Question

Certain amount of an ideal gas are 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 $(\gamma = C_{P}/C_{V})$

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

Answer 1

Answer: (B)

$\frac{Mv^{2}(\gamma - 1)}{2R}$

Answer 2

If m is the total mass of the gas then its kinetic energy$= \frac{1}{2}mv^{2}$

When the vessel is suddenly stopped then total kinetic energy will increase the temperature of the gas (because process will be adiabatic) i.e. $\frac{1}{2}mv^{2} = \mu C_{v}\Delta T = \frac{m}{M}C_{v}\Delta T$

[As $C_{v} = \frac{R}{\gamma - 1}$]

⇒ $\frac{m}{M}\frac{R}{\gamma - 1}\Delta T = \frac{1}{2}mv^{2}$ ⇒ $\Delta T = \frac{Mv^{2}(\gamma - 1)}{2R}$.