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. Heat capacity: Δ Q = m s Δ T . m = 1 kg , s = 386.4 J kg⁻¹ K⁻¹ , Δ T = 50 - 20 = 30 K . Δ Q = 1 × 386.4 × 30 = 11592 J . Using first law ΔU = Q - W, W = ∫ P dV,