A thermally insulated vessel contains an ideal gas of molecular mass $$M$$ and ratio of specific heats $$\gamma .$$ It is moving with speed $$v$$ and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by :

Solution :
Here, work done is zero.
So, loss in kinetic energy = change in internal energy of gas
$$\eqalign{ & \frac{1}{2}m{v^2} = n{C_v}\,\Delta T \cr & = n\frac{R}{{\gamma - 1}}\,\Delta T \cr & \frac{1}{2}m{v^2} = \frac{m}{M}\frac{R}{{\gamma - 1}}\,\Delta T \cr & \therefore \,\,\Delta T = \frac{{M{v^2}\left( {\gamma - 1} \right)}}{{2R}}K \cr} $$