Return a gas to its starting state and obtain net work

Include the return journey

A single expansion can deliver work, but it leaves the gas in a different state. To repeat a cycle, the gas must return to its initial state. Predict whether the return compression cancels all the expansion work before inspecting the default rectangle.

The retained gas is our system. Pressure P is absolute pressure in kilopascals; volume V is in litres. Positive work W is work by the gas, and positive heat Q enters it. A litre-kilopascal equals a joule. We prescribe quasistatic constant-pressure or constant-volume legs, with the external load matched to the gas pressure in the limiting process.

Follow four states

At the default upper pressure 300 kPa and larger volume 30 L, A is (10 L, 100 kPa), B is (10 L, 300 kPa), C is (30 L, 300 kPa), and D is (30 L, 100 kPa). Follow A → B → C → D → A; the diagram is a path in pressure-volume space, not a moving piston animation or time graph.

A → B has fixed volume, so W = 0. B → C expands by 20 L at 300 kPa, giving W = 6000 J. C → D again has fixed volume and zero work. D → A compresses by 20 L at 100 kPa, giving W = −2000 J. The surroundings supply 2000 J during this return leg.

The complete cycle therefore delivers 4000 J net work. The enclosed rectangle has height 300 − 100 = 200 kPa and width 30 − 10 = 20 L. Their product is also 4000 J. It is the area inside the loop, not the entire area below the upper expansion line.

Check the state energy

For the specified one-mole monatomic ideal gas, modeled U = 3PV/2. This gives state energies A: 1500 J, B: 4500 J, C: 13500 J, D: 4500 J. Returning to A restores 1500 J. The four changes are +3000, +9000, −9000, −3000 J, summing to zero.

For each leg, rearrange the first law as Q = ΔU + W. You obtain +3000, +15000, −9000, −5000 J. Their signed sum is +4000 J, exactly the net work. The gas does not supply this work by gradually using up its stored internal energy: that energy is restored every cycle. Net energy arrives as heat.

Check the independent temperature column. For one mole, T = PV/R with R = 8.314 L kPa/(mol K). States B and D have equal PV and equal temperatures despite different pressures and volumes. Equal temperature is not the same complete gas state.

Change the shape, then collapse it

Raise the upper pressure to 400 kPa while keeping 30 L. Before reading the ledger, use the rectangle to predict net work: (400 − 100) × (30 − 10) = 6000 J. Explain which work leg changes and why the return compression stays at −2000 J.

Set upper pressure to 100 kPa. Expansion and compression now retrace the same line: their work cancels. Heat still enters on expansion and leaves on compression. Set larger volume to 10 L with upper pressure above 100 kPa: heating and cooling remain, but every leg has zero volume change and zero work. If both bounds equal their starting values, every state and transfer is unchanged.

If every path leg could be traversed in reverse under suitable controls, the signed work would reverse too. That reversed path needs net work input. A reversed drawing alone does not specify a practical refrigerator, its reservoirs or cooling rate.

Transfer to an unfamiliar rectangle

A quasistatic rectangular cycle expands at 250 kPa from 8 L to 20 L and compresses back at 150 kPa. The connecting legs have fixed volume. Find net work and net heat. Then give the signs if the entire path is reversed.

Check the gas-cycle transfer

Expansion gives 250 × 12 = 3000 J; compression gives 150 × (8−20) = −1800 J. Net work is 1200 J, also (250−150)(20−8). Since the cycle restores the gas state, ΔU = 0 and net heat is +1200 J. The reversed path has −1200 J net work by the gas and −1200 J net heat into it.

The OpenStax thermodynamic-process reference provides background on cyclic and quasistatic processes. The model’s controlled thermal contacts and fixed heat capacities are assumptions, not evidence that a real gas or apparatus follows every displayed state.