 Solution

Let $e=c_{v}T$ be the specific internal energy of a calorically perfect gas and define its specific gas constant $\mathcal{R}=R^{\ast }/\mu$, so $p=\mathcal{R}\rho T$. A reversible adiabatic change has no heat input and obeysUsing $c_p-c_v=\mathcal{R}$ and $\gamma =c_p/c_v$ givesIntegration yields $p/{\rho }^{\gamma }=K$, or $\boxed{p=K{\rho }^{\gamma }}$, with $K$ fixed on that isentrope. Equivalently $s-s_{\mathrm{ref}}=c_v\ln [(p/{\rho }^{\gamma })/K_{\mathrm{ref}}]$, so $K$ specifies entropy as well as composition. In a smooth nondissipative flow this means $D(p{\rho }^{-\gamma })/Dt=0$ along each parcel. A single common $K$ throughout the fluid additionally requires a homentropic flow. No heat transfer alone does not forbid entropy production: different sides of an adiabatic shock generally have different $K$ values. Constant composition and heat capacities are needed for the fixed exponent used here.