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Question

A gas at 3 atm and 300 K has a volume of 10 litres. If the temperature rises to 900 K at constant pressure, what is the new volume?

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Explanation

**Collision frequency** Z = n σ v_rel, σ = π d² cross-section, v_rel = √2 v_avg, so Z ∝ n, for air at STP n≈2.5×10²⁵ m⁻³ d≈3×10⁻¹⁰ m λ≈68 nm, collision frequency ~10⁹ s⁻¹, illustrating frequent collisions at atmospheric pressure. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 10 litres, T₁ = 300 K, T₂ = 900 K.V₂ = V₁ × (T₂)/(T₁) = 10 × (900)/(300) = 30 litres. Substituting values gives 30 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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