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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6 : Change of state"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.1 page no : 194"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"vl = 1; \t\t\t#volume of water in cc\n",
"vs = 1.0908;\t\t\t#volume of ice in cc\n",
"t = 273;\t \t\t#temperature in k\n",
"p = 76*13.6*981;\t\t#pressure in dynes/sq.cm\n",
"l = 80;\t\t\t #latent heat of fusion in cal\n",
"j = 4.2*10**7;\t\t\t#joules consmath.tant in erg/cal\n",
"\n",
"# Calculations\n",
"v = vl-vs;\t\t\t #change in volume\n",
"T = (v*t*p)/(j*l);\t\t\t#change in melting point of water\n",
"\n",
"# Result\n",
"print 'the change in melting point of water is %.5f'%(T)\n",
"print \"there is wrong answer printed in book. Please calculate manually.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the change in melting point of water is -0.00748\n",
"there is wrong answer printed in book. Please calculate manually.\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.2 page no : 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"vv = 1674.;\t\t\t#volume of vapour in cc\n",
"vl = 1.;\t\t\t #volume of liquid in cc\n",
"p = 760.;\t\t\t#pressure of steam and water in mm\n",
"t = 373.;\t\t\t#temperature in K\n",
"p1 = 27.12;\t\t\t#superincumbent pressure in mm\n",
"\n",
"# Calculations\n",
"v = vv-vl; \t\t\t#change in volume\n",
"l = (v*p1*t*0.024203/(p));\t\t\t#latent heat of vapourisation in cal\n",
"\n",
"# Result\n",
"print 'the latent heat of vapourisation is %3.1f cal'%(l)\n",
"print \"Note: Answer is slightly different because of rounding error.\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the latent heat of vapourisation is 539.0 cal\n",
"Note: Answer is slightly different because of rounding error.\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.3 page no : 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"# Variables\n",
"m = 1./(342*100);\t\t\t#molar concentration of water\n",
"t = 289.;\t\t\t #temperature in K\n",
"p = 53.5*13.6*981;\t\t\t#pressure in dynes/sq.cm\n",
"\n",
"# Calculations\n",
"k = p/(t*m);\t\t\t #the value of k in ergs/mol.deg\n",
"\n",
"# Result\n",
"print 'the value of k is %.1e ergs/mol.deg'%(k)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the value of k is 8.4e+07 ergs/mol.deg\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.4 pageno : 196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"p1 = 4.60;\t \t\t #presure at 0deg.C in mm per deg.C\n",
"p2 = 4.94;\t\t \t#pressure at 1deg.C in mm per deg.C\n",
"t = 0.0072;\t\t\t #lowering the melting point in deg.C\n",
"t1 = 7.1563979*10**(-3);\t\t\t#rise in melting point in deg.C\n",
"p = 760;\t\t\t #atmospheric pressure in mm hg\n",
"\n",
"# Calculations\n",
"dp = p2-p1;\t\t\t #rate of increase of pressure in mm per deg.C\n",
"p3 = (t1*p)/t;\t\t\t #pressure in mm\n",
"dt = (755.4-p3)/dp;\t\t\t#tmperature for the triple point in deg.C\n",
"\n",
"# Result\n",
"print 'temperature for the triple point is %3.6f deg.C'%(dt)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"temperature for the triple point is 0.007188 deg.C\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.5 pageno : 196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables\n",
"v = 21*10**4;\t\t\t#change in volume from vapour to liquid in cc\n",
"Ls = 687;\t\t\t#latent heat of sublimation in cal\n",
"lv = 607;\t\t\t#latent heat of vapourisation in cal\n",
"t = 273;\t\t\t#temperature of water in deg.C\n",
"j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
"\n",
"# Calculations\n",
"sv = lv*j/(t*(v));\t\t\t#slope of vapourisation curve at 0 deg.C in dyne/sq.cm/deg.C\n",
"ss = Ls*j/(t*(v));\t\t\t#slope of sublimation curve at 0 deg.C in dyne/sq.cm/deg.C\n",
"\n",
"# Result\n",
"print 'the slope of vapourisation curve is %.2e dyne/sq.cm/deg.C \\\n",
"\\nthe slope of sublimation curve is %.2e dyne/sq.cm/deg.C'%(sv,ss)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the slope of vapourisation curve is 4.45e+02 dyne/sq.cm/deg.C \n",
"the slope of sublimation curve is 5.03e+02 dyne/sq.cm/deg.C\n"
]
}
],
"prompt_number": 10
}
],
"metadata": {}
}
]
}
|
