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{
"metadata": {
"name": "",
"signature": "sha256:2292e5def6e87e01b63e6b748e8fe3955bb5676e5121c51dac319cd9531c4833"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"15: Thermal Properties "
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.1, Page number 323"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"k = 1.38*10**-23; #Boltzmann constant(J/K)\n",
"h = 6.626*10**-34; #Planck's constant(Js)\n",
"f_D = 64*10**11; #Debye frequency for Al(Hz)\n",
"\n",
"#Calculation\n",
"theta_D = h*f_D/k; #Debye temperature(K)\n",
"theta_D = math.ceil(theta_D*10)/10; #rounding off the value of theta_D to 1 decimal\n",
"\n",
"#Result\n",
"print \"The Debye temperature of aluminium is\",theta_D, \"K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Debye temperature of aluminium is 307.3 K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.2, Page number 323"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"N = 6.02*10**26; #Avogadro's number(per kmol)\n",
"k = 1.38*10**-23; #Boltzmann constant(J/K)\n",
"h = 6.626*10**-34; #Planck's constant(Js)\n",
"f_D = 40.5*10**12; #Debye frequency for Al(Hz)\n",
"T = 30; #Temperature of carbon(Ks)\n",
"\n",
"#Calculation\n",
"theta_D = h*f_D/k; #Debye temperature(K)\n",
"C_l = 12/5*math.pi**4*N*k*(T/theta_D)**3; #Lattice specific heat of carbon(J/k-mol/K)\n",
"C_l = math.ceil(C_l*10**3)/10**3; #rounding off the value of C_l to 3 decimals\n",
"\n",
"#Result\n",
"print \"The lattice specific heat of carbon is\",C_l, \"J/k-mol/K\"\n",
"\n",
"#answer given in the book is wrong in the 2nd decimal"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The lattice specific heat of carbon is 7.132 J/k-mol/K\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.3, Page number 323"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"k = 1.38*10**-23; #Boltzmann constant(J/K)\n",
"h = 6.626*10**-34; #Planck's constant(Js)\n",
"theta_E = 1990; #Einstein temperature of Cu(K)\n",
"\n",
"#Calculation\n",
"f_E = k*theta_E/h; #Einstein frequency for Cu(K)\n",
"\n",
"#Result\n",
"print \"The Einstein frequency for Cu is\",f_E, \"Hz\"\n",
"print \"The frequency falls in the near infrared region\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Einstein frequency for Cu is 4.14458194989e+13 Hz\n",
"The frequency falls in the near infrared region\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.4, Page number 323"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"e = 1.6*10**-19; #Energy equivalent of 1 eV(J/eV)\n",
"N = 6.02*10**23; #Avogadro's number(per mol)\n",
"T = 0.05; #Temperature of Cu(K)\n",
"E_F = 7; #Fermi energy of Cu(eV)\n",
"k = 1.38*10**-23; #Boltzmann constant(J/K)\n",
"h = 6.626*10**-34; #Planck's constant(Js)\n",
"theta_D = 348; #Debye temperature of Cu(K)\n",
"\n",
"#Calculation\n",
"C_e = math.pi**2*N*k**2*T/(2*E_F*e); #Electronic heat capacity of Cu(J/mol/K)\n",
"C_V = (12/5)*math.pi**4*(N*k)*(T/theta_D)**3; #Lattice heat capacity of Cu(J/mol/K)\n",
"\n",
"#Result\n",
"print \"The electronic heat capacity of Cu is\",C_e, \"J/mol/K\"\n",
"print \"The lattice heat capacity of Cu is\",C_V, \"J/mol/K\"\n",
"\n",
"#answer for lattice heat capacity given in the book is wrong"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The electronic heat capacity of Cu is 2.52566877726e-05 J/mol/K\n",
"The lattice heat capacity of Cu is 5.76047891492e-09 J/mol/K\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.5, Page number 324"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"T = 1; #For simplicity assume temperature to be unity(K)\n",
"R = 1; #For simplicity assume molar gas constant to be unity(J/mol/K)\n",
"theta_E = T; #Einstein temperature(K)\n",
"\n",
"#Calculation\n",
"C_V = 3*R*(theta_E/T)**2*math.exp(theta_E/T)/(math.exp(theta_E/T)-1)**2; #Einstein lattice specific heat(J/mol/K)\n",
"C_V = C_V/3;\n",
"C_V = math.ceil(C_V*10**3)/10**3; #rounding off the value of C_V to 3 decimals\n",
"\n",
"#Result\n",
"print \"The Einstein lattice specific heat is\",C_V, \"X 3R\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Einstein lattice specific heat is 0.921 X 3R\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 15.6, Page number 324"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"e = 1.6*10**-19; #Energy equivalent of 1 eV(J/eV)\n",
"v = 2; #Valency of Zn atom\n",
"N = v*6.02*10**23; #Avogadro's number(per mol)\n",
"T = 300; #Temperature of Zn(K)\n",
"E_F = 9.38; #Fermi energy of Zn(eV)\n",
"k = 1.38*10**-23; #Boltzmann constant(J/K)\n",
"h = 6.626*10**-34; #Planck's constant(Js)\n",
"\n",
"#Calculation\n",
"N = v*6.02*10**23; #Avogadro's number(per mol)\n",
"C_e = math.pi**2*N*k**2*T/(2*E_F*e); #Electronic heat capacity of Zn(J/mol/K)\n",
"C_e = math.ceil(C_e*10**4)/10**4; #rounding off the value of C_e to 4 decimals\n",
"\n",
"#Result\n",
"print \"The molar electronic heat capacity of zinc is\",C_e, \"J/mol/K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The molar electronic heat capacity of zinc is 0.2262 J/mol/K\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
