PHGN341: Thermal Physics
Exam I - February 22, 2013
NAME: KEY
1. (10) Short answer: a. (2) Define microstate for a system of N spin 1/2 atoms.
Solution: A microstate is given by the precise spin state (up or down) of each atom, e.g. | ↑↑↓ · · · ↑i.
b. (2) Define macrostate for a system of N spin 1/2 atoms.
Solution: A macrostate is given by the total number of spins and the net number of spins in one of their allowed spin
states, e.g. |N N↑ i.
c. (2) State the equal probability a priori hypothesis.
Solution: Every accessible microstate of the system is equally probable.
d. (2) If equilibrium is defined as the most likely macrostate, what does the equal probability hypothesis imply for the
entropy at equilibrium?
Solution: The equilibrium macrostate is the most likely macrostate. Since all accessible microstates are equally likely,
the most likely macrostate is that with greatest number of microstates. Since the entropy is k ln Ω, then the entropy
will be maximal at equilibrium.
e. (2) If a system is not in its equilibrium macrostate, what does the equal probability hypothesis imply for the change
in the entropy as the system spontaneously evolves toward equilibrium?
Solution: If the system is not in equilibrium (i.e. at maximum entropy), the system will evolve spontaneously to more
likely macrostates. By the equal probability hypothesis, the system will therefore evolve to macrostates of greater
multiplicity, i.e. the entropy will increase spontaneously.
2. A volume of a monatomic gas at standard temperature and pressure undergoes an isothermal expansion from its initial
volume to three times the initial volume, i.e. Vi → Vf = 3Vi .
a. How much heat must flow into the gas to keep the process isothermal?
Solution: Since the process is isothermal, by the equipartition theorem the internal energy is constant. Therefore, by
the first law, we have: 0 = ∆U = Q + W ; so Q = −W . The work done in the isothermal expansion is:
Z
W =−
Z
Vf
P dV = −
Vi
N kT
Vf
dV = −N kT ln
= −N kT ln 3.
V
Vi
(1)
where the ideal gas law was used. Therefore the heat that must be added is Q = N kT ln 3.
b. What is the change in the Sackur-Tetrode entropy when a volume of gas undergoes an isothermal expansion to three
times the initial volume? Show all work.
Solution: Again, since the process is isothermal, the internal energy, U remains constant; so the change in the SackurTetrode entropy is
∆S = Sf − Si = S(N, Vf , U ) − S(N, Vi , U ) = N k ln 3.
(2)
c. How are the two expressions related?
Solution: Examining the two expressions, we find: ∆S =
an isothermal process.
1
Q
T
which is the classical result for the change in entropy for
3. (35) Consider a single gas molecule of mass, m, and velocity ~v = {vx , vy , vz }, confined to a cylinder of height, H,
and radius, Rc , that lies with its axis parallel with the z-axis with the bottom end at z=0. Assume the gas molecule
elastically reflects from the walls of the cylinder. Show all work and give your answers in terms of the quantities given
in the problem statement. Ignore gravity. Show all work.
a. (10) Over a long time what is the average force the molecule exerts on the top of the cylinder?
Solution: The force is given by F =
∆p
∆t .
The momentum change in the molecule after a collision with the top of the
2H
vz .
cylinder is ∆p = 2mvz , and the time between collisions is ∆t =
Thus, F =
mvz2
H .
b. (10) Suppose there are now N molecules in the cylinder. What is average pressure they exert on the top of the
cylinder?
Solution: The total force is the sum:
F =
N
X
N
2
X
mvz,i
Fi =
i
i
where
vz2 =
H
=
mN vz2
H
N
1 X 2
v
N i z,i
is the square of the z-component of velocity averaged over all the molecules. The pressure is the force per unit area, or
P =
mN vz2
mN vz2
F
=
=
πRc2
HπRc2
V
where V = HπRc2 is the volume.
c. (10) What is the average kinetic energy of all the molecules in the cylinder as a function of the pressure and volume?
Solution: The average kinetic energy is KE = N 12 mv 2 = N
average vx2 = vy2 = vz2 . Thus, using the result from part (b),
P =
1
2
2 m(vx
+ vy2 + vz2 ) = N
3
2
2 mvz ,
where we used that on
2 KE
N mvz2
=
V
3 V
Thus, KE = 32 P V .
d. (5) Using your result from (c) and the ideal gas model, find the temperature of the gas in terms of the average
kinetic energy?
Solution: Using the ideal gas law, P V = N kT and the result from part (c), we find for the average kinetic energy is
given by KE = 32 N kT (i.e. the equipartition theorem for an ideal gas).
4. A rubber band can be modeled as a one-dimensional collection of N long polymer molecules that can connect at the
ends of each other either pointing to the left or to the right. Each molecule has length, l0 ; so the total length of the
rubber band (assumed pointing to the left) is L ' l0 (NL − NR ), where NL is the number of molecules pointing to the
left and NR is the number of molecules pointing on the right.
(a) What defines a microstate and what defines a macrostate for this rubber band model?
Solution: The microstate is specified by giving the orientation (left or right) of every molecule in the rubber band. The
macrostate is specified by giving the total number of molecules and the net number in one orientation, say left, NL
(which then determines the net number to the right: NR = N − NL .) Equivalently the macrostate is determined by
number of molecules and the overall length, L = l0 (NL − NR ).
(b) Show that for large-N the approximate multiplicity for this model of a rubber band can be written as:
−NR −NL
NR
NL
Ω(N, NL , NR ) '
.
N
N
2
Find the entropy for the rubber band as a function of its net length, L, and the total number of molecules, N .
Solution: From the definition of entropy we have: S = k ln Ω ' −k [NL ln(NL /N ) + NR ln(NR /N )]. Defining f =
NL /N , this gives: S = −kN [f ln f + (1 − f ) ln(1 − f )]. Since the overall length is L = l0 (NL − NR ), we have f =
L
1
L
1
2 (1 + N l0 ) and 1 − f = 2 (1 − N l0 ); so
S
1
L
1
L
1
L
1
L
(1 +
) ln (1 +
) + (1 −
) ln (1 −
)
2
N l0
2
N l0
2
N l0
2
N l0
kN
L
L
L
L
= −
−2 ln 2 + (1 +
) ln(1 +
) + (1 −
) ln(1 −
) .
2
N l0
N l0
N l0
N l0
= −kN
(c) From the entropic definition of force for the rubber band, F = T
∂S
∂L N ,
(3)
find the force as a function of {L, N, T }.
Solution: Taking the derivative, find
∂S
kN T
1
L
1
L
F = T
=−
ln(1 +
)−
ln(1 −
)
∂L N
2
N l0
N l0
N l0
N l0
L
L
kT
ln(1 +
) − ln(1 −
)
= −
2l0
N l0
N l0
(4)
(d) From your result in part (c), find the effective spring constant, ks , for the rubber band. For small L the effective
spring constant is defined as F ' −ks L. Comment on how ks depends on temperature?
Solution: The multiplicity
peaks when NL = NR or when L ∼ 0. This means that L << N l0 , so we can expand the logs
L
kT
2
= − NkTl2 L. The effective spring constant is therefore ks = NkTl2 . (The units of ks are J/m2 .)
giving, F ' − 2l
N l0
0
0
0
Note that the spring constant actually gets larger as the temperature increases which leads to the counter intuitive
result that a rubber band will contract when heated.
3