Question

An insulated container containing $n$ moles of monoatomic gas of molar mass $m$ is moving with a velocity $v_0$. If the container is suddenly stopped, then the change in temperature is

• A. $\frac{m{v}_0^2}{3R}$
• B. $\frac{m{v}_0^2}{3nR}$
• C. $\frac{mn{v}_0^2}{R}$
• D. $\frac{m{v}_0^2}{2R}$

Answer 1

The correct option is A $\frac{m{v}_0^2}{3R}$

There is $n$ moles of the monoatomic gas in the container.

Molar mass of the gas = $m$,
Total mass of the gas in the container, $M=mn$
Change in KE of the gas when the container is suddenly stopped, i.e.,

$\Delta K=(KE{)}_{initial}-(KE{)}_{final}=\frac{1}{2}M{v}_0^2-0=\frac{1}{2}M{v}_0^2=\frac{1}{2}mn{v}_0^2$

This change in kinetic energy $\Delta K$ result in a change in internal energy $\Delta K$ of the gas

$\Delta U=n{C}_v\Delta T=n\left(\frac{3}{2}R\right)\Delta T$
Here, $\Delta T$ is the change in temperature of the gas
As $\Delta U=\Delta K,n\left(\frac{3}{2}R\right)\Delta T=\frac{1}{2}mn{v}_0^2\therefore \Delta T=\frac{m{v}_0^2}{3R}$