March 20, 2007
PHY 3513
Supplement to the Homework from Chapter 8: Ashley Carter
8.8(a). The Helmholtz function is described by df = -sdT –Pdv
Thus the change in f during an isothermal process = df = -∫Pdv. For a van der Waals’ gas, the
pressure is given by P = RT/(v-b) –a/v2.
Therefore
⎡1 1⎤
v −b
a⎤
⎡ RT
) − a⎢ − ⎥
Δf = − ∫ dv ⎢
− 2 ⎥ = − RT ln( 2
v1 − b
⎣v − b v ⎦
⎣ v 2 v1 ⎦
(b) Because they are all state variables, it is useful to know the functional form of these functions.
Thus f(v,T) = -RT ln(v-b) –a/v. This tells us that entropy s = -∂f/∂T |v = Rln(v-b)
Thus the internal energy u = f + Ts = -a/v. The change in the internal energy = as in the book.
8.10 Here the problem is best worked out using the Gibbs free energy. Since I did the original
problem in class, I changed the problem. I will do the revised problem using the Helmholtz
free energy.
I changed the problem to:
Prove that cv = T
∂s
⎛ ∂c
| v and ⎜ v
∂T
⎝ ∂v
∂2P
⎞
=
|
T
|v . Then show that for an ideal gas, cv does not
⎟ T
∂T 2
⎠
depend on v.
cv =
∂q
∂s
|v = T
|v
∂T
∂T
For the second part, consider the Helmholtz free energy df = -sdT –Pdv. Therefore
A second differentiation with respect to T leads to the answer.
Since for an ideal gas P=RT/v,
∂2P
|v =0.
∂T 2
To prove the original problem you have to start with the Gibbs’ free energy g(P,T).
∂s
∂P
|T =
|v
∂T
∂v