PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
Quiz 2
SPRING 2003
Friday, April 18, 2003
NAME: ________SOLUTION___________________________
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. (15%) The Helmholtz Free Energy is defined as F = E – TS, where E is internal energy, T is
F
 , where
temperature, and S is entropy. For a system where dW = pdV, show that S  
 T V , N
N is the number of particles in the system.
dF = dE – TdS – SdT, but the First Law states, dE = TdS – pdV + dN. Then dF becomes,
dF = TdS – pdV + dN – TdS – SdT = – SdT – pdV + dN
 F
 F
F
 dT 
 dV 
 dN
Also, F is a function of T, V, and N, so dF 
 T  V ,N
 V  T ,N
 N  T ,V
A term by term comparison of the two equations for dF shows that S  
F

 T V , N
2. (40%) One mole of an ideal monatomic gas traverses the idealized Diesel cycle shown in the
figure. Process 12 is adiabatic, and process 23 takes place at constant pressure. Process
34 is also adiabatic, and process 41 takes place at constant volume. (Hint: for one mole
of an ideal monatomic gas, the internal energy is E  32 RT .)
The temperatures of the gas at points 1, 2, 3, and 4 are:
T1 = 300 K,
T2 = 1600 K,
T3 = 1800 K,
1
T4 = 365 K
a) (10%) Find V1/V2, the ratio of the volume of the gas at point 1 to the volume at point 2.
(This is the compression ratio for the Diesel cycle.)
 1
For an adiabatic process, T1V1
1
 1
 T2V2
V 
, so  1 
 V2 
 1

T2
, and
T1
1
V1  T2   1  1600 K  5 / 31
    
 5.331.5

V2  T1 
 300 K 
V1/V2 = ___12.3_____
b) (10%) Find the magnitude of the heat added to the gas in the process 23. Express the
answer in terms of the gas constant R.
Q1 = Cp(T3 – T2) =
5
2
R(1800 K  1600 K)
Q1 = _(500 K)R____
c) (5%) Find the magnitude of the work done by the gas in the process 23. Express the
answer in terms of the gas constant R.
W23 = Q1 – E = (500 K)R –
3
2
R(1800 K  1600 K) = (500 K – 300 K)R
W23 = __(200 K)R___
d) (5%) Find the magnitude of the heat removed from gas in the process 41. Express the
answer in terms of the gas constant R.
Q2  32 R T1  T2  32 R 300 K  365 K
Q2 = __(97.5 K)R___
e) (5%) How much work is done by the gas in one complete cycle? Express the answer in
terms of the gas constant R.
W = Q1 – Q2 = (500 K)R – (97.5 K)R
W = __(402.5 K)R___
Or, W = W12 + W23 + W34 + W41. We know W23 = (200 K)R and W41 = 0.
For W12 and W34 , use W = Q – E. Both are adiabatic, so Q = 0, and W = – E.
Then, W12   32 R(1600 K  300 K)  (1950 K)R , and
W34   32 R(365 K  1800 K)  (2152.5 K)R
Finally, W = – (1950 K)R + (200 K)R + (2152.5 K)R = (402.5 K)R
2
f) (5%) Find the efficiency of a heat engine operating with this cycle.

W (402.5 K ) R

Q1
(500 K ) R
 = _0.805 = 80.5%__
3. (15%) A thin walled vessel of volume V contains N particles of gas which slowly leak out
through a small hole of area A. No particles enter the volume through the hole. (You can
imagine the vessel is in outer space.) Find the time required for for the number of particles in
the vessel to decrease to N/2. Express your answer in terms of A, V, and v .
The rate at which particles leave through the hole is: flux  area 
v
4
A
N
dN
v
Nv

A
A
, so the rate of change of the number of particles is:
V
dt
4
V4
vA

t
dN
vA

N . This differential equation can be solved to give: N  N 0e 4V .
dt
4V
vA

t
N 1
4V
vA
1
4V

 , so 
ln 2
t1 / 2  ln   , and t1 / 2 
When N = ½N0, e
vA
N0 2
4V
2

1/ 2
t1 / 2 
3
4V
4V
2.77V
ln 2 
0.693 
vA
vA
vA
4. (30%) Consider a solid consisting of N atoms, each of which has a magnetic dipole moment.
When a magnetic field is applied to the solid, each dipole either points parallel to the
magnetic field or antiparallel to the magnetic field (nothing else is allowed). The interaction
energy of a dipole that is parallel to the field is – H, and one that is antiparallel to the field
is H. The system is contact with a heat reservoir which is at temperature of T.
a.) (5%) What is the partition function  for a single dipole? (Note that it has two energy
levels.)
   e   e  (  H )  e H  e H  e H
n
n
 = eH + e-H = 2 cosh H
b.) (5%) What is the partition function for the entire system of N dipoles? (Note that the N
dipoles are distinguishable because they are in particular positions.) You can give your
answer in terms of  from part (a).
Z   N . This answer is sufficient for this part. If it is completed,
Z  (e H  e H ) N  (2 cosh H ) N
c.) (5%) What is the average energy of the N-dipole system at temperature T?
E 






ln Z  
ln (e H  e H ) N  
N ln( e H  e H )



E  N
H e H  H e H
e H  e H
  NH

e H e H
  NH tanh H
e H  e H
d) (5%)What is the energy of this N-dipole system in the limit of T0? There is no need to
do a calculation in this case. Simply present a well-reasoned argument if you prefer.
As T0,  , and tanh  = 1. Also, as  , the negative exponentials go to zero,
and the ratio of exponentials goes to 1. Then,
ET  0 = _–_NH___________
This result can be obtained by noting that as T0, all N dipoles will be in the lowest state
with  = – H, so E = – NH.
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e) (5%) If the magnetic field is turned off, while the solid is at a temperature close to T = 0,
the energy of the spin system will change. Based on your previous answers, calculate the
change of the energy. Be careful, signs are important. A positive answer means the
energy increases, and a negative answer means the energy decreases.
When H goes to zero, the energy goes to zero.
E = _+_NH__________
f) (5%) As stated earlier, the spin system is in contact with a heat reservoir. If the heat
reservoir is finite, its temperature can change if heat is added or removed. Suppose it is
finite. Based on your work so far, what will happen to the temperature of the reservoir
when the magnetic field is turned off. (Circle the correct answer.)
THE TEMPERATURE WILL INCREASE
THE TEMPERATURE WILL DECREASE
THE TEMPERATURE WILL REMAIN THE SAME
Since energy must be supplied to the spins, energy must be removed from the reservoir.
Therefore, the temperature of the reservoir decreases. This is the principle of “adiabatic
demagnetization” that is used to cool solids to ultra-low temperatures. In that case, the
“reservoir” is the vibrational motion of the atoms of the solid.
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