# Chapter 8 : applications of statistical thermodynamics¶

## Example 8.1 pg : 202¶

In [1]:

# Variables
T = 70. 			#K
Tr = 85.5 			#K

# Calculations
cvrot = 1.1
cvtra = 1.5
cv = cvtra+cvrot

# Results
print "Cv total  =  %.1f R"%(cv)

Cv total  =  2.6 R


## Example 8.2 pg : 202¶

In [2]:

# Variables
T = 2000. 			#K
Tr = 3340. 			#K

# Calculations
print ("From fig 8.2")
cvrot = 0.85
cvtra = 1.5
cvvib = 1.
cv = cvtra+cvrot+cvvib

# Results
print "Cv total  =  %.2f R"%(cv)

From fig 8.2
Cv total  =  3.35 R


## Example 8.3 pg : 208¶

In [3]:

# Variables
T = 200. 			#K
the = 398. 			#K

# Calculations
ratio = T/the
cv = 4.9

# Results
print "Specific heat of aluminium  =  %.1f cal/g mol K"%(cv)

Specific heat of aluminium  =  4.9 cal/g mol K


## Example 8.4 pg : 208¶

In [4]:

# Variables
T = 10. 			#K
td = 315. 			#K

# Calculations
cv = 464.4 *(T/td)**3

# Results
print "specific heat of copper  =  %.5f cal/g mol K"%(cv)

specific heat of copper  =  0.01486 cal/g mol K


## Example 8.5 pg : 213¶

In [6]:

# Variables
import math
N0 = 6.025*10**23
M = 63.57
d = 8.94 			#g/cc
h = 6.624*10**-27
me = 9.1*10**-28

# Calculations
NbyV = N0*d/M
mu0 = h**2 *(3*NbyV/ math.pi)**(2./3) /(8*me)
e0 = 0.6*mu0*10**-7
Teq = 2*e0/(3*1.38*10**-23)

# Results
print "Equivalent temperature  =  %d K"%(Teq)

# check using calculator

Equivalent temperature  =  32681 K


## Example 8.6 pg : 214¶

In [7]:

# Variables
import math
T = 300.         			#K
mu = 1.13*10**-18
k = 1.38*10**-23

# Calculations
cv = math.pi**2 *k*T/(2*mu)

# Results
print "Electron contribution  =  %.4f R"%(cv)

Electron contribution  =  0.0181 R


## Example 8.7 pg : 223¶

In [8]:

# Variables
sig = 5.668*10**-5
T1 = 1000. 			#K
T2 = 2000. 			#K

# Calculations
Eb1 = sig*T1**4 *10**-7
Eb2 = sig*T2**4 *10**-7

# Results
print "total energy emitted in case 1  =  %.3f Watts/cm**2"%(Eb1)
print " total energy emitted in case 2  =  %.3f Watts/cm**2"%(Eb2)

total energy emitted in case 1  =  5.668 Watts/cm**2
total energy emitted in case 2  =  90.688 Watts/cm**2