PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
Quiz 2
SPRING 2001
Friday, April 13, 2001
NAME: ______________ SOLUTION______________________
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (30%) The Gibbs free energy is defined as G = E – TS + pV, where E is internal energy, T is
temperature, S is entropy, p is pressure, and V is volume.
a) Show that: dG = – SdT + Vdp +dN
dG = dE – TdS – SdT + pdV +Vdp, but the First Law states, dE = TdS – pdV +dN.
Then dG becomes,
dG = TdS – pdV +dN – TdS – SdT + pdV +Vdp = – SdT + Vdp +dN
b) Use the equation given in part a) to show that: V 
G is a function of T, p, and N. Therefore, dG 
G

 p T , N
G
G
G 
 dT 
 dp 
 dN
 T  p, N
 p T , N
 N T , p
A term by term comparison with the equation given in part a) shows that V 
G

 p T , N
c) Use the equation given in part a) to derive any Maxwell relation.
Three are possible, one for each pair of terms in the expression for dG. One is shown here.
G 
 . Then,
Using the method of part b), we also see that S  
 T  p , N
S 
V 
V 
 2G
S 
 2G

 

  

, and
. Then, 

 p T , N  T  p , N
 T  p , N  T p
 p T , N
 p T
1
2. (20%) Consider 1 mole of an ideal gas, having 3 degrees of freedom per molecule, initially at
a temperature of 0C (0C = 273 K), pressure of 1 atm and volume of 2.0 m3. If the gas is
expanded adiabatically to 3.0 m3, what will be its final temperature and pressure?
 1
2

1
2 5
  1.67
3 3

 2.0 m3 
V 

p0V0  pV , so p  p0  0   (1 atm) 
3 
V 
 3.0 m 

5/3

 0.509 atm
p = 0.51 atm = 5.1 ×104 N/m²
 1
 1
T0V0
 TV
 1
V 
, so T  T0  0 
V 
 2.0 m3 

 (273 K)
3 
 3.0 m 
2/3
 208 K
T = 208 K = – 65°C
3. (30%) Consider a system consisting of 4 magnetic dipoles lined up in a row. The dipoles are
labeled 1, 2, 3, or 4 depending on their position in the row. Each dipole either points up or
down (nothing else is allowed) and has an energy –  if the ith dipole is pointing up (parallel
to the magnetic field) and an energy  if the ith dipole is pointing down (antiparallel to the
magnetic field). Suppose the spin system is contact with a heat reservoir which is at a
constant temperature of T.
a.) (10%) What is the partition function  for a single dipole? (Note that it has two energy
levels.)
   e   e  (  )  e   e   e 
n
n
It is not necessary, but this could be expressed as:
 = e + e- = 2 cosh 
b.) (5%) What is the partition function for the entire system of four dipoles? (Note that the
four dipoles are distinguishable because they are in particular positions.) You can give
your answer in terms of  from part (a).
Z   N   4   4 . This answer is sufficient for this part. If it is completed,
Z  (e   e  ) N  (e   e  ) 4  16 cosh 4 
c.) (10%) What is the average energy of the 4-dipole system at temperature T?
E 






ln Z  
ln (e   e  ) 4  
4 ln( e   e   )



E  4
2
 e    e 
e   e 

 4
e  e 
 4 tanh 
e   e  
The last step is not necessary.
d) (5%)What is the energy of this 4-dipole system in the limit of T? There is no need to
do a calculation in this case. Simply present a well-reasoned argument if you prefer.
As T,   0, so   0 also. Since tanh 0 = 0, E  0.
Without the mathematics, you should realize that as T, the + and – states will
become equally populated. The average energy of one dipole will then be zero, and
therefore, the average energy of the entire system will be zero.
4. (10%) Explain why increasing the operating temperature of nuclear power plant from 300 C
(using steam) to 1000 C (using liquid sodium) will increase its efficiency. In both cases,
assume that the waste heat is dumped into a river which maintains a constant 25 C.
T2
, where T1 is the
T1
temperature of the high temperature reservoir and T2 is the temperature of the low
temperature reservoir. Increasing T1 will increase the ideal efficiency of the plant from 48%
to 77%.
The maximum efficiency possible is the Carnot efficiency,   1 
5. (10%) An astronomer observes the light from a distant gas cloud in space. A wavelength
analysis of the light reveals two sharp lines which are characteristic of two energy transitions
in a particular molecule. The lower energy line (2.1 × 10-3 eV) corresponds to the energy for
a transition between the first excited state and the ground state while the higher energy line
(3.2 × 10-3 eV) corresponds to the transition between the second excited state and the ground
state. (These energies are determined in the rest frame of the cloud; all relativistic
corrections due to motion of the cloud with respect to earth have already been included. This
is mentioned for those of you that would think of this complication. Please, do not worry
about it here.) Based on the measured intensities of the two lines, the astronomer determines
that there are 5 times more molecules in the first excited state than there are in the second
excited state. Compute the temperature of the gas cloud based on this information.

E1
E 2  E1
kT
 P  E  E1
E  E1
. Then, ln  1   2
, so T  2
P
kT
 P2 
k ln  1 
 P2 
3
3
3.2  10 eV  2.1 10 eV
T
(8.62  105 eV/K) ln 5
P1 e kT
 E e

P2
e kT
2
T = 7.9 K
3