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 . 10 × 5¹.33 = 2 × V₂¹.33 . V₂¹.33 = (10)/(2) × 5¹.33 = 5 × 5¹.33 . 5¹.33 ≈ 9.62 , V₂¹.33 = 5 × 9.62 ≈ 48.1 . V₂ = (48.1)¹/1.33 ≈ 14.5 L . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =