What happens in a reversible adiabatic expansion process?

Answer: Option B

Since for an adiabatic process $$T{V}^{\gamma -1}=\mathrm{CONSTANT}$$     and on integration of this equation we get $$ln\left(\frac{T_2}{T_1}\right)=-\left(\left(\gamma -1\right)ln\frac{V_2}{V_1}\right)$$       since for expansion process $$V_2>V_1$$   Hence $$\Rightarrow T_2<T_1$$
So, cooling takes place.