An insulated container contains a monoatomic gas of molar mass ' $\mathrm{m}$ '. The container is moving…
An insulated container contains a monoatomic gas of molar mass ' $\mathrm{m}$ '. The container is moving with velocity ' $\mathrm{V}$ '. If it is stopped suddenly, the change in temperature of a gas is $[R$ is gas constant]
$\frac{\mathrm{MV}^2}{\mathrm{R}}$
$\frac{\mathrm{MV}^2}{2 \mathrm{R}}$
$\frac{\mathrm{MV}^2}{3 \mathrm{R}}$
$\frac{3 \mathrm{MV}^2}{2 \mathrm{R}}$
Solution
Kinetic energy toss of the gas is
$\Delta \mathrm{E}=\frac{1}{2} \mathrm{MV}^2, \mathrm{n}$...(i)
(where $\mathrm{n}$ is the no. of moles of the gas) Heat gained by the gás due to temperature change $\Delta \mathrm{T}$ is
$\Delta Q=n C V \Delta T$
But, $\mathrm{C}_{\mathrm{V}}=\frac{3}{2} \mathrm{R}$ …(gas is monoatomic)
$\therefore \quad \Delta Q=\frac{3}{2} R \cdot n \Delta T$...(ii)
Equating (i) and (ii),
$\begin{aligned}
& \Delta \mathrm{E}=\Delta \mathrm{Q} \\
& \frac{1}{2} \mathrm{MV}^2 \cdot \mathrm{n}=\frac{3}{2} \mathrm{R} \cdot \mathrm{n} \Delta \mathrm{T} \\
& \Delta \mathrm{T}=\frac{\mathrm{MV}^2}{3 \mathrm{R}}
\end{aligned}$