An ideal gas follows the closed p–V path A→B→C→D→A with (volume in litres, pressure in kPa) coordinates A(1.8, 180), B(1.8, 280), C(3.8, 280), D(3.8, 180). All legs are straight. Take work done by the gas as positive and use ΔU = Q − Wby. (a) Calculate work on each leg and net work. (b) Determine net heat input per cycle and explain why internal-energy change is zero. Use 1 L = 10⁻³ m³.

Question

Accepted Answer

WAB = 0; WBC = 560 J; WCD = 0; WDA = −360 J; net work = 200 J and net heat input = 200 J. Work done by the gas is the area under a process on a pressure–volume plot, with expansion positive. On the vertical A→B and C→D legs volume is fixed, so both work terms are zero regardless of pressure change. During B→C, volume increases by 2.00 L = 0.00200 m³ at 280 kPa. Hence WBC = 280000 × 0.00200 = 560 J. One kPa litre is exactly one joule. During D→A the same volume change is negative at 180 kPa, giving WDA = −360 J. Net work is 560 − 360 = 200 J, equal to the enclosed clockwise rectangle's area. The gas returns to its starting thermodynamic state, so its internal energy returns to its initial value and total ΔU is zero. Applying ΔU = Q − Wby over the whole cycle gives net heat input Q = 200 J. Individual legs can still have nonzero internal-energy changes.

Energy transfer around a gas cycle 9

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Gas-cycle states

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Gas-cycle states — supplied values
Volume / L	Pressure / kPa
1.8	180
1.8	280
3.8	280
3.8	180
1.8	180