Work by a Gas

In the previous section, you learned to read PV diagrams. Now we put them to work — literally. The area under a process curve on a PV diagram tells you the work done. And the direction of a cycle tells you whether you’re looking at an engine (which makes useful work) or a refrigerator (which uses work to move heat).

This is one of the most underrated MCAT shortcuts: many work questions can be answered just by eyeballing the area on a PV diagram — no calculation required.

Work as Area Under the Curve

The fundamental relationship: work done by a gas equals the area under the process path on a PV diagram. For an isobaric (constant pressure) process, this simplifies beautifully:

For non-isobaric processes, the area under the curve still equals the work, but you may need to estimate the area geometrically (the MCAT usually gives you enough information to do this, or just asks you to compare areas).

Sign Conventions for Work

The sign of the work tells you the direction of energy flow:

Thermodynamic Cycles

A cycle is a process that returns to its starting state. On a PV diagram, it appears as a closed loop. Since the system returns to its original state, ΔU = 0 for the complete cycle. By the first law:

ΔU = Q - W = 0, so $Q_{\text{net}}$ = $W_{\text{net}}$ for any complete cycle.

The direction of the cycle matters enormously:

Clockwise Cycle = Heat Engine

A clockwise loop on a PV diagram means the expansion (rightward) happens at higher pressure than the compression (leftward). The expansion does more work than the compression takes, so the net work is positive - the system outputs useful work.

This is a heat engine: it takes in heat from a hot source, converts some to work, and dumps the rest into a cold sink.

Counterclockwise Cycle = Refrigerator/Heat Pump

A counterclockwise loop means the compression happens at higher pressure than the expansion. More work goes into the system than comes out, so the net work is negative - work is done on the system.

This is a refrigerator: it uses work input to move heat from cold to hot (the opposite of the natural direction).

Calculating Work for Each Process Type

Worked Example

A gas undergoes an isobaric expansion at P = 2 x $10^5$ Pa from $V_1$ = 0.01 m³ to $V_2$ = 0.03 m³.

W = PΔV = (2 x $10^5$)(0.03 - 0.01) = (2 x $10^5$)(0.02) = 4000 J

The gas does 4000 J of work on the surroundings by expanding.