Practice question
Question
What is the total internal energy of 1.5 moles of a monatomic gas at 250 K? (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. Monatomic gas: U = (3)/(2) μ R T.U = (3)/(2) × 1.5 × 8.31 × 250 = 4674.375 J ≈ 4.67 kJ. Substituting values gives 4.67 kJ, 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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