Question

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 $\bigg(\gamma =\frac{C_p}{C_v}\bigg)$

• A. $\frac{3\gamma +5 }{6}$
• B. $\frac{\gamma +1}{2}$
• C. $\frac{3 \gamma-5}{6}$
• D. $\frac{\gamma-1}{2}$

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relationship between the average time of collision between gas molecules and the volume during an adiabatic expansion of an ideal gas.

In an adiabatic process, the following relation holds between pressure $P$ and volume $V$:

P V^{\gamma} = \text{constant},

where \gamma is the adiabatic index, which is the ratio of specific heats \gamma = \frac{C_p}{C_v}.

The mean free path \lambda of the gas molecules is given by:

\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P},

where k_B is the Boltzmann constant, T is the temperature, and d is the diameter of a molecule.

Since the number density n is inversely proportional to volume n \propto \frac{1}{V}, and for adiabatic processes T \propto V^{\gamma - 1}, the pressure P = n k_B T then transforms as follows:

P \propto \frac{1}{V} \times V^{\gamma - 1} = V^{\gamma - 2}.

Thus, the mean free path becomes proportional to:

\lambda \propto \frac{T}{P} \propto \frac{V^{\gamma - 1}}{V^{\gamma - 2}} = V.

The average time of collision \tau is proportional to:

\tau \propto \frac{\lambda}{v_{\text{rms}}},

where v_{\text{rms}} \propto \sqrt{T} \propto V^{(\gamma - 1)/2}.

Therefore, the average time of collision scales as:

\tau \propto \frac{V}{V^{(\gamma - 1)/2}} = V^{1 - \frac{(\gamma - 1)}{2}} = V^{(\gamma + 1)/2}.

Thus, the value of q is:

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

Therefore, the correct answer is: \frac{\gamma + 1}{2}.