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Question

In an isobaric process, 1.6 moles of an ideal gas expand from 6 L to 12 L at 400 K . What is the work done by the gas? ( R = 8.3 J mol⁻¹ K⁻¹ )

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Explanation

**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. W = P Δ V , P V = μ R T . Δ V = 12 - 6 = 6 L . P = (μ R T)/(V₁) = (1.6 × 8.3 × 400)/(6) = 885.33 atm (unit correction needed).Directly: W = μ R T ((V₂ - V₁)/(V₁)) , but W = P Δ V . W = 1.6

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