Practice question
Question
What is the volume of 0.6 moles of an ideal gas at 3 atm and 527°C? (R = 8.31 J mol⁻¹ K⁻¹)
Explanation
**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. PV = μ R T, V = (μ R T)/(P).T = 527 + 273 = 800 K, P = 3 × 1.01 × 10⁵ = 3.03 × 10⁵ Pa.V = (0.6 × 8.31 × 800)/(3.03 × 10⁵) = 1.317 × 10⁻² m³ ≈ 13.17 litres. Substituting values gives 13.2 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
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.