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{
"metadata": {
"name": "",
"signature": "sha256:7cd73e591684a9b750091e874ca08a72b326eee20de62e33004e879fa5300f72"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 08: Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example Problem 8.2, Page Number 186 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import log, exp\n",
"\n",
"#Varialble Declaration\n",
"Tn = 353.24 #normal boiling point of Benzene, K\n",
"pi = 1.19e4 #Vapor pressure of benzene at 20\u00b0C, Pa\n",
"DHf = 9.95 #Latent heat of fusion, kJ/mol\n",
"pv443 = 137. #Vapor pressure of benzene at -44.3\u00b0C, Pa\n",
"R = 8.314 #Ideal Gas Constant, J/(mol.K)\n",
"Pf = 101325 #Std. atmospheric pressure, Pa\n",
"T20 = 293.15 #Temperature in K\n",
"P0 = 1.\n",
"Pl = 10000.\n",
"Ts = -44.3 #Temperature of solid benzene, \u00b0C\n",
"\n",
"#Calculations\n",
"Ts = Ts + 273.15\n",
"#Part a\n",
"\n",
"DHv = -(R*log(Pf/pi))/(1./Tn-1./T20)\n",
"#Part b\n",
"\n",
"DSv = DHv/Tn\n",
"DHf = DHf*1e3\n",
"#Part c\n",
"\n",
"Ttp = -DHf/(R*(log(Pl/P0)-log(pv443/P0)-(DHv+DHf)/(R*Ts)+DHv/(R*T20)))\n",
"Ptp = exp(-DHv/R*(1./Ttp-1./Tn))*101325\n",
"\n",
"#Results\n",
"print 'Latent heat of vaporization of benzene at 20\u00b0C %4.1f kJ/mol'%(DHv/1000)\n",
"print 'Entropy Change of vaporization of benzene at 20\u00b0C %3.1f J/mol'%DSv\n",
"print 'Triple point temperature = %4.1f K for benzene'%Ttp\n",
"print 'Triple point pressure = %4.2e Pa for benzene'%Ptp"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Latent heat of vaporization of benzene at 20\u00b0C 30.7 kJ/mol\n",
"Entropy Change of vaporization of benzene at 20\u00b0C 86.9 J/mol\n",
"Triple point temperature = 267.3 K for benzene\n",
"Triple point pressure = 3.53e+03 Pa for benzene\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example Problem 8.3, Page Number 191"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import cos, pi\n",
"\n",
"#Varialble Declaration\n",
"gama = 71.99e-3 #Surface tension of water, N/m\n",
"r = 1.2e-4 #Radius of hemisphere, m\n",
"theta = 0.0 #Contact angle, rad\n",
"\n",
"#Calculations\n",
"DP = 2*gama*cos(theta)/r\n",
"F = DP*pi*r**2\n",
"\n",
"#Results\n",
"print 'Force exerted by one leg %5.3e N'%F"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Force exerted by one leg 5.428e-05 N\n"
]
}
],
"prompt_number": 40
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example Problem 8.4, Page Number 191"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import cos\n",
"\n",
"#Varialble Declaration\n",
"gama = 71.99e-3 #Surface tension of water, N/m\n",
"r = 2e-5 #Radius of xylem, m\n",
"theta = 0.0 #Contact angle, rad\n",
"rho = 997.0 #Density of water, kg/m3\n",
"g = 9.81 #gravitational acceleration, m/s2\n",
"H = 100 #Height at top of redwood tree, m\n",
"\n",
"#Calculations\n",
"h = 2*gama/(rho*g*r*cos(theta))\n",
"\n",
"#Results\n",
"print 'Height to which water can rise by capillary action is %3.2f m'%h\n",
"print 'This is very less than %4.1f n, hence water can not reach top of tree'%H"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Height to which water can rise by capillary action is 0.74 m\n",
"This is very less than 100.0 n, hence water can not reach top of tree\n"
]
}
],
"prompt_number": 41
}
],
"metadata": {}
}
]
}
|
