Practice question

**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. First Law: Δ Q = Δ U + Δ W . Δ Q = -700 J (heat released), Δ W = 250 J (work by system). -700 = Δ U + 250 ⇒ Δ U = -700 - 250 = -950 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η =