Photon gas: thermodynamics

1. Nov 28, 2007

ed321

1. The problem statement, all variables and given/known data
I’m struggling to reconcile two results about the behaviour of a photon gas, any help would be appreciated:
First of all the Gibbs free energy=0, which means that dG=0=Vdp-SdT
But also p=1/3 U/V and S=4/3 U/T which means p=1/4 ST/V. Now if we call the entropy per unit volume S/V=s, p=1/4sT and differentiating both sides with respect to p at constant s we get 1=1/4s dp/dT which doesn’t agree with the result from the Gibb’s function.

2. Relevant equations
(My expressions for G, p and S agree with wikipedia: http://en.wikipedia.org/wiki/Photon_gas)

3. The attempt at a solution
I think I must not be differentiating p=1/4sT correctly, but I can’t see what I need to do differently.

Thanks.

2. Nov 29, 2007

ed321

Would this be more appropriate somewhere else?

3. Nov 29, 2007

ozymandias

Hey ed,

I'll rephrase your question for you.
Why do you think you can transform

$$0 = V dp - SdT$$

into

$$\frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s$$

(which is really what you're comparing your derivation to - correct me if I'm mistaken.)

--------
Assaf
http://www.physicallyincorrect.com/" [Broken]

Last edited by a moderator: May 3, 2017
4. Dec 1, 2007

ed321

I thought $$\frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s$$
followed straight away from $$0 = V dp - SdT$$ by "dividing" by dT at constant s. Maybe this step isn't valid since you can't always treat dT like a ordinary number, but I can't see any reason this would be the case this time?

$$dG = 0 = V dp - SdT$$ gives
$$\frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_G$$, not
$$\frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s$$.