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. First Law: Δ Q = Δ U + Δ W . Given Δ Q = 600 J , Δ W = 150 J (work by system). 600 = Δ U + 150 ⇒ Δ U = 600 - 150 = 450 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric