Question

Certain amount of an ideal gas of molecular mass M is contained in a closed vessel. If the vessel is moving with a constant velocity v, then the rise in temperature of the gas when the vessel is suddenly stopped will be
(Take $\gamma =\frac{C_P}{C_V}$)

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

Answer 1

The correct option is C $\frac{M{v}^2(\gamma -1)}{2R}$

If m is the total mass of the gas then its kinetic energy $=\frac{1}{2}m{v}^2$ when the vessel is suddenly stopped, then the total kinetic energy will increase the temperature of the gas. Hence $\frac{1}{2}m{v}^2=\mu C_v△T$
or $\frac{1}{2}m{v}^2=\frac{m}{M}{C}_V△T$
But $C_V=\frac{R}{\lambda -1}$
$\therefore \frac{1}{2}m{v}^2=\frac{m}{M}\times \frac{R}{\lambda -1}\times △T$
$△T=\frac{M{v}^2}{2R}(\lambda -1)$