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 to τ ₂ . If Cₚ/Cᵥ = γ for this gas then a good estimate for (τ ₂)/(τ ₁) is given by :

Solution

Physics sir · 4 steps

Collision time is path over speed: the path grows with V, and the speed falls as the gas cools adiabatically.

1. Collision time. The mean collision time is the mean free path divided by the average speed, τ = λ / v̄. Find how each one depends on the volume V.
2. Mean free path. λ = 1/(√2 π d² n), where n = N/V is the number of molecules per unit volume. With N fixed, λ ∝ V.
3. Speed after expansion. v̄ ∝ √T. For an adiabatic process TV^(γ - 1) is constant, so T ∝ V^(1-γ) and v̄ ∝ V^((1-γ)/2).

   τ ∝ V/V^((1-γ)/2) = V^((1+γ)/2)

4. So. Doubling the volume gives τ₂/τ₁ = 2^((1+γ)/2), which is option (a).

The idea to keep. For collision time, track two things: the mean free path λ ∝ V/N and the speed v̄ ∝ √T. Then use the process equation to write T in terms of V.

Answer from the NTA final answer key. Solution by PrepWiser.