Q. Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity $v$. The rise in temperature of the gas when the vessel is suddenly stopped is ($M$ is molecular mass, $\gamma = C_{p}/C_{v}$)
A) $\frac{M\ v^{2}(\gamma - 1)}{2R}$
B) $\frac{M\ v^{2}(\gamma + 1)}{2R}$
C) $\frac{M\ v^{2}}{2R_{\gamma}}$
D) $\frac{M\ v^{2}}{2R(\gamma + 1)}$
Explanation
(a) KE of the vessel $= \frac{1}{2}Mv^{2}$
When the vessel is suddenly stopped, the ordered motion of the gas molecules is converted into disordered motion of the molecules increasing thereby the internal energy of the gas. Thus,
$\mathrm{\Delta}U = nC_{v}\mathrm{\Delta}T = \frac{1}{2}mv^{2} = \frac{1}{2}(nM)v^{2}$
Where $n$ is number of moles of the gas in the vessel and $M$ is molecular weight of the gas.
$\therefore\ \ \mathrm{\Delta}T = \frac{Mv^{2}}{2C_{v}}$
As $C_{v} = \frac{R}{\gamma - 1}$
$\therefore\ \ \mathrm{\Delta}T = \frac{Mv^{2}(\gamma - 1)}{2R}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.