A gas is taken through the cycle A → B → C → A as shown in the P–V graph. A = (2 × 10⁻³ m³, 2 × 10⁵ Pa), B = (7 × 10⁻³ m³, 6 × 10⁵ Pa), and C = (7 × 10⁻³ m³, 2 × 10⁵ Pa). What is the net work done by the gas?
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Question

A gas is taken through the cycle A → B → C → A as shown in the P–V graph. A = (2 × 10⁻³ m³, 2 × 10⁵ Pa), B = (7 × 10⁻³ m³, 6 × 10⁵ Pa), and C = (7 × 10⁻³ m³, 2 × 10⁵ Pa). What is the net work done by the gas?

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G1 Physics · Accepted Answer

Step 1: For a full cycle, net work is the signed area enclosed by the P–V loop. Step 2: The path encloses a triangle with horizontal base ΔV = (7−2) × 10⁻³ = 5 × 10⁻³ m³. Step 3: Vertical height ΔP = (6−2) × 10⁵ = 4 × 10⁵ Pa. Step 4: Area = (1/2)ΔPΔV = (1/2)(4 × 10⁵)(5 × 10⁻³) = 1000 J. Step 5: The path A→B→C→A is clockwise; hence work done by gas is positive. Final answer: (B) 1000 J.