= Solution
Continuity gives $D\log\rho/Dt=-\nabla\cdot\mathbf u$. Using $p=\rho RT$, the temperature equation gives $D\log T/Dt=-(R/c_v)\nabla\cdot\mathbf u$. Therefore
$$
\frac{D\log p}{Dt}=\frac{D\log\rho}{Dt}+\frac{D\log T}{Dt}
=-\gamma\nabla\cdot\mathbf u,\qquad\gamma=1+R/c_v.
$$
Consequently
$$
\boxed{\frac{DS}{Dt}=c_v\left(\frac{D\log p}{Dt}-\gamma\frac{D\log\rho}{Dt}\right)=0.}
$$
Each smooth parcel evolves isentropically under these adiabatic inviscid equations; the argument does not cover <entropy>-producing shocks.
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