In our week on radiation, we saw that the Helmholtz free energy of a
box of radiation at temperature \(T\) is \begin{align}
F &= -8\pi \frac{V(kT)^4}{h^3c^3}\frac{\pi^4}{45}
\end{align} From this we also found the internal energy and entropy
\begin{align}
U &= 24\pi \frac{(kT)^4}{h^3c^3}\frac{\pi^4}{45} V \\
S &= 32\pi kV\left(\frac{kT}{hc}\right)^3 \frac{\pi^4}{45}
\end{align} Given these results, let us consider a Carnot engine that
uses an empty metalic piston (i.e. a photon gas).
Given \(T_H\) and \(T_C\), as well as \(V_1\) and \(V_2\) (the two
volumes at \(T_H\)), determine \(V_3\) and \(V_4\) (the two volumes
at \(T_C\)).
What is the heat \(Q_H\) taken up and the work done by the gas
during the first isothermal expansion? Are they equal to each other,
as for the ideal gas?
Does the work done on the two isentropic stages cancel each other,
as for the ideal gas?
Calculate the total work done by the gas during one cycle. Compare
it with the heat taken up at \(T_H\) and show that the energy
conversion efficiency is the Carnot efficiency.