A thermally insulated vessel contains an ideal gas of molecular mass $$M$$ and ratio of specific heats $$1.4$$. Vessel is moving with speed $$v$$ and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by :
($$R$$ = universal gas constant)

A thermally insulated vessel contains an ideal gas with molecular mass $$M$$ and $$\gamma = 1.4$$. The vessel moves with speed $$v$$ and is suddenly brought to rest. Since the vessel is insulated, the kinetic energy of the gas converts entirely into internal energy, causing a temperature rise.

For one mole of gas, the kinetic energy of the vessel is given by $$KE = \frac{1}{2}Mv^2$$. The increase in internal energy is $$\Delta U = nC_v \Delta T$$, and for one mole ($$n = 1$$) with $$C_v = \frac{R}{\gamma - 1}$$, this becomes $$\Delta U = \frac{R}{\gamma - 1} \Delta T$$.

Equating kinetic energy to the change in internal energy yields $$\frac{1}{2}Mv^2 = \frac{R}{\gamma - 1} \Delta T$$, which gives $$\Delta T = \frac{Mv^2(\gamma - 1)}{2R}$$. Substituting $$\gamma = 1.4$$ leads to $$\Delta T = \frac{Mv^2(1.4 - 1)}{2R} = \frac{Mv^2 \times 0.4}{2R} = \frac{0.4 \cdot Mv^2}{2R} = \frac{Mv^2}{5R}$$.

Therefore, the correct answer is Option B.