\n", + "V_w = n2*mol_V2;\n", + "\n", + "# Results\n", + "print 'Volume of methanol to be taken is %f cubic m'%V_m\n", + "print ' Volume of water to be taken is %f cubic m'%V_w\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume of methanol to be taken is 1.017111 cubic m\n", + " Volume of water to be taken is 1.052866 cubic m\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.4" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To calculate the volume of water to be added and volume of dilute alcohol solution\n", + "\n", + "# Variables\n", + "V1_w = 0.816*10**-3; \t\t\t#partial molar volume of water in 96% alcohol solution\n", + "V1_e = 1.273*10**-3; \t\t\t#partial molar volume of ethanol in 96% alcohol solution\n", + "V2_w = 0.953*10**-3; \t\t\t#partial molar volume of water in 56% alcohol solution\n", + "V2_e = 1.243*10**-3; \t\t\t#partial molar volume of ethanol in 56% alcohol solution\n", + "d = 0.997*10**3; \t\t\t #density of water (kg/m**3)\n", + "\n", + "\n", + "# Calculations\n", + "#To calculate the volume of water to be added and volume of dilute alcohol solution\n", + "#Basis: \n", + "V = 2*10**-3; \t\t\t #volume of alcohol solution (m**3)\n", + "V_sp = (0.96*V1_e)+(0.04*V1_w); \t\t\t#volume of 1 kg of laboratory alcohol\n", + "m_e = V/V_sp; \t\t\t #mass of 2*10**-3 m**3 alcohol \n", + "\n", + "#(a).\n", + "#Let mass of water added be m kg\n", + "#Taking an alcohol balance\n", + "m = (m_e*0.96)/0.56 - m_e;\n", + "v = m/d;\n", + "\n", + "# Results\n", + "print ' (a).'\n", + "print ' Mass of water added is %f kg'%m\n", + "print ' Volume of water added is %4.3e cubic m'%v\n", + "\n", + "#(b)\n", + "m_sol = m_e + m; \t\t\t #mass of alcohol solution obtained\n", + "sp_vol = (0.56*V2_e)+(0.44*V2_w); \t\t\t#specific volume of 56% alcohol\n", + "V_dil = sp_vol*m_sol; \t\t\t #volume of dilute alcohol solution\n", + "print ' (b)'\n", + "print ' Volume of dilute alcohol solution is %5.4e cubic m'%V_dil\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " (a).\n", + " Mass of water added is 1.138558 kg\n", + " Volume of water added is 1.142e-03 cubic m\n", + " (b)\n", + " Volume of dilute alcohol solution is 3.0479e-03 cubic m\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.6" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To determine enthalpies of pure components and at infinite dilution\n", + "\n", + "# Variables\n", + "x1= .3\n", + "x2 = .7\n", + "\n", + "# Calculations and Results\n", + "H = 400.*x1 + 600*x2 + x1*x2*(40*x1+20*x2)\n", + "H1 = 420.-60+40;\n", + "#Using eq. 7.28 (Page no. 264)\n", + "#H = H2_bar as x2 = 1\n", + "H2 = 600.;\n", + "print ' (b).'