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Homework Help: Specific Heat at constant pressure for photon gas

  1. Mar 17, 2015 #1
    1. The problem statement, all variables and given/known data
    A blackbody photon gas is contained within an evacuated cavity ([tex]V = 0.01 m^3[/tex]).
    Calculate [tex]C_p[/tex] for the photon gas at T = 1000K

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

    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
    [tex] C_p - C_v = T(\frac{\partial S} {\partial V}) (\frac{\partial V}{\partial T})[/tex]
    [tex] C_v = T(\frac{\partial S} {\partial T}) [/tex]
    [tex] C_p = T(\frac{\partial S} {\partial T}) + T(\frac{\partial S} {\partial V}) (\frac{\partial V}{\partial T})[/tex]

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

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

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

    which lead to
    [tex]C_p = \frac{4}{3} \frac{\partial U}{\partial T} = \frac{4}{3} C_v[/tex]

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

    I am not sure why the two results do not agree. Any help will be appreciated. Thanks!
  2. jcsd
  3. Mar 17, 2015 #2
    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
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