= Solution
With a thermally insulating boundary, $\delta Q=0$. If the slow piston motion is frictionless and reversible, it is an <isentropic process>:
$$
\boxed{S=\mathrm{constant}.}
$$
The <first law of thermodynamics> gives $dU=-p\,dV$, so expansion lowers the <internal energy> and <temperature>. For constant <heat capacity ratio> $\gamma$, $pV^\gamma$ and $TV^{\gamma-1}$ are constant combinations of <state functions>. Insulation alone does not guarantee constant <entropy>: an irreversible <adiabatic process> can produce <entropy>. The conclusion uses the intended reversible, quasistatic piston idealization.
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