Solve this:
A box of negligible mass containing 2 moles of an ideal gas of molar mass M and adiabatic exponent $\gamma$ moves with constant speed $v$ on a smooth horizontal surface. If the box suddenly stops, then change in temperature of gas will be 
 $$\begin{gathered} \left(1\right) \frac{(\gamma -1) M{v}^2}{4R} \left(2\right) \frac{\gamma M{v}^2}{2R} \\ \left(3\right) \frac{M{v}^2}{2(\gamma -1)R} \left(4\right) \frac{(\gamma -1)M{v}^2}{2R} \end{gathered}$$