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Question

In an adiabatic process, 1 mole of gas at 600 K does 2000 J of work. What is the final temperature? ( gamma = 1.4 , 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 = (μ R (T₁ - T₂))/(γ - 1) . 2000 = (1 × 8.3 × (600 - T₂))/(1.4 - 1) . 2000 = (8.3 × (600 - T₂))/(0.4) . 2000 × 0.4 = 8.3 × (600 - T₂) ⇒ 800 = 8.3 × (600 - T₂) . 600 - T₂ = (800)/(8.3) ≈ 96.39 . T₂ = 600

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