Practice question

**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.9 , R = 8.3 , T₁ = 820 , T₂ = 410 , γ = 1.67 . W = (0.9 × 8.3 × (820 - 410))/(1.67 - 1) = (7.47 × 410)/(0.67) ≈ 4570.15 J ≈ 4570 J . Using first law ΔU = Q - W, W = ∫ P