Specific Heat at constant pressure for photon gas

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1. Mar 17, 2015

sy7kenny

1. The problem statement, all variables and given/known data
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

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

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

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

$$C_p = \frac{4}{3} \frac{\partial U}{\partial T} = \frac{4}{3} C_v$$

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!

2. Mar 17, 2015

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.

Last edited: Mar 17, 2015