PV Diagrams & Reading Work off a Graph
- Find the work done by or on a gas as the area under a PV curve
- Determine the sign of work from the direction of the process
- Compute the net work of a cycle as the enclosed area
Work is the area under the curve
When a gas changes volume, the work involved is the area under the process path on a PV diagram. This is the same slope-and-area logic as every other graph in physics: pressure times volume change has units of energy, so the area is an energy. It also means work depends on the path, not just the endpoints — two processes joining the same two states can enclose very different areas and therefore involve different work.
The sign convention, and why it trips people
A gas that expands pushes outward on its surroundings and therefore does positive work on them: W_by gas > 0. A gas that is compressed has work done on it, so W_by gas < 0. The confusion arises because the first law can be written with either sign convention. The AP convention is ΔU = Q + W where W is work done ON the gas — so a compression gives positive W in that equation. Read the subscript before substituting.
A cycle encloses its net work
A cyclic process returns the gas to its starting state, so ΔU = 0 and the first law reduces to Q = −W_on = W_by. The net work done by the gas over the cycle is the enclosed area of the loop. If the loop runs clockwise, the expansion happens at higher pressure than the compression, so net work by the gas is positive — that is a heat engine. Counterclockwise gives negative net work, which is a refrigerator or heat pump.
A gas expands at a constant pressure of 2.0 × 10⁵ Pa from 0.010 m³ to 0.030 m³, then is cooled at constant volume, then compressed at 1.0 × 10⁵ Pa back to 0.010 m³, then heated back to the start. Find the net work done by the gas per cycle.
- 1.Expansion at 2.0 × 10⁵ Pa: W = PΔV = (2.0 × 10⁵)(0.030 − 0.010) = +4000 J done by the gas.
- 2.Constant volume: ΔV = 0, so no work.
- 3.Compression at 1.0 × 10⁵ Pa: W = (1.0 × 10⁵)(0.010 − 0.030) = −2000 J.
- 4.Constant volume again: no work.
- 5.Net = 4000 − 2000 = +2000 J. Equivalently, the enclosed area: (2.0 − 1.0) × 10⁵ × 0.020 m³.
A constant-volume process does no work however much the pressure or temperature changes — there is no area under a vertical line. Heat still flows and internal energy still changes; only work is zero.
A gas is compressed at constant pressure. The work done BY the gas is:
A gas undergoes a clockwise cycle on a PV diagram enclosing an area of 800 J. Over one complete cycle:
Two different processes connect the same initial and final states. Compared with each other, they must have the same:
Label the axes P and V and mark the direction of travel with an arrow. Rubrics award the direction, and it is what determines whether the cycle is an engine or a refrigerator.
Answer the 3 checkpoints as you read.
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