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. P₁ V₁^γ = P₂ V₂^γ . 2 × 4¹.33 = 1 × V₂¹.33 . V₂¹.33 = 2 × 4¹.33 . V₂ = (2 × 4¹.33)¹/1.33 = 2¹/1.33 × 4 . 2⁰.7519 ≈ 1.681 , V₂ ≈ 1.681 × 4 ≈ 6.724 L ≈ 6.7 L . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,