Certain amount of an ideal gas are contained in a closed vessel. The vessel is moving with a constant velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is suddenly stopped is  $(\mathrm{\gamma }={\mathrm{C}}_{\mathrm{P}}/{\mathrm{C}}_{\mathrm{V}})$

If m is the total mass of the gas then its kinetic energy $=\frac{1}{2}{\mathrm{mv}}^2$
When the vessel is suddenly stopped then total kinetic energy will increase the temperature of the gas. Hence 
$\begin{array}{l}\left.\frac{1}{2}{\mathrm{mv}}^2={\mathrm{\mu C}}_{\mathrm{v}}\mathrm{\Delta T}=\frac{\mathrm{m}}{\mathrm{M}}{\mathrm{C}}_{\mathrm{v}}\mathrm{\Delta T}\text{ [As }{\mathrm{C}}_{\mathrm{v}}=\frac{\mathrm{R}}{\mathrm{\gamma }-1}\right]\\ \Rightarrow \frac{\mathrm{m}}{\mathrm{M}}\frac{\mathrm{R}}{\mathrm{\gamma }-1}\mathrm{\Delta T}=\frac{1}{2}{\mathrm{mv}}^2\Rightarrow \mathrm{\Delta T}=\frac{{\mathrm{Mv}}^2(\mathrm{\gamma }-1)}{2\mathrm{R}}\end{array}$