 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 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 leavesNow $p=\rho RT$ and $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$ implyThus $\boxed{DS/Dt=0}$: smooth adiabatic inviscid trajectories are isentropic. Shocks need an entropy inequality rather than this smooth equality.

Adding $\nabla \cdot (k\nabla T)$ models Fourier heat conduction, with heat flux $-k\nabla T$. The same subtraction gives $\rho c_{v}DT/Dt+p\nabla \cdot \mathbf{u}=\nabla \cdot (k\nabla T)$, henceFor a fixed container use impermeability $\mathbf{u}\cdot \mathbf{n}=0$, insulation $k\nabla T\cdot \mathbf{n}=0$, and mass conservation. ThenThe boundary term vanishes and $\nabla (1/T)=-\nabla T/T^2$ supplies the positive sign. For $T>0$, $k\ge 0$, total entropy cannot decrease: this is entropy production by Fourier heat conduction.