= Solution
The first <energy> term is the rate of change of internal plus <kinetic energy> per unit volume; the second transports that <energy> with the moving gas; the <pressure> term is the mechanical <work> <flux> across the boundary. There is no <heat flux> or viscous <work> in the stated inviscid <adiabatic process>[adiabatic] model.
Use mass conservation to replace a conservative derivative of $\rho e$ by $\rho De/Dt$. Dotting the momentum equation with $\mathbf u$ gives $\rho D(|\mathbf u|^2/2)/Dt=-\mathbf u\cdot\nabla p$. Subtraction from total <energy> leaves
$$
\rho c_v\frac{DT}{Dt}+p\nabla\cdot\mathbf u=0.
$$
Now $p=\rho RT$ and $D\rho/Dt=-\rho\nabla\cdot\mathbf u$ imply
$$
\frac{DS}{Dt}=c_v\left(\frac1p\frac{Dp}{Dt}-\frac\gamma\rho\frac{D\rho}{Dt}\right)=\frac{c_v}T\frac{DT}{Dt}+R\nabla\cdot\mathbf u=0.
$$
Thus $\boxed{DS/Dt=0}$: smooth <adiabatic process>[adiabatic] inviscid trajectories are <isentropic flow>[isentropic]. <Shocks> need an <Lagrangian entropy inequality>[entropy inequality] rather than this smooth equality.
Adding $\nabla\cdot(k\nabla T)$ models <Fourier's law>[Fourier heat conduction], with <heat flux> $-k\nabla T$. The same subtraction gives $\rho c_vDT/Dt+p\nabla\cdot\mathbf u=\nabla\cdot(k\nabla T)$, hence
$$
\boxed{T\frac{DS}{Dt}=\frac1\rho\nabla\cdot(k\nabla T).}
$$
For a fixed container use impermeability $\mathbf u\cdot\mathbf n=0$, insulation $k\nabla T\cdot\mathbf n=0$, and mass conservation. Then
$$
\frac d{dt}\int_D\rho S\,dV=\int_D\frac1T\nabla\cdot(k\nabla T)\,dV=\int_D\frac{k|\nabla T|^2}{T^2}\,dV.
$$
The boundary term vanishes and $\nabla(1/T)=-\nabla T/T^2$ supplies the positive sign. For $T>0$, $k\geq0$, total <entropy> cannot decrease: this is <entropy production by Fourier heat conduction>.
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