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Photon gas: thermodynamics

  1. Nov 28, 2007 #1
    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.

  2. jcsd
  3. Nov 29, 2007 #2
    Would this be more appropriate somewhere else?
  4. Nov 29, 2007 #3
    Hey ed,

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

    [tex]0 = V dp - SdT[/tex]


    [tex] \frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s[/tex]

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

    Physically Incorrect
    Last edited: Nov 29, 2007
  5. Dec 1, 2007 #4
    Thanks for the reply.

    I thought [tex] \frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s[/tex]
    followed straight away from [tex]0 = V dp - SdT[/tex] 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?

    Any further advice appreciated.
  6. Apr 7, 2009 #5
    dG = 0 = V dp - SdT
    [/tex] gives
    \frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_G
    [/tex], not
    \frac{S}{V} = \left( \frac{\partial p}{\partial T}\right)_s
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