NAME: KEY
PHGN341: Thermal Physics
Exam I - February 19, 2010
1. Suppose an air bubble of initial volume 10 cm3 (10−5 m3 ) and temperature of 20 C (298 K) rises from the bottom of
a 5 m deep lake. Assume the water temperature is a uniform 20 C. The pressure depends on depth according to the
relation: P = P0 + ρgd, where P0 is the pressure at the surface, P0 = 1.0 × 105 Pa, ρ = 103 kg/m3 is the density of
water, g = 9.8 m/s2 is the acceleration of gravity, and d is the depth. Show all work. (kB = 1.38 × 10−23 J/K, R = 8.31
J/K-mol)
(a) What is the pressure at the bottom of the lake?
Solution: Plugging into the given equation for the pressure, find:
3
2
P = P0 + ρgd = 1.0 × 105 Pa + (103 kg/m )(9.8 m/s )(20 m) = 1.49 × 105 Pa.
(b) How many air molecules are in the air bubble?
Solution: Using the ideal gas law, P V = N kT , and using the pressure from part a, find N = (P V )/(kT ) = 3.68 × 1020 .
(c) Suppose the bubble rises fast enough such that no heat has time to enter the bubble before it reaches the surface;
that is, its expansion is adiabatic. What is the volume of the air bubble when it reaches the surface?
Solution: The adiabatic condition is P V γ = constant, where γ = (1 + 2/f ). For this case (diatomic below vibrational
1/γ
threshold), f = 5; so γ = 7/5. The constant is given by the initial conditions; therefore, Vf = ((Pi Viγ )/Pf )
=
−5
3
1.33 × 10 m .
(d) For the situation described in part (c), what is the temperature of the air in the bubble when it reaches the surface?
Solution: Using the results from above, the ideal gas law gives the final temperature: Tf = Pf Vf /(N k) = 261.6 K.
2. Consider two magnetic systems in thermal contact with each other but otherwise thermally isolated. The first, system
A, consists of four spin-1/2 atoms. Each atom in system A can have two possible values for its magnetic quantum
number: mA = +1/2, −1/2. The second, system B, consists of three spin-1 atoms. Each of these atoms can have three
possible values for its magnetic quantum number: mB = +1, 0, −1.
(a) How many microstates does each system have separately and together?
Solution: For system A there are 24 = 16 microstates while for system B there are 33 = 27 microstates. Thus, there
are 16 × 27 = 432 microstates total.
(b) Define the macroscopic magnetization for each system as the net value of the magnetic quantum number, i.e.
MA = [(+ 21 ) nA (+ 12 ) + (− 12 ) nA (− 12 )] and MB = [(+1) nB (+1) + (0) nB (0) + (−1) nB (−1)], where nA (mA ) (i.e. the
function,“nA -of-mA ”) is the number of atoms in system A with magnetic quantum number, mA = ± 21 , and similarly
for system B but with mB = −1, 0, +1. Let ΩA (NA = 4, MA ) be the multiplicity for system A with magnetization MA
and similarly for system B. Let’s treat the case where the total magnetization is fixed at M = MA + MB = +1. Finish
filling out the multiplicity table below for the M = +1 case where M = MA + MB for each row adds up to M = +1.
(The MB = −3 and MB = −2 cases don’t contribute to the M = +1 case under study because it would require a
magnetization for system A that cannot be reached with 4 spin-1/2 atoms.)
Solution: For system A, you could just write out all 16 possibilities and count how many combinations there
arefor
N↑
each magnetization, or you could use the multiplicity formula for the two state system which is: Ω(N, N↑ ) =
=
N
N!
N↑ !(N −N↑ )! . For this case N↑ is given by the magnetization, MA as N↑ = MA + NA /2. For system B, you just have to
write out 27 possibilities and count how many ways each magnetization can be constructed. The results are given in
the table below.
1
MA
MB (= 1 − MA )
ΩA (4, MA )
0×1=0
-2
3
0×3=0
=1
-1
6
1×6=6
=4
0
7
4 × 7 = 28
=6
+1
6
6 × 6 = 36
=4
+2
3
4 × 3 = 12
=1
+3
1
1×1=1
0 (not possible)
-3
0 (not possible)
+2
+1
0
-1
-2
TOTAL:
4
4
3
4
2
4
1
4
0
4
ΩA (4, MA )ΩB (3, MB = 1 − MA )
1
-4
ΩB (3, MB )
-3
16
TOTAL: 27
TOTAL: 83
(c) What is the probability that system A has MA = +1? Explain your reasoning.
Solution: The probability of magnetization, MA , is the number of ways that magnetization occurs divided by the total
multiplicity. For M= + 1, we have P (MA = +1) = 28/83 = 33.7%.
(d) What is the equilibrium value of MA ? Explain your reasoning.
Solution: The equilibrium value of the magnetization, MA , is that which has the greatest multiplicity, i.e. MA = 0
with a total multiplicity of 36.
3. (a) Consider a closed gas cylinder of fixed total volume, V . A moveable leak-proof piston separates the cylinder into
two systems, A and B. The piston is a good thermal conductor so the two systems are in thermal equilibrium (i.e.
TA = TB ) at all times. Recall that when two systems are in mechanical equilibrium, the composite multiplicity is
maximal with respect to variation of the volume of either one of the systems. Suppressing the energy and particle
number dependence of the multiplicity, the entropy of the two systems is:
SAB (V, VA ) = SA (VA ) + SB (VB ) = k [ln ΩA (VA ) + ln ΩB (VB = V − VA )] ,
where V = VA + VB is the total volume (fixed). Show that when the two systems are in mechanical equilibrium
∂SB
A
under these conditions, the entropies satisfy the condition: ∂S
∂VA = ∂VB . (Note that the systems are always in thermal
equilibrium so the internal energies are not changed as the volume changes.)
Solution: In equilibrium the total entropy is maximal with respect to variations in the volume of system A: 0 =
∂SA
∂SB
∂SA
∂SB
∂
∂
∂VA + ∂VA . Since the total volume is fixed, ∂VA = − ∂VB . Thus, ∂VA − ∂VB = 0.
∂SAB
∂VA
=
(b) Recall that the pressure is the “thing” that is the same when two systems are in mechanical equilibrium. Based on
the result from part (a), one can define the “entropic pressure”, PA , for system A to be:
PA = TA
∂SA
.
∂VA
Use this definition to derive an expression for the pressure of a monatomic ideal gas as a function of volume and
temperature. The Sakur-Tetrode expression for the multiplicity of NA gas atoms of mass, m, confined to volume, VA ,
is
2
"
VA
Ω(NA , VA , UA ) =
NA
4πmUA
3h2 NA
#NA
32
e
5
2
,
where h is Planck’s constant, VA is the volume, and, UA is the internal energy which, by the equipartion theorem, can
be written, UA = 32 NA kT . Show all work.
∂
A
Solution: The pressure is given by PA = TA ∂S
∂VA = kTA ∂VA (ln Ω(NA , VA , UA )). Using the above expression for the
multiplicity, find:
"
3/2 #
∂
e5/2 4πmUA
1
PA = NA kT
ln VA + ln
= NA kTA
2
∂VA
NA 3h NA
VA
which is the ideal gas law.
3