Practice question
Question
The work done in the reversible adiabatic expansion of 1 mol of an ideal gas from 10 L to 20 L is ( γ = 1.4 ) at 300 K. Calculate w . ( R = 8.314 J/mol·K )
Explanation
For adiabatic reversible expansion, T₂ = T₁ (V₁/V₂)γ⁻¹ = 300 × (10/20)⁰.⁴ ≈ 300 × 0.7579 = 227.4 K . Then, Cv = R/(γ-1) = 8.314 / 0.4 = 20.785 J/mol·K . Work done: w = nCvΔ T = 1 × 20.785 × (227.4 - 300) ≈ -1509.2 J .
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