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{
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 09: Ideal and Real Solutions"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.2, Page Number 202"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "\n",
      "#Variable Declaration\n",
      "nb = 5.00      #Number of moles of Benzene, mol\n",
      "nt = 3.25      #Number of moles of Toluene, mol\n",
      "T = 298.15     #Temperature, K\n",
      "P = 1.0        #Pressure, bar\n",
      "R = 8.314      #Ideal Gas Constant, J/(mol.K)\n",
      "\n",
      "#Calculations\n",
      "n = nb + nt\n",
      "xb = nb/n\n",
      "xt = 1. - xb\n",
      "dGmix = n*R*T*(xb*log(xb)+xt*log(xt))\n",
      "dSmix = -n*R*(xb*log(xb)+xt*log(xt))\n",
      "\n",
      "#Results\n",
      "print 'Gibbs energy change of mixing is %4.3e J'%dGmix\n",
      "print 'Gibbs energy change of mixing is < 0, hence the mixing is spontaneous'\n",
      "print 'Entropy change of mixing is %4.2f J/K'%dSmix"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Gibbs energy change of mixing is -1.371e+04 J\n",
        "Gibbs energy change of mixing is < 0, hence the mixing is spontaneous\n",
        "Entropy change of mixing is 45.99 J/K\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.3, Page Number 205"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "nb = 5.00      #Number of moles of Benzene, mol\n",
      "nt = 3.25      #Number of moles of Toluene, mol\n",
      "T = 298.15     #Temperature, K\n",
      "R = 8.314      #Ideal Gas Constant, J/(mol.K)\n",
      "P0b = 96.4     #Vapor pressure of Benzene, torr\n",
      "P0t = 28.9     #Vapor pressure of Toluene, torr\n",
      "\n",
      "#Calculations\n",
      "n = nb + nt\n",
      "xb = nb/n\n",
      "xt = 1. - xb\n",
      "P = xb*P0b + xt*P0t\n",
      "y = (P0b*P - P0t*P0b)/(P*(P0b-P0t))\n",
      "yt = 1.-yb\n",
      "\n",
      "#Results\n",
      "print 'Total pressure of the vapor is %4.1f torr'%P\n",
      "print 'Benzene fraction in vapor is %4.3f '%yb\n",
      "print 'Toulene fraction in vapor is %4.3f '%yt"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total pressure of the vapor is 69.8 torr\n",
        "Benzene fraction in vapor is 0.837 \n",
        "Toulene fraction in vapor is 0.163 \n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.4, Page Number 206"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from sympy import symbols, solve\n",
      "#Variable Declaration\n",
      "nb = 5.00      #Number of moles of Benzene, mol\n",
      "nt = 3.25      #Number of moles of Toluene, mol\n",
      "T = 298.15     #Temperature, K\n",
      "R = 8.314      #Ideal Gas Constant, J/(mol.K)\n",
      "P0b = 96.4     #Vapor pressure of Benzene, torr\n",
      "P0t = 28.9     #Vapor pressure of Toluene, torr\n",
      "nv = 1.5       #moles vaporized, mol\n",
      "\n",
      "#Calculations\n",
      "n = nb + nt\n",
      "nl = n - nv\n",
      "zb = nb/n\n",
      "\n",
      "x,y, P = symbols('x y P')\n",
      "e1 = nv*(y-zb)-nl*(zb-x)\n",
      "print 'Mass Balance:', e1\n",
      "e2 = P - (x*P0b + (1-x)*P0t)\n",
      "print 'Pressure and x:',e2\n",
      "e3 = y - (P0b*P - P0t*P0b)/(P*(P0b-P0t))\n",
      "print 'Pressure and y:', e3\n",
      "equations = [e1,e2,e3]\n",
      "sol = solve(equations)\n",
      "\n",
      "#Results\n",
      "for i in sol:\n",
      "    if ((i[x] > 0.0 and i[x] <1.0) and (i[P] > 0.0) and (i[y]>zb and i[y]<1.0)):\n",
      "        print 'Pressure is %4.1f torr' %i[P]\n",
      "        print 'Mole fraction of benzene in liquid phase %4.3f' %i[x]\n",
      "        print 'Mole fraction of benzene in vapor phase %4.3f' %i[y]\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Mass Balance: 6.75*x + 1.5*y - 5.0\n",
        "Pressure and x: P - 67.5*x - 28.9\n",
