A thermodynamic system is taken through the rectangular cycle ABCDA shown in the P–V diagram, with A = (V, P), B = (3V, P), C = (3V,2P), and D = (V,2P). The net heat rejected by the gas during the cycle is:
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Question

A thermodynamic system is taken through the rectangular cycle ABCDA shown in the P–V diagram, with A = (V, P), B = (3V, P), C = (3V,2P), and D = (V,2P). The net heat rejected by the gas during the cycle is:

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G1 Physics · Accepted Answer

Step 1: The system returns to its initial state after one cycle; hence total ΔU = 0. Step 2: For a cycle the first law becomes Qₙₑₜ = Wₙₑₜ. Step 3: The rectangle has pressure difference ΔP = 2P − P = P and volume difference ΔV = 3V − V = 2V. Step 4: Enclosed area = ΔPΔV = 2PV. Step 5: The directed path A→B→C→D→A is COUNTERCLOCKWISE, so work done by gas is negative: Wₙₑₜ = −2PV. Step 6: Therefore Qₙₑₜ = −2PV, i.e. the gas rejects net heat of magnitude 2PV. Final answer: (A) 2PV.

A thermodynamic system is taken through the rectangular cycle ABCDA shown in the P–V diagram, with A = (V,P), B = (3V,P), C = (3V,2P), and D = (V,2P). The net heat rejected by the gas during the cycle is: