Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an 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 $$\left(\gamma = \frac{C_p}{C_v}\right)$$

Let the average time between two successive molecular collisions be denoted by $$\tau$$. By kinetic theory, the collision frequency (collisions per unit time) is proportional to the number density of molecules multiplied by their average speed. Stating this idea in symbols:

$$\text{collision frequency}\;z \;\propto\; n\,\bar v$$

where

$$n=\frac{N}{V}$$

is the number density (here $$N$$ is the fixed number of molecules in the isolated chamber) and $$\bar v$$ is the mean molecular speed.

The average time between collisions is the reciprocal of the collision frequency, so

$$\tau \;=\;\frac{1}{z}\;\propto\;\frac{1}{n\,\bar v}.$$

Substituting $$n = \dfrac{N}{V}$$ we have

$$\tau \;\propto\;\frac{1}{(N/V)\,\bar v}\;=\;\frac{V}{N\,\bar v}.$$

The system is closed, hence $$N$$ is constant, and we may absorb it into the proportionality constant. Thus

$$\tau \;\propto\; \frac{V}{\bar v}.$$

For an ideal gas the average molecular speed varies with absolute temperature as

$$\bar v \;\propto\;\sqrt{T}.$$

Using this in the previous relation gives

$$\tau \;\propto\;\frac{V}{\sqrt{T}} \;=\;V\,T^{-1/2}.$$

Now an adiabatic process for an ideal gas satisfies the well-known relation

$$T\,V^{\gamma-1} = \text{constant},$$

where $$\gamma=\dfrac{C_p}{C_v}$$. From this equation we can express temperature in terms of volume:

$$T \;\propto\; V^{1-\gamma}.$$

Taking the square root on both sides, we obtain

$$\sqrt{T} \;\propto\; V^{\tfrac{1-\gamma}{2}}.$$

Substituting this result back into $$\tau \propto V\,T^{-1/2}$$ yields

$$\tau \;\propto\; V\,\bigl(V^{\tfrac{1-\gamma}{2}}\bigr)^{-1} \;=\;V\,V^{\tfrac{\gamma-1}{2}} \;=\;V^{\,1+\tfrac{\gamma-1}{2}}.$$

Adding the exponents explicitly,

$$1+\frac{\gamma-1}{2} \;=\; \frac{2+\gamma-1}{2} \;=\;\frac{\gamma+1}{2}.$$

Hence

$$\tau \;\propto\; V^{\tfrac{\gamma+1}{2}}.$$

Comparing with the statement of the problem, the exponent $$q$$ is

$$q=\frac{\gamma+1}{2}.$$

Hence, the correct answer is Option D.