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: ( \gamma = C_p \over C_v )

Heat and Thermodynamics PYQ — JEE Main 2015 · Shift 0 · Q63 with Solution

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Correct answer: A

For an ideal gas, the average time between two successive collisions for a molecule is $au \sim \frac{\lambda}{\bar v},$ where:

We need how $\tau$ varies with volume $V$ during adiabatic expansion.

For an ideal gas, $\lambda = \frac{1}{\sqrt{2}\,n\sigma},$ where $n=N/V$ is number density and $\sigma$ is collision cross-section.

Since $N$ is constant in the closed chamber, $n \propto \frac{1}{V} \quad \Rightarrow \quad \lambda \propto V.$

Average molecular speed scales as $\bar v \propto \sqrt{T}.$

So, $\tau \propto \frac{V}{\sqrt{T}}.$

For an adiabatic process of an ideal gas, $TV^{\gamma-1}=\text{constant}.$

Hence, $T \propto V^{-(\gamma-1)}.$

Therefore, $\sqrt{T} \propto V^{-\frac{\gamma-1}{2}}.$

Substitute into the expression for $\tau$ :

=V^{1+\frac{\gamma-1}{2}}.$$ Simplifying, $$1+\frac{\gamma-1}{2} = \frac{2+\gamma-1}{2}=\frac{\gamma+1}{2}.$$ Thus, $$\tau \propto V^{\frac{\gamma+1}{2}}.$$ So, $$q=\frac{\gamma+1}{2}.$$ --- 5. **Option check** - **A:** $\dfrac{\gamma+1}{2}$ ✅ - **B:** $\dfrac{\gamma-1}{2}$ ❌ - **C:** $\dfrac{3\gamma+5}{6}$ ❌ - **D:** $\dfrac{3\gamma-5}{6}$ ❌ Therefore, the correct option is **A**.