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. Δ U = μ C_v Δ T . μ = 0.6 , C_v = 20.8 , Δ T = 320 - 270 = 50 . Δ U = 0.6 × 20.8 × 50 = 624 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W