\n", + "print ' Pure state enthalpies are:'\n", + "print ' H1 = %i J/mol'%H1\n", + "print ' H2 = %i J/mol'%H2\n", + "\n", + "\t\t\t#(c).\n", + "\t\t\t#H1_inf = H1_bar as x1 = 0, so from eq. 7.27\n", + "H1_inf = 420.;\n", + "\t\t\t#H2_inf = H2_bar as x2 = 0. so from eq 7.28\n", + "H2_inf = 640.;\n", + "print ' (c).'\n", + "print ' At infinite dilution:'\n", + "print ' H1 = %i J/mol'%H1_inf\n", + "print ' H2 = %i J/mol'%H2_inf\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " (b).\n", + " Pure state enthalpies are:\n", + " H1 = 400 J/mol\n", + " H2 = 600 J/mol\n", + " (c).\n", + " At infinite dilution:\n", + " H1 = 420 J/mol\n", + " H2 = 640 J/mol\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.7" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To calculate the partial molar volume of the components\n", + "\n", + "def V(m):\n", + " y = 1.003*10**-3 + 0.1662*10**-4*m + 0.177*10**-5*m**1.5 + 0.12*10**-6*m**2\n", + " return y\n", + "\n", + "# Variables\n", + "m = 0.1; \t\t\t#molality of solution (mol/kg)\n", + "\n", + "# Calculations\n", + "#To calculate the partial molar volume of the components\n", + "#Differentiating Eq. 7.29 with reference to m, we get\n", + "V1_bar = 0.1662*10**-4 + 0.177*1.5*10**-5*m**0.5 + 0.12*2*10**-6*m;\n", + "V_sol = V(m) \t\t\t#volume of aqueous soluttion\n", + "n1 = m;\n", + "n2 = 1000./18;\n", + "V2_bar = (V_sol - n1*V1_bar)/n2;\n", + "\n", + "# Results\n", + "print 'Partial molar volume of water = %4.3e cubic m/mol'%V2_bar\n", + "print ' Partial molar volume of NaCl = %4.3e cubic m/mol'%V1_bar\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Partial molar volume of water = 1.805e-05 cubic m/mol\n", + " Partial molar volume of NaCl = 1.748e-05 cubic m/mol\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To estimate the solubility of oxygen in water at 298K\n", + "\n", + "# Variables\n", + "K = 4.4*10**4; \t\t\t#Henry's law constant (bar)\n", + "pp = 0.25; \t\t\t #partial pressure of oxygen in bar\n", + "M_O2 = 32.; \t\t\t#molecular wt of oxygen\n", + "M_water = 18.; \t\t\t#molecular wt of water\n", + "\n", + "# Calculations and Results\n", + "#To estimate the solubility of oxygen in water at 298 K\n", + "#Using eq. 7.72 (Page no. 275)\n", + "x_O2 = pp/K; \t\t\t#mole fraction of O2\n", + "print 'Solubility of oxygen is %5.4e moles per mole of water'%x_O2\n", + "\n", + "#In mass units\n", + "sol_O2 = (x_O2*M_O2)/M_water;\n", + "print ' Solubility of oxygen in mass units is %4.3e kg oxygen per kg water'%sol_O2\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Solubility of oxygen is 5.6818e-06 moles per mole of water\n", + " Solubility of oxygen in mass units is 1.010e-05 kg oxygen per kg water\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To confirm that mixture conforms to Raoults Law and to determine Henrys Law constant?