{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 13 : Introduction to statistical thermodynamics" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.1 page no : 474" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "p1 = 1./6;\t\t\t#probability for the first throw gives 6\n", "p2 = 1./6;\t\t\t#probability for the first throw gives 5\n", "n = 2;\t \t\t#the no.of dice are two\n", "\n", "# Calculations\n", "p = p1*p2*n;\t\t\t#the required probability is\n", "\n", "# Result\n", "print 'the required probability is %3.2f'%(p)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the required probability is 0.06\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.2 pageno : 474" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "p1 = 4./52;\t\t\t#the probability for getting ace in first draw is\n", "p2 = 3./51;\t\t\t#the probability for getting ace in second draw is\n", "p3 = 2./50;\t\t\t#the probability for getting ace in third draw is\n", "p4 = 1./49;\t\t\t#the probability for getting ace in fourth draw is\n", "\n", "# Calculations\n", "p = p1*p2*p3*p4;\t\t\t#total probability is\n", "\n", "# Result\n", "print 'total probability is %3.9f'%(p)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "total probability is 0.000003694\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.3 page no : 475" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "n = 12.;\t\t\t#no.of particles\n", "n1 = 8.;\n", "n2 = 4.;\n", "\n", "# Calculations\n", "p = n*(n-1)*(n-2)*(n-3)/(n2*(n2-1)*(n2-2)*(2**n));\t\t\t#probability of distribution (8,4)\n", "\n", "# Result\n", "print 'probability of distribution 8(4) is %3.5f'%(p)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "probability of distribution 8(4) is 0.12085\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 13.4 page no: 475" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "m = 32;\t\t\t#mass of the oxygen molecule in gm\n", "n = 1.67*10**-27;\t\t\t#mass of one electron\n", "k = 1.38*10**-23;\t\t\t#boltzzmann consmath.tant in ergs/cal\n", "t = 200;\t\t\t#temperature of the oxygen in K\n", "c = (100.+101)/2;\t\t\t#average speed of the oxygen molecule in m/s\n", "\n", "# Calculations\n", "a = m*n/(2*3.14*k*t);\n", "p = 4*3.14*(a**(3./2))*(c**2)*(2.303**(-a));\t\t\t#probability that the oxygen speed is lies between in m/sec\n", "\n", "# Result\n", "print 'probability that the oxygen speed is lies between is %3.6e m/sec'%(p)\n", "print \"Note : answer is slightly different because of rounding error.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "probability that the oxygen speed is lies between is 6.867794e-04 m/sec\n", "Note : answer is slightly different because of rounding error.\n" ] } ], "prompt_number": 10 } ], "metadata": {} } ] }