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{
"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": {}
}
]
}
|
