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(A) When the container is stopped suddenly, the kinetic energy of the container is converted into the internal energy of the gas.Let $n$ be the number

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$\frac{MV^2}{3R}$

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(A) When the container is stopped suddenly, the kinetic energy of the container is converted into the internal energy of the gas.Let $n$ be the number of moles of the gas. The kinetic energy of the container is $K = \frac{1}{2} M_{total} V^2$, where $M_{total} = nM$.So, $K = \frac{1}{2} n M V^2$.This kinetic energy increases the internal energy of the gas: $\Delta U = n C_v \Delta T$.For a monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.Equating the kinetic energy to the change in internal energy: $\frac{1}{2} n M V^2 = n (\frac{3}{2} R) \Delta T$.Solving for $\Delta T$: $\Delta T = \frac{\frac{1}{2} n M V^2}{n (\frac{3}{2} R)} = \frac{MV^2}{3R}$.

An insulated container contains a monoatomic gas of molar mass $M$. The container is moving with velocity $V$. If it is stopped suddenly, the change in temperature of the gas is ($R$ = gas constant).

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