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Physics 4230, Fall 2008, Midterm Exam 2, Solutions
1. (a) System B is at a higher temperature than system A (TB> TA). Recall the definition of temperature in
terms of entropy:
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T=∂S
∂U N ,V = slope of S(U). (1)
Since system B has a smaller slope of S(U), it has a larger temperature.
(b) Here, Qis the heat flowing from the hotter system to the colder one. So, it’s the heat flowing from B to A.
We define QAto be the heat added to A, and QBthe heat added to B. We have QA=Q, and QB=−Q.
The change in entropy of system A is
∆SA=h∂S
∂U iU=UA
QA=h∂S
∂U iU=UA
Q. (2)
The partial derivative is evaluated at constant Nand V(not explicitly written above), and it’s evaluated
at U=UA.
Similarly, the change in entropy of B is
∆SB=h∂S
∂U iU=UB
QB=h∂S
∂U iU=UB
(−Q). (3)
So the total change in entropy is
∆S= ∆SA+ ∆SB= h∂S
∂U iU=UA
−h∂S
∂U iU=UB!Q. (4)
2. (a) ∆U= 0. To see this, first focus on the gas on the right of the wall. Its temperature before and after the
process is T, and its energy is U= (5/2)Nk T . So its energy doesn’t change. The same is true for the gas
on the left of the wall, and therefore the total energy of all the gas doesn’t change.
(b) Let’s start with the thermodynamic identity:
dU =T dS −P dV . (5)
Now, since the temperature is held constant during this process, the energy Udoesn’t change, since the
energy of a diatomic ideal gas is just U= (5/2)N kT . Therefore we can set dU = 0. Since we’re interested
in the change in entropy, we can solve the thermodynamic identity for dS:
dS =P
TdV =1
T
NkT
VdV =Nk
VdV . (6)
In the second equality above, I used the ideal gas law to plug in P=N kT/V .
Let’s get the change in entropy for the gas on the left and right separately (∆SLand ∆SR), and then add
them up to get the final answer.
We find the entropy change by integrating dS from the initial to the final volume. For the gas on the left,
the initial volume is Vand the final volume is 5V, so
∆SL=ZVf
Vi
dS =Z5V
V
Nk
VdV =N k Z5V
V
dV
V=N k ln(V)
5V
V(7)
=Nkhln(5V)−ln(V)i=N k ln 5V
V=N k ln5. (8)
