Question

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes adiabatic expansion, the average time of collision between molecules increases as $ {V^q} $ , where $ V $ is the volume of the gas. The value of $ q $ is: $ \left( {\gamma = \dfrac{{{C_P}}}{{{C_V}}}} \right) $
(A) $ \dfrac{{3\gamma + 5}}{6} $
(B) $ \dfrac{{3\gamma - 5}}{6} $
(C) $ \dfrac{{\gamma + 1}}{2} $
(D) $ \dfrac{{\gamma - 1}}{2} $

Answer 1

Hint : We need to find the relation between pressure, volume and temperature of an adiabatic process. Then we need to draw a relation between the average time between collisions and the volume of gas from there to find the answer.

Complete step by step answer
For an adiabatic process, $ P = {V^{ - \gamma }} $ where $ P $ is the pressure, $ V $ is the volume and $ \gamma $ is the specific heat ratio of the gas.
It can be written that, $ {\text{T}}{{\text{V}}^{\gamma {\text{ - 1}}}}{\text{ = constant}} $ .
Also, $ \tau = \dfrac{1}{{n\pi \sqrt 2 {V_{rms}}{d^2}}} $ where the number of molecules per unit volume is $ n $ , $ {V_{rms}} $ is the root mean square velocity of molecules and $ d $ is the diameter of the collision tube.
The average time between collisions is also given by, the ratio of mean free path and the root mean square velocity.
The root mean square velocity is the square root of the average of the square of the velocity. As such, it has units of velocity. The reason we use the rms velocity instead of the average is that for a typical gas sample the net velocity is zero since the particles are moving in all directions.
As $ n\alpha \dfrac{1}{V} $ and $ {V_{rms}}\alpha \sqrt T $ ,
thus, $ \tau \alpha \dfrac{V}{{\sqrt T }} $ .
It can be written that, $ n = {k_1}{V^{ - 1}} $ and $ {V_{rms}} = {K_2}{T^{{1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}} $ where $ {k_1} $ and $ {K_2} $ are constants.
For an adiabatic process, $ {\text{T}}{{\text{V}}^{\gamma {\text{ - 1}}}}{\text{ = y}} $ where $ y $ is a constant.
 $ \tau \alpha V{T^{ - {1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}}\alpha V{\left( {{V^{1 - \gamma }}} \right)^{ - {1 \mathord{\left/
 {\vphantom {1 2}} \right.} 2}}} $
Thus, $ \tau \alpha {V^{\dfrac{{\gamma + 1}}{2}}} $ .
Thus, $ q = \dfrac{{\gamma + 1}}{2} $ .
Hence, the correct answer is Option C.

Note
The mean free path is the distance that a molecule travels between collisions. The mean free path is determined by the criterion that there is one molecule within the "collision tube" that is swept out by a molecular trajectory. The mean free path equation depends upon the temperature and pressure as well as the molecular diameter. The constant $ \tau $ is the average time between collisions. Suppose, for example, that in an hour there are 60 collisions; then $ \tau $ is one minute. We would then say that $ \tau $ (one minute) is the average time between the collisions.
 $ t = \dfrac{1}{{\dfrac{{\pi {d^2}{N \mathord{\left/
 {\vphantom {N V}} \right.} V}}}{{\sqrt {\dfrac{{3RT}}{M}} }}}} $ ; $ t = \dfrac{{CV}}{{\sqrt T }} $ where $ C $ is a constant.