{ "metadata": { "celltoolbar": "Raw Cell Format", "name": "", "signature": "sha256:58e66a1486b17622aacd34bb93b225c18134442ddc1f2fbedecd0abdd0c9b88e" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "A TEXTBOOK OF PHYSICAL CHEMISTRY BY K.I. KAPOOR" ] }, { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "CHAPTER NUMBER 1 : EQUILIBRIUM BETWEEN PHASES" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE NUMBER 1.5.1 : PAGE NUMBER 10" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "T1 = 234.5 # Temperature in K\n", "P = 1 # Pressure in atm\n", "rho1 = 14.19 # Density of solid Hg in g/(cm**3)\n", "rho2 = 13.70 # Density of liquid Hg in g/(cm**3)\n", "V = 200.59 # volume of liquid and solid in g/mol\n", "delV = ((V/rho2)-(V/rho1))*(10**-3)# in dm**3/mol\n", "delTdelP = 0.0051 # K/atm\n", "R1 = 8.314 # in J\n", "R2 = 0.082 # in (dm)**3/atm\n", "delH = ((delV*T1)/(delTdelP))*(R1/R2)*10**-3;#molar heat of fusion in kJ/mol\n", "print \" delH = \",round(delH,4),\"(KJ)/mol \"\n", "T2 = 273# in K\n", "delP = (delH*(R2/R1)*(T2-T1))/(delV*T1)*10**3;#pressure required to raise melting point to T2 in atm\n", "print \" delP = \",round(delP,4),\"atm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " delH = 2.3571 (KJ)/mol \n", " delP = 7549.0196 atm\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE NUMBER 1.5.2 : PAGE NUMBER 15" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math;\n", "T1=373.15;#in K\n", "P=1;#atm\n", "Vv=1674;#in cm**3/gm\n", "delPdelT=27.12;#in torr/K\n", "R1=8.314;#in J\n", "R2=0.082;#in atm/(dm)**3\n", "delH=((delPdelT)/760)*T1*((Vv*10**-3)*18)*(R1/R2)\n", "print \" delH = \",round(delH,4),\" J/mol \"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " delH = 40680.2549 J/mol \n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE NUMBER 1.5.3 : PAGE NUMBER 16" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math;\n", "T1=313.75;#in K\n", "P1=59.1;#in torr\n", "T2=353.15;#in K\n", "P2=298.7;#in torr\n", "R=2.303*8.314;#in J/(K*mol)\n", "delH=R*math.log10(P2/P1)*((T2*T1)/(T2-T1))\n", "print \" delH= \",round(delH,4),\" J/mol \"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " delH= 37888.375 J/mol \n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE NUMBER 1.5.4 : PAGE NUMBER 16" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "T1=325.15;#in K\n", "T2=338.15;#in K\n", "P2=760;#in torr\n", "DelHm_v=10.5;#\n", "P1=P2/(10**((DelHm_v/2.303)*((T2/T1)-1)));#in torr\n", "print \" P1= \",round(P1,4),\"torr \"\n", "P=200;#in torr\n", "T=T2/(1+((2.303/10.5)*math.log10(P2/P)));#in K\n", "print \" T =\",round(T,4),\" K \"\n", "I=math.log10(P2)-(((DelHm_v*T2)/2.303)*(-1/T2));#\n", "print \" I = \",round(I,4)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " P1= 499.4901 torr \n", " T = 306.1154 K \n", " I = 7.4401\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE NUMBER 1.5.5 : PAGE NUMBER 17" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math;\n", "P=760;#in torr\n", "dP=52;#in torr\n", "dT=2;#in K\n", "DelH_RTb=10.5;#Trouton rule\n", "Tb=(DelH_RTb*P)/(dP/dT)\n", "print \" Tb = \",round(Tb,4),\" K\"\n", "R=8.314;#in J/Kmol\n", "DelH_v=(DelH_RTb*R*Tb)\n", "print \" DelH_v = \",round(DelH_v,4),\"J/mol \"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Tb = 306.9231 K\n", " DelH_v = 26793.4638 J/mol \n" ] } ], "prompt_number": 16 } ], "metadata": {} } ] }