{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 14 - Equilibrium and the third law" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1 - Pg 372" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the equlibrium constant\n", "#Initialization of variables\n", "import math\n", "n1=0.95\n", "n2=0.05\n", "n3=0.025\n", "P=147 #psia\n", "pa=14.7 #psia\n", "#calculations\n", "n=n1+n2+n3\n", "p1=n1/n *P/pa\n", "p2=n2/n *P/pa\n", "p3=n3/n *P/pa\n", "Kp1= p1/(p2*math.pow(p3,0.5))\n", "Kp2= p1*p1 /(p2*p2 *p3)\n", "#results\n", "print '%s %.1f' %(\"In case 1, Equilibrium constant = \",Kp1)\n", "print '%s %.1f' %(\"\\n In case 2, Equilibrium constant = \",Kp2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "In case 1, Equilibrium constant = 38.5\n", "\n", " In case 2, Equilibrium constant = 1480.1\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2 - Pg 373" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the degree of dissociation\n", "#Initialization of variables\n", "kp=5. \n", "import numpy\n", "from numpy import roots\n", "#calculations\n", "vec=numpy.roots([24,0, 3,-2])\n", "x=vec[2]\n", "vec2=numpy.roots([249,0,3,-2])\n", "y=vec2[2]\n", "#results\n", "print '%s %.2f' %(\"\\n degree of dissociation = \",x)\n", "print '%s %.2f' %(\"\\n If pressure =10 . degree of dissociation =\",y)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " degree of dissociation = 0.34\n", "\n", " If pressure =10 . degree of dissociation = 0.18\n" ] }, { "output_type": "stream", "stream": "stderr", "text": [ "-c:12: ComplexWarning: Casting complex values to real discards the imaginary part\n", "-c:13: ComplexWarning: Casting complex values to real discards the imaginary part\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3 - Pg 373" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the degree of dissociation\n", "#Initialization of variables\n", "k=5.\n", "import numpy\n", "from numpy import roots\n", "#calculations\n", "p=[k*k-1, k-5*k*k,7*k*k, -3*k*k]\n", "vec=roots(p)\n", "x=vec[2]\n", "#results\n", "print '%s %.2f' %(\"degree of dissociation = \",x)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "degree of dissociation = 0.78\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4 - Pg 395" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the work done in all the cases\n", "import math\n", "#Initialization of variables\n", "T=77+460. #R\n", "x1=0.21\n", "x2=1-x1\n", "G=-169557 #Btu/mole\n", "n1=1\n", "n2=3.76\n", "R0=1.986\n", "v=0.0885\n", "pi=14.7\n", "J=778.\n", "#calculations\n", "dg1=-n1*R0*T*math.log(x1)\n", "dg2=-n2*R0*T*math.log(x2)\n", "dg=dg1+dg2\n", "dG=dg+G\n", "W=-dG\n", "W2=-G\n", "p=0.0004 #atm\n", "G1=-n1*R0*T*math.log(1./p)\n", "W3= -(dg1+G+G1)\n", "dgf=v*pi*144/J\n", "#results\n", "print '%s %d %s' %(\"In case 1,Work done =\",W,\"Btu/mole C\")\n", "print '%s %d %s' %(\"\\n In case 2,Work done =\",W2,\"Btu/mole C\")\n", "print '%s %d %s' %(\"\\n In case 3,Work done =\",W3,\"Btu/mole C\")\n", "print '%s %.2f %s' %(\"\\n In case 4,Work done =\",dgf,\" Btu/mole C\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "In case 1,Work done = 166947 Btu/mole C\n", "\n", " In case 2,Work done = 169557 Btu/mole C\n", "\n", " In case 3,Work done = 176236 Btu/mole C\n", "\n", " In case 4,Work done = 0.24 Btu/mole C\n" ] } ], "prompt_number": 5 } ], "metadata": {} } ] }