Answer
No.
Work Step by Step
From the entropy relation for an ideal gas:
$\Delta s=c_p\ln{\dfrac{T_2}{T_1}}-R\ln{\dfrac{P_2}{P_1}}$
Since the initial and final pressures are the same, and the process is steady and adiabatic:
$s_{gen}=c_p\ln{\dfrac{T_2}{T_1}}\geq0$
Therefore $T_2\geq T_1$ and cooling is not possible.
