An ideal gas follows the clockwise rectangular cycle A(1.0 L, 100 kPa) → B(1.0 L, 300 kPa) → C(3.0 L, 300 kPa) → D(3.0 L, 100 kPa) → A. During B→C it absorbs 900 J, during D→A it rejects 300 J, during A→B it absorbs 100 J and during C→D it rejects 300 J. Use 1 kPa·L = 1 J. Determine the work on every leg, net work, changes in internal energy on B→C and D→A, total heat over the cycle, and efficiency. Explain the clockwise sign.

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Leg works: 0, +600 J, 0, −200 J; net work = +400 J. ΔU_BC = +300 J; ΔU_DA = −100 J. Net heat = +400 J and efficiency = 40.0% for total input 1000 J. Isochoric legs A→B and C→D have ΔV = 0, so W_by = 0. At constant pressure, B→C gives W = 300(3−1) = +600 J. For D→A, W = 100(1−3) = −200 J. Net work is therefore +400 J, also the rectangular area (300−100)(3−1) in kPa·L. Using ΔU = Q − W_by, B→C has ΔU = 900−600 = +300 J. D→A has Q = −300 J and W = −200 J, giving ΔU = −100 J. A complete cycle returns to the initial thermodynamic state, so total ΔU = 0 and Q_net = W_net = +400 J. This also constrains the combined heat on the two vertical legs. The stated remaining absorption makes total heat input 900 + 100 = 1000 J. Efficiency is W_net/Q_in = 400/1000 = 40.0%. Clockwise traversal means high-pressure expansion and low-pressure compression, hence positive work by the gas.

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