c
If $m$ is the total mass of the gas then its kinetic energy$ = \frac{1}{2}m{v^2}$
When the vessel is suddenly stopped then total kinetic energy will increase the temperature of the gas. Hence
$\frac{1}{2}m{v^2} = \mu \,{C_v}\Delta T$$ = \frac{m}{M}{C_v}\Delta T$ [As ${C_v} = \frac{R}{{\gamma - 1}}$]
==>$\frac{m}{M}\frac{R}{{\gamma - 1}}\Delta T = \frac{1}{2}m{v^2}$ ==> $\Delta T = \frac{{M{v^2}(\gamma - 1)}}{{2R}}$.