JEE Main 2015 · 4 Apr · Shift 1 · Physics Q11

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PhysicsSingle correct+4 / -1Question 11 of 90

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 :
$\left(\gamma=\frac{C_p}{C_v}\right)$

\dfrac{3\gamma+5}6

\dfrac{3\gamma-5}6

\dfrac{\gamma+1}2

\dfrac{\gamma-1}2

Answer and worked solution

Official answer: C

The mean free time is $\tau=\dfrac\lambda{\bar v}$ , with $\lambda\propto V$ (fixed number of molecules) and $\bar v\propto\sqrt T$ . So $\tau\propto VT^{-1/2}$ .

Adiabatic: $T\propto V^{1-\gamma}$ , so $\tau\propto V\cdot V^{(\gamma-1)/2}=V^{(\gamma+1)/2}$ , and $q=\dfrac{\gamma+1}2$ .

Topics:Mean free path Thermodynamic processes

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 11