{
 "metadata": {
  "name": "",
  "signature": "sha256:8929370eda354c28fce9b829f100333cf5fedfea917f7bb51452b097667996b2"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 14 : The Properties of Solution"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.1 Page No : 322"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "import math \n",
      "M2 = 92. \t\t\t#gms\n",
      "M1 = 78. \t\t\t#gms\n",
      "pb = 118.2 \t\t\t#mm\n",
      "pt = 36.7 \t\t\t#mm\n",
      "\n",
      "# Calculations\n",
      "n1 = M2/(M1+M2)\n",
      "n2 = 1-n1\n",
      "p1 = n1*pb\n",
      "p2 = n2*pt\n",
      "w = p1*M1/(p2*M2)\n",
      "\n",
      "# Results\n",
      "print  'partial pressure of benzene  = %.f mm'%(p1)\n",
      "print  ' partial pressure of toulene  = %.1f mm'%(p2)\n",
      "print  ' weight proportions  = %.2f '%(w)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "partial pressure of benzene  = 64 mm\n",
        " partial pressure of toulene  = 16.8 mm\n",
        " weight proportions  = 3.22 \n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.2 Page No : 325"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "vpe = 42 \t\t\t#atm\n",
      "p2 = 1. \t\t\t#atm\n",
      "\n",
      "# Calculations\n",
      "N2 = p2/vpe\n",
      "\n",
      "# Results\n",
      "print  'Ideal solubility of ethane  = %.3f mole fraction'%(N2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Ideal solubility of ethane  = 0.024 mole fraction\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.3 Page No : 325"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "import math \n",
      "p1 = 25.7 \t\t\t#atm\n",
      "p2 = 11.84 \t\t\t#atm\n",
      "T1 = 173. \t\t\t#K\n",
      "T2 = 153. \t\t\t#K\n",
      "T3 = 25. \t\t\t#C\n",
      "\n",
      "# Calculations\n",
      "dH = math.log10(p1/p2)*4.579*T1*T2/(T1-T2)\n",
      "p =  p1*10**((dH/4.576)*(273+T3-T1)/((273+T3)*T1))\n",
      "s = 1./p\n",
      "\n",
      "# Results\n",
      "print  'Heat of reaction  = %d cal mole**-1'%(dH)\n",
      "print  ' pressure  = %d atm'%(p)\n",
      "print  ' Solubility of methane  = %.5f '%(s)\n",
      "\n",
      "# note : rounding error is there. please check. "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat of reaction  = 2039 cal mole**-1\n",
        " pressure  = 309 atm\n",
        " Solubility of methane  = 0.00323 \n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.4 Page No : 329"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T1 = 20. \t\t\t#C\n",
      "T2 = 80. \t\t\t#C\n",
      "H1 = 4540. \t\t\t#cal mole**-1\n",
      "\n",
      "# Calculations\n",
      "n = 10**(H1*(-T2+T1)/(4.576*(273+T1)*(273+T2)))\n",
      "\n",
      "# Results\n",
      "print  'ideal solubility of napthalene  = %.3f '%(n)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "ideal solubility of napthalene  = 0.266 \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14.5 Page No : 342"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "R = 1.987 \t\t\t#cal mole**-1 K**-1\n",
      "T = 278.6 \t\t\t#K\n",
      "dH = 30.2 \t\t\t#cal g**-1\n",
      "m = 6.054 \t\t\t#gms\n",
      "a = 0.1263 \t\t\t#degrees\n",
      "\n",
      "# Calculations\n",
      "l = R*T**2/(1000*dH)\n",
      "m1 = a/l\n",
      "M2 = m/m1\n",
      "\n",
      "# Results\n",
      "print  'molal depression consmath.tant  = %.2f '%(l)\n",
      "print  ' molality  = %.4f '%(m1)\n",
      "print  ' molecular weight of solute  = %.f gms'%(M2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "molal depression consmath.tant  = 5.11 \n",
        " molality  = 0.0247 \n",
        " molecular weight of solute  = 245 gms\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}