Calculate gas work along different paths

Same destination, different work

One mole of classical monatomic ideal gas begins at 200 kPa absolute and 20 L. Select a final state of 300 kPa and 30 L. Path A first changes pressure at fixed volume, then expands at the final pressure. Path B first expands at the initial pressure, then changes pressure at fixed volume. Predict which path transfers more energy as work before inspecting the ledger.

These are prescribed quasistatic paths: successive states remain close to equilibrium and the external load is matched to the gas pressure in the limiting process. Appropriate heat exchange and load control are assumed. They are not predictions of a freely moving piston, rapid compression or how long an apparatus takes to respond.

Derive the rectangular area

For a piston of area A moving outward by distance Δx at constant pressure P, force is PA and the volume change is ΔV = AΔx. Work by the gas is force times displacement: W = PAΔx = PΔV. A is piston area here, not the path label A.

We take W as positive when the gas transfers energy outward by doing work. Expansion has positive ΔV and positive W. Compression has negative ΔV and negative W: the surroundings do positive work on the gas. The pressure itself remains positive in both cases.

With P in kPa and V in L, the product is in joules: 1 kPa L = 1 J. At fixed volume, ΔV is zero, so pressure-volume work is zero even when pressure and temperature change. This does not mean no heat is exchanged or no energy changes.

The OpenStax work and internal-energy reference develops this boundary-work relation. For a continuously varying quasistatic pressure, one sums many thin pressure-volume strips, written W = ∫P dV. No calculus is needed for the horizontal and vertical legs here.

Calculate both default paths

For path A, the fixed-volume leg does no work. The horizontal leg expands from 20 to 30 L at 300 kPa, so WA = 300 × 10 = 3000 J.

For path B, expansion takes place at 200 kPa: WB = 200 × 10 = 2000 J. Its later pressure change at fixed 30 L adds zero pressure-volume work. Both processes end at 300 kPa, 30 L, but their total work differs by 1000 J.

The shaded rectangular areas use identical scales in the two diagrams. The text below each diagram states travel order so the direction remains clear even when points overlap. Shading shows magnitude; the ledger supplies the sign. The area between the horizontal legs is (300 − 200) × (30 − 20) = 1000 J, exactly the work difference.

Test compression and degeneracies

Choose final pressure 100 kPa and volume 10 L. Path A compresses at 100 kPa and does −1000 J of work. Path B compresses at 200 kPa and does −2000 J. More energy is transferred into the gas as work along B. Calling the shaded positive geometric area the signed work would miss this direction.

Set final volume to 20 L. Both volume-changing legs shrink to zero length and both works become zero. Set final pressure to 200 kPa: both paths share the same horizontal segment and their work agrees. Set both controls to the initial state: both paths have no change, and all transfers in this particular construction vanish. Identical endpoints can support other nontrivial cycles, but no such cycle is drawn here.

The general difference is WA − WB = (Pf − Pi)(Vf − Vi). Either factor can reverse its sign. Compare the physical paths before declaring one always does more work.

Keep state and path separate

The ideal-gas equation determines temperature at every selected point: T = PV/(nR), using R = 8.314 L kPa/(mol K). It does not determine heat or work from endpoints alone. The path specification supplies the missing pressure-volume history. The next lesson adds an internal-energy model and the first law to determine heat.

Do not apply an equilibrium gas-pressure area indiscriminately to a strongly irreversible expansion. Boundary work then needs the actual external resistance and a suitable process model. The present diagrams do not include friction, piston acceleration, leaks or any other kind of work.

Transfer to a new rectangular comparison

A gas starts at 120 kPa, 15 L and ends at 180 kPa, 25 L. Compare pressure-change-first and volume-change-first paths. Find both works and their difference. Then describe the sign if each path is traversed exactly in reverse.

Check the gas-work transfer

Pressure-change-first gives 180 × (25 − 15) = 1800 J. Volume-change-first gives 120 × 10 = 1200 J. The difference is 600 J. Traversing each same path in reverse changes each work's sign, giving −1800 J and −1200 J. Those signs describe compression work by the gas, not negative absolute pressure.