15: Thermal Properties

Example number 15.1, Page number 323

In [2]:
#importing modules
import math
from __future__ import division

#Variable declaration
k = 1.38*10**-23;      #Boltzmann constant(J/K)
h = 6.626*10**-34;      #Planck's constant(Js)
f_D = 64*10**11;         #Debye frequency for Al(Hz)

#Calculation
theta_D = h*f_D/k;     #Debye temperature(K)
theta_D = math.ceil(theta_D*10)/10;     #rounding off the value of theta_D to 1 decimal

#Result
print "The Debye temperature of aluminium is",theta_D, "K"

The Debye temperature of aluminium is 307.3 K

Example number 15.2, Page number 323

In [3]:
#importing modules
import math
from __future__ import division

#Variable declaration
N = 6.02*10**26;    #Avogadro's number(per kmol)
k = 1.38*10**-23;    #Boltzmann constant(J/K)
h = 6.626*10**-34;    #Planck's constant(Js)
f_D = 40.5*10**12;     #Debye frequency for Al(Hz)
T = 30;        #Temperature of carbon(Ks)

#Calculation
theta_D = h*f_D/k;      #Debye temperature(K)
C_l = 12/5*math.pi**4*N*k*(T/theta_D)**3;       #Lattice specific heat of carbon(J/k-mol/K)
C_l = math.ceil(C_l*10**3)/10**3;     #rounding off the value of C_l to 3 decimals

#Result
print "The lattice specific heat of carbon is",C_l, "J/k-mol/K"

#answer given in the book is wrong in the 2nd decimal

The lattice specific heat of carbon is 7.132 J/k-mol/K

Example number 15.3, Page number 323

In [4]:
#importing modules
import math
from __future__ import division

#Variable declaration
k = 1.38*10**-23;       #Boltzmann constant(J/K)
h = 6.626*10**-34;      #Planck's constant(Js)
theta_E = 1990;        #Einstein temperature of Cu(K)

#Calculation
f_E = k*theta_E/h;     #Einstein frequency for Cu(K)

#Result
print "The Einstein frequency for Cu is",f_E, "Hz"
print "The frequency falls in the near infrared region"

The Einstein frequency for Cu is 4.14458194989e+13 Hz
The frequency falls in the near infrared region

Example number 15.4, Page number 323

In [5]:
#importing modules
import math
from __future__ import division

#Variable declaration
e = 1.6*10**-19;     #Energy equivalent of 1 eV(J/eV)
N = 6.02*10**23;      #Avogadro's number(per mol)
T = 0.05;       #Temperature of Cu(K)
E_F = 7;       #Fermi energy of Cu(eV)
k = 1.38*10**-23;     #Boltzmann constant(J/K)
h = 6.626*10**-34;     #Planck's constant(Js)
theta_D = 348;      #Debye temperature of Cu(K)

#Calculation
C_e = math.pi**2*N*k**2*T/(2*E_F*e);     #Electronic heat capacity of Cu(J/mol/K)
C_V = (12/5)*math.pi**4*(N*k)*(T/theta_D)**3;      #Lattice heat capacity of Cu(J/mol/K)

#Result
print "The electronic heat capacity of Cu is",C_e, "J/mol/K"
print "The lattice heat capacity of Cu is",C_V, "J/mol/K"

#answer for lattice heat capacity given in the book is wrong

The electronic heat capacity of Cu is 2.52566877726e-05 J/mol/K
The lattice heat capacity of Cu is 5.76047891492e-09 J/mol/K

Example number 15.5, Page number 324

In [6]:
#importing modules
import math
from __future__ import division

#Variable declaration
T = 1;      #For simplicity assume temperature to be unity(K)
R = 1;      #For simplicity assume molar gas constant to be unity(J/mol/K)
theta_E = T;    #Einstein temperature(K)

#Calculation
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)
C_V = C_V/3;
C_V = math.ceil(C_V*10**3)/10**3;     #rounding off the value of C_V to 3 decimals

#Result
print "The Einstein lattice specific heat is",C_V, "X 3R"

The Einstein lattice specific heat is 0.921 X 3R

Example number 15.6, Page number 324

In [8]:
#importing modules
import math
from __future__ import division

#Variable declaration
e = 1.6*10**-19;      #Energy equivalent of 1 eV(J/eV)
v = 2;     #Valency of Zn atom
N = v*6.02*10**23;      #Avogadro's number(per mol)
T = 300;     #Temperature of Zn(K)
E_F = 9.38;     #Fermi energy of Zn(eV)
k = 1.38*10**-23;    #Boltzmann constant(J/K)
h = 6.626*10**-34;    #Planck's constant(Js)

#Calculation
N = v*6.02*10**23;      #Avogadro's number(per mol)
C_e = math.pi**2*N*k**2*T/(2*E_F*e);    #Electronic heat capacity of Zn(J/mol/K)
C_e = math.ceil(C_e*10**4)/10**4;     #rounding off the value of C_e to 4 decimals

#Result
print "The molar electronic heat capacity of zinc is",C_e, "J/mol/K"

The molar electronic heat capacity of zinc is 0.2262 J/mol/K
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