Specific Heat at constant pressure for photon gas

[sy7kenny]

Homework Statement

A blackbody photon gas is contained within an evacuated cavity (V = 0.01 m^3).
Calculate C_p for the photon gas at T = 1000K

Homework Equations

C_p - C_v = T(\frac{\partial S} {\partial V}) (\frac{\partial V}{\partial T})
C_v = T(\frac{\partial S} {\partial T})
S = \frac{16}{3} \frac{\sigma}{c} V T^3
p = \frac{1} {3} \frac{U}{V}
dU = đ Q - p dV

The Attempt at a Solution

I got two different answers for the following methods, please let me know if I am neglecting any assumptions.

Method 1 Maxwell Relation
C_p - C_v = T(\frac{\partial S} {\partial V}) (\frac{\partial V}{\partial T})
C_v = T(\frac{\partial S} {\partial T})
C_p = T(\frac{\partial S} {\partial T}) + T(\frac{\partial S} {\partial V}) (\frac{\partial V}{\partial T})

then by using the Maxwell relation,
(\frac{\partial V} {\partial T}) = -(\frac{\partial S}{\partial P})
C_p = T(\frac{\partial S} {\partial T}) -T(\frac{\partial S} {\partial V})(\frac{\partial S}{\partial P})
but S does not depend on pressure thus (\frac{\partial S}{\partial P}) = 0
leaving C_p = T(\frac{\partial S} {\partial T}) = C_v = 16\frac{\sigma}{c}V T^3

Method 2,
Using p = \frac{1} {3} \frac{U}{V},
U = 3pV
C_v = \frac{\partial U}{\partial T}
dU = đ Q - p dV

(đ Q)_p = dU + p dV = 4 p dV = \frac{4}{3} dU
C_p = (\frac{\partial Q}{\partial T})_p

which lead to
C_p = \frac{4}{3} \frac{\partial U}{\partial T} = \frac{4}{3} C_v

in short,
method 1 says C_p = C_v
method 2 says C_p = \frac{4}{3} C_v

I am not sure why the two results do not agree. Any help will be appreciated. Thanks!

 

[QuasiParticle]

The heat capacity of photon gas at constant pressure is a trick question. It's not well defined, since temperature and pressure cannot be varied independently of each other in this system.