More practice questions

4. Q4. Find the mean free path at STP (n = 2.7 × 10^²⁵ m^⁻³) for molecules of diameter 3 × 10^⁻¹⁰ m. How many diameters is that?

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   Here d^² = 9 × 10^⁻²⁰ m^², and √2nπd^² = √2 × 2.7 × 10^²⁵ × 3.1416 × 9 × 10^⁻²⁰ = 1.08 × 10^⁷ m^⁻¹, so l = 1/(1.08 × 10^⁷ m^⁻¹), about 9.3 × 10^⁻⁸ m. Dividing by d: about 310 diameters.

5. Q5. In air at STP (n = 2.7 × 10^²⁵ m^⁻³), compare the average spacing between molecules, about n^−1/3, with the mean free path 2.1 × 10^⁻⁷ m.

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   n^−1/3 = (2.7 × 10^²⁵)^−1/3, about 3.3 × 10^⁻⁹ m. The mean free path is about 2.1 × 10^⁻⁷/3.3 × 10^⁻⁹, roughly 60 times the spacing.

6. Q6. In the fixed-diameter hard-sphere model, a gas in a sealed rigid container is heated until the average speed of its molecules doubles. What happens to the mean free path and to the average time between collisions?

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   The container is sealed and rigid, so n does not change and l = 1/(√2 nπd^²) stays the same. The time between collisions, τ = l/⟨v⟩, halves.