 Solution

With a thermally insulating boundary, $\delta Q=0$. If the slow piston motion is frictionless and reversible, it is an isentropic process:The first law of thermodynamics gives $dU=-pdV$, so expansion lowers the internal energy and temperature. For constant heat capacity ratio $\gamma$, $p{V}^{\gamma }$ and $T{V}^{\gamma -1}$ 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.