\n", + "\n", + "%pylab inline\t\t\t\n", + " \n", + "from numpy import *\n", + "from matplotlib.pyplot import *\n", + "import numpy\n", + "\n", + "# Variables\n", + "xb = array([0, 0.2, 0.4, 0.6, 0.8, 1.0])\n", + "pa_bar = [0.457 ,0.355, 0.243, 0.134, 0.049, 0];\n", + "pb_bar = [0,0.046, 0.108, 0.187, 0.288, 0.386];\n", + "\n", + "\n", + "#To confirm mixture conforms to Raoult's Law and to determine Henry's law constant\n", + "xa = 1 - xb\n", + "plot(xa,pa_bar)\n", + "plot(xa,pb_bar)\n", + "\n", + "# Calculations and Results\n", + "#For Henry's law plotting\n", + "x = [0,0.2, 0.4 ,0.6 ,0.8 ,1.0];\n", + "#Form the partial presures plot of component A and B\n", + "yh1 = [0,0,0,0,0,0]\n", + "yh1[0] = 0; \n", + "yh1[1] = 0.049; \t\t\t#For component A\n", + "for i in range(2,6):\n", + " yh1[i] = yh1[i-1]+(x[i]-x[i-1])*((yh1[1]-yh1[0])/(x[1]-x[0]))\n", + "\n", + "yh_2 = [0,0,0,0,0,0]\n", + "yh_2[5] = 0; \n", + "yh_2[4] = 0.046; \t\t\t#For component B\n", + "i = 3;\n", + "while (i>=0):\n", + " yh_2[i] = yh_2[i+1] + (x[i]-x[i+1])*((yh_2[5]-yh_2[4])/(x[5]-x[4]))\n", + " i = i-1;\n", + "\n", + "plot(x,yh1)\n", + "plot(x,yh_2)\n", + "\t\t\t#legend(\"Partial pressure \",\" \",\"Raoults law\",\" \",\"Henrys Law\"\n", + "show()\n", + "#(a)\n", + "print 'From the graph it can be inferred that, in the region where Raoults law is obeyed by A, the Henrys law is obeyed by B, and vice versa'\n", + "\n", + "#(b)\n", + "#Slope of Henry's law\n", + "print ' For component A, Ka = %f bar'%yh1[5]\n", + "print ' For component B, Kb = %f bar'%yh_2[0]\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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OE7kxkuRzycwMmslE/4nUsrfcdbl8WQ3yAwfU9ou/v8U+2vYZjeos85gY9Uf9\n+jB+vBroXbpImFeSBLuVOnblSuno4UN5eQx2dSXEw4NRbm541IC7/OVJ/CuR17a+RsaVDP478L+M\nu3ec2XcdHTigtl4GDlS3NMoWaBMoKYGdO9VV+Xffqf2sa+NvO3WSML8LEuzVQEZxMRszM4nNyGBr\nVhZdnJxKRw+3q189tgOamqIofH/ie17b+hq17GsxZ9AchrQeYvKANxrhww/hvffg00/VRaS4C3o9\nbN+uhvm6ddCiRVmYt2+vdXU2Q4K9mik0GNianV26mnd1cCD4asumZ6NGNe5IQKNiJPpwNK9vfx2v\nhl7MGTyHXt633j5bWefPw6RJ6sC/r74CHx+TvG3NU1SkDtOKjlZnmbdtWxbmMozeLCTYqzGjonDg\n8mV19LBOx8XiYsZcvfk6xNWV+jWoL19iLGHFLyuY9dMsujXrxuxBs+ns2bnK7xcfr+5LnzIF3nhD\nPYtYVEJBgTr2NjpaHYPbuXPZwRS3eFhRmJYEuw05UVDA+qs3X5MuX2agiwshHh6MdnfHs4b05QtL\nClm0fxHv7nqXoW2GMitoFq1dW1f49xcXw2uvwddfw8qV6rwXUUF5ebB5sxrm8fHqDpbQUPXBoWbN\ntK6uRpFgt1GZej2bdDridDq2ZGbi26BB6WlRHerXt/kRB7lFuczfM5+P933MBN8JzOg/g2YNyw+X\nP/5Qb5B6e8PSpeDubqFiq7PcXNiwQQ3zrVuhd281zMeOhcaNta6uxpJgrwGKjEYSsrNLT4uqX6tW\n6c3XPs7ONt2Xz7iSwTs732H5oeU82e1JXr7/ZVzr3TiXSFHU1fkLL8CsWRARIRsyypWZCXFx6rbE\nn35SH+EPDYXgYHCTM22tgQR7