Question

Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity $v$. The molecular mass of gas is $M$. The rise in temperature of the gas when the vessel is suddenly stopped is $\left(\gamma=C_{P} / C_{V}\right)$

• A. $\frac{M v^{2}(\gamma-1)}{2 R(\gamma+1)}$
• B. $\frac{M v^{2}(\gamma-1)}{2 R}$
• C. $\frac{M v^{2}}{2 R(\gamma+1)}$
• D. $\frac{M v^{2}}{2 R(\gamma-1)}$

Answer 1

Answer: (B)

If $m$ is the total mass of the gas, then its kinetic energy $=1 / 2 m v^{2}$. When the vessel is suddenly stopped, total kinetic energy will increase the temperature of the gas (because process will be adiabatic), i.e.,
$ \frac{1}{2} m v^{2}=\mu C_{v} \Delta T=\frac{m}{M} C_{v} \Delta T$
$\Rightarrow \frac{m}{M} \frac{R}{\gamma-1} \Delta T=\frac{1}{2} m v^{2} \left(\text { As } C_{v}=\frac{R}{\gamma-1}\right)$
$\Rightarrow \Delta T=\frac{M v^{2}(\gamma-1)}{2 R}$