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{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 16 - Molecular substances"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E1 - Pg 385"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the net dipole moment\n",
      "#Initialization of variables\n",
      "import math\n",
      "e=1.609*math.pow(10,-19) #C\n",
      "#calculations\n",
      "mux=(-0.36*e*(-0.8) + 0.45*e*(2.1) )*math.pow(10,-12) /(3.33564*math.pow(10,-30))\n",
      "muy=-0.96\n",
      "muz=0\n",
      "mux=-1.1\n",
      "mu=math.sqrt(mux*mux+muy*muy+muz*muz)\n",
      "#results\n",
      "print '%s %.1f %s' %(\"Net dipole moment =\",mu,\"D\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Net dipole moment = 1.5 D\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E2 - Pg 390"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the potential energy of the system\n",
      "#Initialization of variables\n",
      "import math\n",
      "Na=6.023*math.pow(10,23) # /mol\n",
      "e=1.60228*math.pow(10,-19) #C\n",
      "e0=8.85419*math.pow(10,-12) #C^2/J m\n",
      "#calculations\n",
      "factor=Na*e*e /(4*math.pi*e0)\n",
      "#Multiply by Z^2/R to get the value of potential energy. Plot the graph\n",
      "#results\n",
      "print '%s %.3e %s' %(\"Potential energy =\",factor,\" Z*Z/R  kJ/mol\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Potential energy = 1.390e-04  Z*Z/R  kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example I1 - Pg 383"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calcualte the predicted dipole moment\n",
      "#Initialization of variables\n",
      "EH=2.1\n",
      "EBr=2.8\n",
      "#calculations\n",
      "diff=-EH+EBr\n",
      "#results\n",
      "print '%s %.1f %s' %(\"Prediced dipole moment =\",diff,\"D\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Prediced dipole moment = 0.7 D\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example I2 - Pg 387"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the potential energy\n",
      "#Initialization of variables\n",
      "import math\n",
      "mu1=1.4 #D\n",
      "mu2=1.4 #D\n",
      "angle=180. #degrees\n",
      "d=3 #nm\n",
      "D=4.7*math.pow(10,-30) #C m\n",
      "#calculations\n",
      "Vmol=D*D*(1-3*math.cos(angle*math.pi/180.)*math.cos(angle*math.pi/180.))/(4*math.pi*8.854*math.pow(10,-12) *math.pow((d*math.pow(10,-9)),3))\n",
      "V=Vmol*(6.023*math.pow(10,23))\n",
      "#results\n",
      "print '%s %.1f %s' %(\"Potential energy =\",V,\" J/mol\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Potential energy = -8.9  J/mol\n"
       ]
      }
     ],
     "prompt_number": 4
    }
   ],
   "metadata": {}
  }
 ]
}