{ "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": {} } ] }