Two moles of a monatomic ideal gas execute cycle A→B→C→A on a P–T graph: A=(300 K, 2000 Pa), B=(600 K, 2000 Pa), C=(600 K, 1000 Pa), with C→A following PT=constant. If efficiency is (12ln2 − x)/(12ln2 + 15), find x.
Physics · Mixed Chapter
JEE Mains

Question

Two moles of a monatomic ideal gas execute cycle A→B→C→A on a P–T graph: A=(300 K, 2000 Pa), B=(600 K, 2000 Pa), C=(600 K, 1000 Pa), with C→A following PT=constant. If efficiency is (12ln2 − x)/(12ln2 + 15), find x.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Read the P–T diagram: A(300 K, 2000 Pa), B(600 K, 2000 Pa), C(600 K, 1000 Pa). Number of moles = 2. Step 2 — A → B is isobaric heating. Work W₁ = nR(T₂ − T₁) = 2R(600 − 300) = 600R. Heat absorbed Q₁ = nCₚΔT = 2 × (5R/2) × 300 = 1500R. Step 3 — B → C is isothermal expansion at 600 K. Since V ∝ 1/P at fixed T, V(C)/V(B) = P(B)/P(C) = 2000/1000 = 2. W₂ = Q₂ = nRT ln2 = 2 × R × 600 × ln2 = 1200R ln2. Step 4 — C → A satisfies PT = constant, so P ∝ 1/T. Combining with PV = nRT gives V ∝ T².…