 Solution

Use the printed stellar gas-pressure fraction, $\beta =P_g/P$, with $P=P_g+P_{\mathrm{rad}}$, and keep composition fixed. For a monatomic perfect gas plus equilibrium blackbody radiation, the specific heats of a monatomic gas-radiation mixture follow from its specific internal energy and specific enthalpy:At fixed total pressure, differentiating the equation of state givesIn particular, $\beta$ is not held constant during that derivative. The specific heat capacity at constant pressure is $c_P=(\partial h/\partial T{)}_P$, soSince $a{T}^3/(3\rho )=\mathcal{R}(1-\beta )/\beta$, this simplifies toThe pure-gas limit is $5\mathcal{R}/2$. For a gas with unspecified molecular degrees of freedom, replace $3\mathcal{R}/2$ in $u$ by its gas heat capacity $c_{V,g}$: then $c_P=c_{V,g}+\mathcal{R}+4\mathcal{R}(1-\beta )(4+\beta )/{\beta }^2$. The boxed formula uses the conventional monatomic stellar-gas interpretation.