 Solution

For zero chemical potential, each species contributes $S_i=(U_i+P_{i}V)/T=4U_i/(3T)$. ThusEntropy density scales as $T^3$ at fixed relativistic particle content; total entropy also includes the changing proper volume. An isolated adiabatically expanding comoving gas has conserved total entropy and $aT$ constant, rather than its total entropy falling as $T^3$.

At equal temperature, single electrons and photons each have two spin/polarization states. In the ultrarelativistic approximation,Positrons are a separate species with the same contribution, so $(S_{e^{-}}+S_{e^{+}})/S_{\gamma }=7/4$. This distinction is needed for the heating count in part d. At the exact supplied temperature the electron rest mass gives corrections to the ideal $7/8$ ratio; it is the limiting relativistic value requested by the problem.

There is also a sign convention issue in the introductory entropy formula. With the chemical-potential convention used in the supplied distribution, the Euler relation is $U=TS-PV+{\sum }_i{\mu }_i{N}_i$, so $S=(U+PV-{\sum }_i{\mu }_i{N}_i)/T$. Its chemical-potential term has the opposite sign to the printed one. For a concrete counterexample, a dilute monatomic ideal gas with one internal state and $n{\lambda }_T^3=e^{-1}$, where ${\lambda }_T$ is the thermal de Broglie wavelength, has $\mu =-k_{\mathrm{B}}T$, $U=3N{k}_{\mathrm{B}}T/2$ and $PV=N{k}_{\mathrm{B}}T$. Its entropy is $7N{k}_{\mathrm{B}}/2$; the printed plus sign would instead give $3N{k}_{\mathrm{B}}/2$. All chemical potentials have been neglected here, so this source error does not change any of the above results.