MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
Problem Set 9 Solutions
1. Given the partition function of a monatomic gas equation:
⎡ 2π kT ⎤
℘=V ⎢
⎣ m ⎥⎦
use the partition function and your knowledge of statistical thermodynamics to derive
the following in terms of n (or No, the total number of particles in the system) AND
R:
a. ln[℘]
⎛ d ln[℘] ⎞
b. ⎜
⎟
⎝ dT ⎠v
3/ 2
⎛ d 2 ln[℘] ⎞
c. ⎜
⎟
2
⎝ dT
⎠v
d. Helmholtz free energy, F
e. Entropy, S
f. Internal free energy, U
g. Heat capacity, Cv
h. Comment on your findings in part g relative to the value obtained for Cv using
classical thermodynamics
1 of 4
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
h. Comment on your findings in part g relative to the value obtained for Cv using
classical thermodynamics:
Cp – Cv = R
Where Cp = 5/2R and Cv = 3/2 R. Thus, both Classical and Statistical
thermodynamics give the same answer!
2 of 4
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
2. Using the Classical Thermodynamics definition of Helmholtz free energy, F, find the
coefficient relationship for pressure. Use your answer and your answer in problem 1a to
derive an equation for pressure. Comment on your answer.
3 of 4
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
3. Define configurational entropy and thermal entropy. Give examples that differentiate the
two from one another.
The total entropy of the system is a combination of Thermal Entropy and
Configurational entropy. The statistical thermodynamics definition of total
entropy is given by:
STotal = Sthermal + Sconfigurational
= k ln Ωthermal + k ln Ω configurational
= k ln Ωthermal Ω configurational
given on page 82 of Gaskell.
Configurational entropy considers the distribution, arrangement or
configuration of particles in space. The more ordered particles are in space
(high symmetry), the fewer the number of distributions in space (Ω) hence
the lower the entropy. For a perfect crystal, Ω = 1. Hence, the entropy is
zero. This cannot be, so there must always be defects in the crystal.
Thermal energy describes the distribution of energy among particles. The
greater the number of ways the thermal energy can be distributed, the
greater the entropy.
4 of 4