        "Pressure and y: y - 0.0148148148148148*(96.4*P - 2785.96)/P"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "Pressure is 66.8 torr"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "Mole fraction of benzene in liquid phase 0.561\n",
        "Mole fraction of benzene in vapor phase 0.810\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.6, Page Number 212"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "m = 4.50        #Mass of substance dissolved, g\n",
      "ms = 125.0      #Mass of slovent (CCl4), g\n",
      "TbE = 0.65      #Boiling point elevation, \u00b0C\n",
      "Kf, Kb = 30.0, 4.95    #Constants for freezing point elevation \n",
      "                        # and boiling point depression for CCl4, K kg/mol\n",
      "Msolvent = 153.8  #Molecualr wt of solvent, g/mol\n",
      "#Calculations\n",
      "DTf = -Kf*TbE/Kb\n",
      "Msolute = Kb*m/(ms*1e-3*TbE)\n",
      "nsolute = m/Msolute\n",
      "nsolvent = ms/Msolvent \n",
      "x = 1.0 -  nsolute/(nsolute + nsolvent)\n",
      "\n",
      "#Results\n",
      "print 'Freezing point depression %5.2f K'%DTf\n",
      "print 'Molecualr wt of solute %4.1f g/mol'%Msolute\n",
      "print 'Vapor pressure of solvent is reduced by a factor of %4.3f'%x"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Freezing point depression -3.94 K\n",
        "Molecualr wt of solute 274.2 g/mol\n",
        "Vapor pressure of solvent is reduced by a factor of 0.980\n"
       ]
      }
     ],
     "prompt_number": 69
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.7, Page Number 214"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "csolute = 0.500 #Concentration of solute, g/L\n",
      "R = 8.206e-2    #Gas constant L.atm/(mol.K)\n",
      "T = 298.15      #Temperature of the solution, K\n",
      "\n",
      "#Calculations\n",
      "pii = csolute*R*T\n",
      "\n",
      "#Results\n",
      "print 'Osmotic pressure %4.2f atm'%pii\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Osmotic pressure 12.23 atm\n"
       ]
      }
     ],
     "prompt_number": 70
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.8, Page Number 220"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "xCS2 = 0.3502   #Mol fraction of CS2, g/L\n",
      "pCS2 = 358.3    #Partial pressure of CS2, torr\n",
      "p0CS2 = 512.3   #Total pressure, torr\n",
      "\n",
      "#Calculations\n",
      "alpha = pCS2/p0CS2\n",
      "gama = alpha/xCS2\n",
      "\n",
      "#Results\n",
      "print 'Activity of CS2 %5.4f atm'%alpha\n",
      "print 'Activity coefficinet of CS2 %5.4f atm'%gama"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Activity of CS2 0.6994 atm\n",
        "Activity coefficinet of CS2 1.9971 atm\n"
       ]
      }
     ],
     "prompt_number": 72
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.9, Page Number 220"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "xCS2 = 0.3502   #Mol fraction of CS2, g/L\n",
      "pCS2 = 358.3    #Partial pressure of CS2, torr\n",
      "kHCS2 = 2010.   #Total pressure, torr\n",
      "\n",
      "#Calculations\n",
      "alpha = pCS2/kHCS2\n",
      "gama = alpha/xCS2\n",
      "\n",
      "#Results\n",
      "print 'Activity of CS2 %5.4f atm'%alpha\n",
      "print 'Activity coefficinet of CS2 %5.4f atm'%gama"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Activity of CS2 0.1783 atm\n",
        "Activity coefficinet of CS2 0.5090 atm\n"
       ]
      }
     ],
     "prompt_number": 73
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.10, Page Number 221"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "rho = 789.9     #Density of acetone, g/L\n",
      "n = 1.0         #moles of acetone, mol\n",
      "M = 58.08       #Molecular wt of acetone, g/mol\n",
      "kHacetone = 1950 #Henrys law constant, torr\n",
      "#Calculations\n",
      "H = n*M*kHacetone/rho\n",
      "\n",
      "#Results\n",
      "print 'Henrys constant = %5.2f torr'%H"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Henrys constant = 143.38 torr\n"
       ]
      }
