- #1
sy7kenny
- 5
- 0
Homework Statement
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
Homework 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]
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!