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Question

What is the volume of 0.3 moles of an ideal gas at 2.5 atm and 127°C? (R = 8.31 J mol⁻¹ K⁻¹)

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Explanation

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. PV = μ R T, V = (μ R T)/(P).T = 127 + 273 = 400 K, P = 2.5 × 1.01 × 10⁵ = 2.525 × 10⁵ Pa.V = (0.3 × 8.31 × 400)/(2.525 × 10⁵) = 3.95 × 10⁻³ m³ ≈ 3.95 litres. Substituting values gives 3.95 litres, which matches expected kinetic theory

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