     ],
     "prompt_number": 76
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.11, Page Number 221"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "m = 0.5         #Mass of water, kg\n",
      "ms = 24.0       #Mass of solute, g\n",
      "Ms = 241.0      #Molecular wt of solute, g/mol\n",
      "Tfd = 0.359     #Freezinf point depression, \u00b0C or K\n",
      "kf = 1.86       #Constants for freezing point depression for water, K kg/mol\n",
      "\n",
      "#Calculations\n",
      "msolute = ms/(Ms*m)\n",
      "gama = Tfd/(kf*msolute)\n",
      "\n",
      "#Results\n",
      "print 'Activity coefficient = %4.3f'%gama"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Activity coefficient = 0.969\n"
       ]
      }
     ],
     "prompt_number": 81
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.12, Page Number 223"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable Declaration\n",
      "m = 70.0        #Mass of human body, kg\n",
      "V = 5.00        #Volume of blood, L\n",
      "HN2 = 9.04e4    #Henry law constant for N2 solubility in blood, bar\n",
      "T = 298.0       #Temperature, K\n",
      "rho = 1.00      #density of blood, kg/L\n",
      "Mw = 18.02      #Molecualr wt of water, g/mol\n",
      "X = 80          #Percent of N2 at sea level\n",
      "p1, p2 = 1.0, 50.0  #Pressures, bar\n",
      "R = 8.314e-2       #Ideal Gas constant, L.bar/(mol.K)\n",
      "#Calculations\n",
      "nN21  = (V*rho*1e3/Mw)*(p1*X/100)/HN2\n",
      "nN22  = (V*rho*1e3/Mw)*(p2*X/100)/HN2\n",
      "V = (nN22-nN21)*R*T/p1\n",
      "#Results\n",
      "print 'Number of moles of nitrogen in blood at 1 and 50 bar are %3.2e,%3.3f mol'%(nN21,nN22)\n",
      "print 'Volume of nitrogen released from blood at reduced pressure %4.3f L'%V"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Number of moles of nitrogen in blood at 1 and 50 bar are 2.46e-03,0.123 mol\n",
        "Volume of nitrogen released from blood at reduced pressure 2.981 L\n"
       ]
      }
     ],
     "prompt_number": 90
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example Problem 9.14, Page Number 226"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from numpy import arange,array,ones,linalg, divide\n",
      "from matplotlib import pyplot\n",
      "%matplotlib inline\n",
      "#Variable Declaration\n",
      "cCB = array([1e-6,2e-6,3e-6,5e-6,10e-6])\n",
      "nu = array([0.006,0.012,0.018,0.028,0.052])\n",
      "y = nu/cCB\n",
      "print y\n",
      "xlim(0.0, 0.06)\n",
      "ylim(5000,6300)\n",
      "#Calculations\n",
      "A = array([ nu, ones(size(nu))])\n",
      "print A\n",
      "# linearly generated sequence\n",
      "\n",
      "w = linalg.lstsq(A.T,y)[0] # obtaining the parameters\n",
      "print 'slope %8.1f'%w[0]\n",
      "print 'Intercept %8.1f' %w[1]\n",
      "# Use w[0] and w[1] for your calculations and give good structure to this ipython notebook\n",
      "# plotting the line\n",
      "line = w[0]*nu+w[1] # regression line\n",
      "  \n",
      "#Results\n",
      "plot(nu,line,'r-',nu,y,'o')\n",
      "xlabel('$ \\overline{\\upsilon} $')\n",
      "ylabel('$ \\overline{\\upsilon}/C_{CB},  M^{-1} $')\n",
      "#ylabel('$ \\dfrac{\\overline{\\upsilon}}{C_{CB}} $')\n",
      "show()"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "[ 6000.  6000.  6000.  5600.  5200.]\n",
        "[[ 0.006  0.012  0.018  0.028  0.052]\n",
        " [ 1.     1.     1.     1.     1.   ]]"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "slope -19188.2\n",
        "Intercept   6205.2\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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095XAcqC/mbUDdnL3Wcl+jwOnpx+2iMg2dO0KEydCRQWVF13BSXsOoaz/NZx0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\n9ASq8UBFKwAAAABJRU5ErkJggg==\n",
       "text": [
        "<matplotlib.figure.Figure at 0x5851530>"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}