DaMoCsl5ecRlZBCbkcGZ4mJGXR1xMNTNjQY22pdPz03nrZ/e4rsj\n3/F8r+d59r5naVCnAbm5EBmpTnRdswb8/LSu1EpdulQ2y3zPHhgyRO2ZjxoFzjXniMPqQoK9hjtd\nWFi6w2Zvbi79nZ0J9vBgjLs7zepq93SnuRzTHeON7W/w0+mf+Nc901k7/QkeGFSXDz9Un4kR1zl3\nrmyWeVISDB+uhvmIEeBUvWfm2zoJdlEqW68nPjOTWJ2O+MxM2terV9qy8W3QwGb68kYjPP/eQRYf\nm45L2995b+RM/uX/L4s+xWq10tLKZpmnpqor8rAw9QxQmWVUbUiwi1sqNhrZkZ1N3NWBZQ52dqU3\nX/s6O+OmC7MjAAAd0klEQVRQTYcvnTsHjz6qzpL68ks4rexk2tZpZBVmMXvgbMZ2HGsz38Aq7NSp\nsrksf/yh9srDwtR2iw3+q60mkGAXd6QoCin5+aU3X08VFjLS3Z3gq0cCNqwmm7w3b1b3poeHw4wZ\nZXvTFUVh8/HNvLb1Neo61GXOoDkMbj1Y22LN7dixslnmaWnqLpawMHVmgoySrvYk2EWlpRcWsv7q\nSn53bi73OzuXHiTiZYUrvKIimDZNzbGVK9VNHLdiVIx889s3vL79dXycfZgzeA49vXpatlhzURT1\nmLhrYX7pUtks83795AksGyPBLu5KbkkJ32dmEqfTsUmno5Wjo9qy8fDA3wr68seOqQdLt2wJn39e\nsd14eoOe5b8s560dbxHYPJDZA2fj6+lr9lpNTlHg0KGyMM/LKxt/26ePzDK3YRLswmT0RiO7cnKI\nvbqaNyoKwVdvvg5wcaG2BYNEUWDFCnjpJfjvf9X2S2W/xxToC1i4fyHv7X6P4W2HMytoFi1dWpql\nXpNRFHUU5bUwNxrL5rL06CFhXkNIsAuzUBSF3/LzS2++HisoYLibGyHu7oxwd8fZjP/0z8lRHzJK\nSVH3pneu+sgYQH2K9cM9H/LJvk94pPMjTO8/naZOTU1TrCkYjfDzz2Vh7uhYFuYBAfK0VQ0kwS4s\n4lxRUWlffmdODvc1akSIuztjPDzwcTTdWaZ796pjAYYNgw8+MO3e9Ev5l3gn8R1WHFpBePdwXr7/\nZVwcXUz3AZVhMEBiohrma9eCq6sa5mFh4OsrYV7DSbALi8srKWHL1SMBN2Zm4l23buno4QAnpyr1\n5Y1GdWb6/PmweLE6d8pc0nLSmPXTLGKPxvJC7xeY2nMqDepY4OQNvR4SEtRV+XffQfPmZSvzjh3N\n//mi2pBgF5oqMRrZc92RgIVGY+kOmyAXF+pWoCd87hxMnKhOZly1Cu65xwKFA0czjvL69tdJ/CuR\n6f2m80T3J6hTy8RTNIuL1VnmMTEQGwutW5eFeZs2pv0sYTMk2IXVUBSFI1euEKfTEZeRwW/5+Qy9\nOsdmhJsbbrfYX71xozoz/amnYPp0bXbtJZ9LZvq26RzNOMqsoFn8w+8fd/cUa0EBbNmihvmGDeox\ncaGh6vZEOe1DVIAEu7BaF4qL2XA15LdnZ9O9YUNCrrZsmtvX45VX1I7EqlXqVmyt7Ti9g2lbp5FT\nmMPbg94muENwxdtK+fk3zjIPCCibZe7lZd7Chc2RYBfVwhWDgR+zstSWzUUd+el18DnjzsePeDDY\nqyH2VnKzUFEUNv6xkenbplPPoR5zBs9hUKtBt35xbq76T47oaLXdct99ZbPMmzSxbOHCpkiwi2rD\nYIAlS2DGmwqT38vFrq86lTKnpIQxV1fyg1xccLSC0cNGxcjXqV/z+vbXaeXaijmD5tDDqwdkZann\nfkZHqzdC+/VTe+bBwXKyhzAZCXZRLWzdCv/5jzr6e9EidUffNceuXCkdPXwoL48hrq4Ee3gwys0N\nD42PBNQb9Hy1/SN+XTKbfxyrS5cT+dQaNFgN89GjwUWj7ZLCpkmwC6t27Bi8+CL89pu6nXHcuPK3\naF8qLmZTZiaxGRlszcqii5NT6ejhdpYcuH7hQtks8/37MQwZzMYu9XjW7nuC/Mfw5oA3rf8pVlFt\nSbALq5SZCW+9pd4YfeUVeOaZyk+QLTQY2JqdXbqad3VwKN0vf1+jRqbvy585UzbL/NAhGDlSXZkP\nH176pFROYQ4f7PmAT/d/yj/9/sn0ftNp4iT9dGFaEuzCquj1aqtl9mz1PuKsWeDpeffva1QUDly+\nrI4e1um4VFzM6Kv75Ye4ulK/qn3506fLZpkfOaL2ykND4YEH1Ef7b+Ni/kXm7JzDypSVRARG8GKf\nF7V7ilXYHAl2YRUURd0g8uKL6gNGH3549zNeynOioID1Vx+KSrp8mYEuLoR4eDDa3R3PO/Xljx8v\nm8vy55/qLpbQUBg0CCrZ0z+dfZq3fnqL9cfW82KfF3m659PUry1n9Im7I8EuNPfrr+qN0bQ0db7L\nyJGWHXWSqdezSacjTqdjS2Ymvg0alJ4W1bHB1VEBv/9eFubnz5fNMu/f3yRPRR3JOMLr219nd9pu\nZvSbwb+7/dv0T7GKGkOCXWjm4kV4/XX1HuPrr6tPj2p9eE+R0UhCdrbasjl3jgZ5eQTv2UPwnj30\n6dyZWtdmmZtpS2XS2SRe2/YaxzOPMytoFo90fkTOYhWVJsEuLK6wEBYsgPffV2e8vPGGOpxQc4oC\nycmlK3NFryd58mTigoKIdXTkTHExo66OOBjq5kYDM+6XT/gzgWlbp5FXnMfbg95mTPsxmh9aIqoP\nCXZhMYqidjJefhn8/NRgb99e46KMRnXWb0yM+sPBoWz8bbduN/SEThcWlu6w2ZubS39nZ4I9PBjj\n7k4zMxwJqCgKG45tYPq26TjVcWLO4DkEtQwy+ecI2yPBLiziwAG1j56To94YHazlWdEGA+zeXdYz\nd3Yum5jo51ehBn+2Xk98ZiaxOh3xmZm0r1ePkKtHAnaqX9+kq2uD0cCa1DW8kfAGbd3aMmfQHLo3\n726y9xe2R4JdmNWZM/Daa+qwwv/+Fx5/3Gzt6fKVlMBPP6lBvnYtNG1aFub33ntXb11sNLIjO5vY\nqwPLHOzsSh+K6uvsjIOJjqMrNhSzNHkps3fOprd3b2YPmk1HD5nDLm5Wley849/S+Ph4/Pz86NSp\nE3Pnzr3p148cOULv3r1xdHTkgw8+qNSHi+ohP1/dg+7vrw4nPHZMHa9r0VAvLlYnJT7xBDRrBq++\nqo69TUyEX36BGTPuOtQB6tjbM8TNjY/btePPXr2I8fXFxcGBF06coMnu3Uz8/XeiL17kcknJ3X1O\nrTpE9Ijgj6l/0NOrJ/2/6M/4b8ez7dQ2WQCJu1buir2oqIiOHTuSmJhIkyZN6N27N5999hkBAQGl\nr7l06RKnT59m3bp1uLq68sILL9z6g2TFXu0YjfDll+oq/f774d13oWVLCxZQWAg//KC2WTZsgA4d\n1JX5uHEWLkSVXlhYOl9+d24u9zs7lx4k4nWXffnLRZdZlbKKT/d/ikExEBEYwaNdHpUHnYTpWzE7\nduzgvffeY8OGDQDMmzePwsJCZsyYcdNrZ82ahZOTkwS7jdi1C55/Xm1Rz5+v7gq0iCtX1JV5dLQ6\n07xLl7KDKaxolnluSQnfZ2YSp9OxSaejlaOj2rLx8MC/QYMq9+UVRSHxr0QWHlhI/PF4xncaT0Rg\nBAHNAu78m4VNqkp2lvs0Rnp6Oi1atCj92tvbm4SEhCoVJ6qHU6fUeS579sA778A//gEmaivfXl5e\n2SzzLVugZ081zD/8UO2fW6FGDg6M9/RkvKcneqORxJwc4nQ6HkxNxago6kNRHh70d3amdiUuoJ2d\nHf18+tHPpx8X8i6w9OBSQtaE4NXIi8jASMb7jsfRwXQHhgvbVG6wm3qv7cyZM0t/HhQURFBQkEnf\nX1Rdbi7MmaPOSH/2WVi+vHTWlXnk5JTNMt++Xe31hIaqw2U8PMz4waZX296ega6uDHR15cM2bfgt\nP59YnY7XTp7kj4IChru5Eez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+ "text": [ + "<matplotlib.figure.Figure at 0x2769cd0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "From the graph it can be inferred that, in the region where Raoults law is obeyed by A, the Henrys law is obeyed by B, and vice versa\n", + " For component A, Ka = 0.245000 bar\n", + " For component B, Kb = 0.230000 bar\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To calculate activity and activity coeffecient of chloroform\n", + "\n", + "from numpy import array\n", + "\n", + "\n", + "# Variables\n", + "xa = array([0, 0.2, 0.4, 0.6, 0.8, 1.0]);\n", + "Pa_bar = [0 ,0.049, 0.134, 0.243, 0.355, 0.457];\n", + "Pb_bar = [0.386 ,0.288, 0.187, 0.108, 0.046, 0];\n", + "\n", + "# Calculations and Results\n", + "#To calculate activity and activity coeffecient of chloroform\n", + "xb = 1-xa;\n", + "Pbo = 0.386; \t\t\t#vapour pressure of pure chloroform\n", + "#(a). Based on standard state as per Lewis-Randall rule\n", + "\n", + "print 'Based on Lewis Randall Rule'\n", + "print ' Activity Activity coeffecient'\n", + "a = [0,0,0,0,0,0]\n", + "ac = [0,0,0,0,0,0]\n", + "for i in range(6):\n", + " a[i] = Pb_bar[i]/Pbo;\n", + " print ' %f'%a[i],\n", + " if(xb[i]==0):\n", + " print ' Not defined',\n", + " else:\n", + " ac[i] = a[i]/xb[i];\n", + " print ' %f'%ac[i]\n", + "\n", + "#(b). Based on Henry's Law \n", + "Kb = 0.217; \t\t\t#bar (From Example 7.11 Page no. 276)\n", + "\n", + "print '\\n\\n Based on Henrys Law'\n", + "print ' Activity Activity coeffecient'\n", + "for i in range(6):\n", + " a[i] = Pb_bar[i]/Kb;\n", + " print ' %f'%a[i],\n", + " if(xb[i]==0):\n", + " print ' Not defined',\n", + " else:\n", + " ac[i] = a[i]/xb[i];\n", + " print ' %f'%ac[i]\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Based on Lewis Randall Rule\n", + " Activity Activity coeffecient\n", + " 1.000000 1.000000\n", + " 0.746114 0.932642\n", + " 0.484456 0.807427\n", + " 0.279793 0.699482\n", + " 0.119171 0.595855\n", + " 0.000000 Not defined \n", + "\n", + " Based on Henrys Law\n", + " Activity Activity coeffecient\n", + " 1.778802 1.778802\n", + " 1.327189 1.658986\n", + " 0.861751 1.436252\n", + " 0.497696 1.244240\n", + " 0.211982 1.059908\n", + " 0.000000 Not defined\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To determine fugacity fugacity coeffecient Henrys Law constant and activity coeffecient\n", + "\n", + "# Variables\n", + "P = 20.; \t\t\t#pressure in bar\n", + "\n", + "# Calculations and Results\n", + "def f1(x1):\n", + " y = (50*x1)-(80*x1**2)+(40*x1**3)\n", + " return y\n", + "\n", + "#To determine fugacity fugacity coeffecient Henry's Law constant and activity coeffecient\n", + "\n", + "#(a)\n", + "#Fugacity of component in solution becomes fugacity of pure component when mole fraction approaches 1 i.e. \n", + "x1 = 1.;\n", + "f1_pure = f1(x1);\n", + "print '(a). Fugacity f1 of pure component 1 is %i bar'%f1_pure\n", + "\n", + "#(b)\n", + "phi = f1_pure/P;\n", + "print ' (b). Fugacity coeffecient is %f'%phi\n", + "\n", + "#(c)\n", + "#Henry's Law constant is lim (f1/x1)and x1 tends to 0 \n", + "x1 = 0;\n", + "K1 = 50 - (80*x1) + (40*x1**2);\n", + "print ' (c). Henrys Law constant is %i bar'%K1\n", + "\n", + "#(d)\n", + "print ' (d). This subpart is theoretical and does not involve any numerical computation'\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a). Fugacity f1 of pure component 1 is 10 bar\n", + " (b). Fugacity coeffecient is 0.500000\n", + " (c). Henrys Law constant is 50 bar\n", + " (d). This subpart is theoretical and does not involve any numerical computation\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To determine enthalpies at infinite dilution\n", + "\n", + "# Variables\n", + "H1_pure = 400.; \t\t\t#enthalpy of pure liquid 1 at 298 K and 1 bar (J/mol)\n", + "H2_pure = 600.; \t\t\t#enthalpy of pure liquid 2 (J/mol)\n", + "x1 = 0.;\n", + "\n", + "# Calculations\n", + "delH1_inf = 20*((1-x1)**2)*(2*x1+1);\n", + "H1_inf = H1_pure + delH1_inf; \t\t\t#(J/mol)\n", + "\n", + "#For infinite dilution of 2, x1 = 1 and delH2_inf = H2_bar\n", + "x1 = 1.;\n", + "delH2_inf = 40.