Practice question

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. Isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.15 , T = 320 , V₂ = 9 , V₁ = 3 . W = 0.15 × 8.3 × 320 × ln((9)/(3)) = 398.4 × 1.0986 ≈ 438 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =