Question

One mole of a mono-atomic ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity $v$ and the molecular mass of the gas is $M$. The rise in temperature of the gas when the vessel is suddenly stopped is given by:
$[\gamma =\frac{C_p}{C_v}]$. Assume that the walls of the container are adiabatic.

• A. $\frac{M{v}^2}{R}$
• B. $\frac{M{v}^2}{2R}$
• C. $\frac{2M{v}^2}{R}$
• D. $\frac{M{v}^2}{3R}$

Answer 1

The correct option is D $\frac{M{v}^2}{3R}$

Due to sudden stopping of the vessel, the internal energy of the gas increases due to increase in randomness. Since the walls are adiabatic, the kinetic energy is completely converted to internal energy.
$\Delta U=\frac{1}{2}m{v}^2-0$
$\Rightarrow n{C}_V\Delta T=\frac{1}{2}\left(nM\right)v^2.....\left(1\right)$
[Where $n=$ number of moles of the gas]

From $\left(1\right)$, we can say that
$\Delta T=\frac{M{v}^2}{2C_V}=\frac{M{v}^2}{2\left(\frac{R}{\gamma -1}\right)}$ $[\because C_V=\frac{R}{\gamma -1}]$
or $\Delta T=\frac{M{v}^2(\gamma -1)}{2R}$
For a monoatomic gas, $\gamma =\frac{5}{3}$
Thus. we get $\Delta T=\frac{M{v}^2}{3R}$