1. Given data

• Ideal gas in a thermally insulated vessel
• Molecular mass of gas $= M$
• Ratio of specific heats $\gamma = 1.4 = \dfrac{7}{5}$
• Vessel initially moves with speed $v$ and is suddenly brought to rest
• No heat is exchanged with surroundings: $Q=0$

We need the increase in temperature of the gas.

2. Physical idea

When the vessel is suddenly stopped, the macroscopic kinetic energy of the gas (due to common motion with the vessel) gets converted into internal energy of the gas.

Since the vessel is insulated,

$Q=0$

So the loss in bulk kinetic energy of the gas becomes gain in internal energy.

For 1 mole of gas, initial bulk kinetic energy is

$\frac{1}{2} M v^2$

because molar mass is $M$.

Thus,

$\Delta U = \frac{1}{2} M v^2$

3. Relating internal energy change to temperature change

For 1 mole of ideal gas,

$\Delta U = C_V \Delta T$

So,

$C_V \Delta T = \frac{1}{2} M v^2$

Hence,

$\Delta T = \frac{M v^2}{2C_V}$

4. Find $C_V$ using $\gamma$

We know

$\gamma = \frac{C_P}{C_V} = \frac{7}{5}$

and

$C_P - C_V = R$

Using

$C_V = \frac{R}{\gamma - 1}$

So,

$C_V = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5R}{2}$

5. Substitute into temperature rise formula

$\Delta T = \frac{M v^2}{2 \cdot \frac{5R}{2}}$

$\Delta T = \frac{M v^2}{5R}$

6. Match with options

The increase in temperature is

$\boxed{\frac{M v^2}{5R}}$

So the correct option is B.

7. Verification with stored answer

Stored correct answer: B

Our derived answer: B

They agree.