Derive the Bose-Einstein expression for the mean number of Bose particles $\bar{n}$ occupying a particular single-particle quantum state of energy $\varepsilon$, when the chemical potential is $\mu$ and the temperature is $T$ in energy units.

Why is the chemical potential for a gas of photons given by $\mu=0$ ?

Show that, for black-body radiation in a cavity of volume $V$ at temperature $T$, the mean number of photons in the angular frequency range $(\omega, \omega+d \omega)$ is

$$\frac{V}{\pi^{2} c^{3}} \frac{\omega^{2} d \omega}{e^{\hbar \omega / T}-1}$$

Hence, show that the total energy $E$ of the radiation in the cavity is

$$E=K V T^{4}$$

where $K$ is a constant that need not be evaluated.

Use thermodynamic reasoning to find the entropy $S$ and pressure $P$ of the radiation and verify that

$$E-T S+P V=0$$

Why is this last result to be expected for a gas of photons?