Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $\tau_{1}$ to $\tau_{2}$. If $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ for the gas then a good estimate for $\frac{\tau_{2}}{\tau_{1}}$ is given by

(C) $\left(\frac{1}{2}\right)^{\gamma}$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Heat & Thermodynamics

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from

τ_{1}

τ_{2}

. If

\frac{C _{p}}{C _{v}} = γ

for the gas then a good estimate for

\frac{τ _{2}}{τ _{1}}

is given by

(A)2
(B) $2^{\frac{γ + 1}{2}}$ ✓ Correct
(C) $(\frac{1}{2})^{γ}$
(D) $\frac{1}{2}$

Answer: (B)

Solution

Mean collision time

τ = λ / \overset{v}{ˉ}

, with mean free path

λ \propto V

(since

n = N / V

) and

\overset{v}{ˉ} \propto T

, so

τ \propto V / T

. For an adiabatic process

T V^{γ - 1} = const \Rightarrow T \propto V^{1 - γ}

, giving

τ \propto V / V^{(1 - γ) /2} = V^{(γ + 1) /2}

. Doubling the volume,

\frac{τ _{2}}{τ _{1}} = 2^{(γ + 1) /2}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option

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