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Question

A diatomic gas undergoes an adiabatic expansion from 860 K to 430 K with 0.5 moles . What is the work done? ( R = 8.3 J mol⁻¹ K⁻¹ , gamma = 1.4 )

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Explanation

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.5 , R = 8.3 , T₁ = 860 , T₂ = 430 , γ = 1.4 . W = (0.5 × 8.3 × (860 - 430))/(1.4 - 1) = (4.15 × 430)/(0.4) = 4467.5 J ≈ 4468 J . Using first law ΔU = Q - W,

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