*x1**3;\n", + "H2_inf = delH2_inf + H2_pure; \t\t\t#(J/mol)\n", + "\n", + "# Results\n", + "print 'Enthalpy at infinite dilution for component 1 is %i J/mol'%H1_inf\n", + "print ' Enthalpy at infinite dilution for component 2 is %i J/mol'%H2_inf\n", + "\n", + " \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Enthalpy at infinite dilution for component 1 is 420 J/mol\n", + " Enthalpy at infinite dilution for component 2 is 640 J/mol\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.19 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To determine change in entropy for the contents of the vessel\n", + "\n", + "# Variables\n", + "R = 8.314; \t\t\t#ideal gas constant\n", + "n1 = 100.; \t\t\t#moles of nitrogen\n", + "n2 = 100.; \t\t\t#moles of oxygen\n", + "\n", + "# Calculations\n", + "#To determine the change in entropy of the contents of the vessel\n", + "x1 = n1/(n1+n2);\n", + "x2 = n2/(n1+n2);\n", + "import math\n", + "\n", + "#Using eq. 7.122 (Page no. 292)\n", + "S = -R*(x1*math.log (x1) + x2*math.log (x2));\n", + "S_tot = S*(n1+n2);\n", + "\n", + "# Results\n", + "print 'Change in entropy of components are %f J/K'%S_tot\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in entropy of components are 1152.565132 J/K\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To determine heat of formation of LiCl in 12 moles of water\n", + "\n", + "# Variables\n", + "Hf = -408.610; \t\t\t#heat of formation (kJ)\n", + "#For reaction 2\n", + "#LiCl + 12H2O --> LiCl(12H2O)\n", + "H_sol = -33.614; \t\t\t#heat of solution (kJ)\n", + "\n", + "#To determine heat of formation of LiCl in 12 moles of water\n", + "#Adding reaction 1 and 2% we get\n", + "\n", + "# Calculations\n", + "#Li + 1/2Cl2 + 12H2O --> LiCl(12H2O)\n", + "H_form = Hf+H_sol;\n", + "\n", + "# Results\n", + "print 'Heat of formation of LiCl in 12 moles of water is %f kJ'%H_form\n", + "\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat of formation of LiCl in 12 moles of water is -442.224000 kJ\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To calculate the free energy of mixing\n", + "\n", + "# Variables\n", + "R = 8.314; \t\t\t#ideal gas constant\n", + "n1 = 3.; \t\t\t#moles of hydrogen\n", + "n2 = 1.; \t\t\t#moles of nitrogen\n", + "T = 298.; \t\t\t#temperature in K\n", + "P1 = 1.; \t\t\t#pressure of hydrogen in bar\n", + "P2 = 3.; \t\t\t#pressure of nitrogen in bar\n", + "import math\n", + "\n", + "# Calculations\n", + "#To calculate the free energy of mixing\n", + "V1 = (n1*R*T)/(P1*10**5); \t\t\t#volume occupied by hydrogen\n", + "V2 = (n2*R*T)/(P2*10**5); \t\t\t#volume occupied by nitrogen\n", + "V = V1+V2; \t\t\t#total volume occupied\n", + "P = ((n1+n2)*R*T)/(V*10**5); \t\t\t#final pressure attained by mixture (bar)\n", + "G1 = R*T*(n1*math.log(P/P1) + n2*math.log(P/P2));\n", + "\n", + "#For step 2, using eq. 7.121 (Page no. 292)\n", + "x1 = n1/(n1+n2);\n", + "x2 = n2/(n1+n2);\n", + "G2 = (n1+n2)*R*T*(x1*math.log (x1) + x2*math.log (x2));\n", + "G = G1+G2; \t\t\t#free energy in J\n", + "\n", + "# Results\n", + "print 'The free energy of mixing when partition is removed is %f kJ'%(G/1000)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The free energy of mixing when partition is removed is -6.487935 kJ\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To calculate the mean heat capacity of 20 mol percent solution\n", + "\n", + "# Variables\n", + "C_water = 4.18*10**3; \t\t\t#heat capacity of water (J/kg K)\n", + "C_ethanol = 2.58*10**3; \t\t#heat capacity of ethanol (J/kg K)\n", + "G1 = -758.; \t\t\t#heat of mixing 20 mol percent ethanol water at 298 K(J/mol)\n", + "G2 = -415.; \t\t\t#heat of mixing 20 mol percent ethanol water at 323 K (J/mol)\n", + "n_wat = 0.8; \t\t\t#moles of water\n", + "n_eth = 0.2; \t\t\t#moles of ethanol\n", + "T1 = 323.; \t\t\t#initial temperature in K\n", + "T2 = 298.; \t\t\t#final temperature in K\n", + "\n", + "# Calculations\n", + "#Step 1: Water is cooled from 323 K t0 298 K\n", + "H1 = n_wat*18*C_water*(T2-T1)/1000; \t\t\t#(J)\n", + "\n", + "#Step 2: Ethanol is cooled from 323 to 298 K\n", + "H2 = n_eth*46*C_ethanol*(T2-T1)/1000; \t\t\t#(J)\n", + "\n", + "#Step 3: 0.8 mol water and 0.2 mol ethanol are mixed at 298 K\n", + "H3 = G1; \t\t\t#(J)\n", + "\n", + "#Step 4: \n", + "H = G2;\n", + "Cpm = (H-H1-H2-H3)/(T1-T2);\n", + "\n", + "# Results\n", + "print 'Mean heat capacity of solution is %f J/mol K'%Cpm\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean heat capacity of solution is 97.648000 J/mol K\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 7.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# To find the final temperature attained\n", + "\n", + "# Variables\n", + "To = 298.; \t\t\t#initial temperature (K)\n", + "Cpm = 97.65; \t\t\t#Mean heat capacity of solution (J/mol K)\n", + "Hs = -758.; \t\t\t#heat of mixing (J/mol)\n", + "H = 0.;\n", + "\n", + "# Calculations\n", + "T = (H-Hs)/Cpm + To;\n", + "\n", + "# Results\n", + "print 'The final temperature attained by the mixing is %f K'%T\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The final temperature attained by the mixing is 305.762417 K\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +}
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