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Why $C_p$ is zero for photon gas?

Physics Asked on February 21, 2021

For photon gas, We know that
$$U=u(T)V, p =frac{u(T)}{3}$$
As $$G=U+pV-TS=uV+frac{u}{3}V-frac{4u}{3}V=0$$
$$dG=-SdT+Vdp=0$$
$$Rightarrow s=frac{dp}{dT}not=0$$
$$dp=0Rightarrow dT=0$$
which means that $C_p=infty$.

which is obvious as pressure is a function of energy density which is a function of temperature and thus the pressure is a function of temperature. So I can’t vary one of them keeping another fix.

But Why? I’m able to understand it on a thermodynamic level but not at the microscopic level.

So What’s the microscopic point of view explanation for $C_p$ to be infinite for ideal gas?

One Answer

Isobar heat capacity, volume expansivity, isothermal compressibility are not meaningfully defined quantities for photon gas, exactly because P and T is not independent. This causes the partial derivatives in their definition to be meaningless, and the heuristic infinite values that one may associate with them is not particularly useful in my opinion.

Nevertheless, if you really want to, you can interpret it the following way: There is no finite amount of energy, that could increase the temperature while the pressure is fixed.

Answered by Gomboc on February 21, 2021

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