Area of a $pV$ diagram [duplicate]

Question

On a $pV$ graph (where the Pressure is on the Y axis, and the Volume is on the X axis), I am conceptually confused on how the area under the graph is equivalent to the work done by the system (assuming positive change in volume = the system increasing in volume). I understand it mathematically (P times V = Work done). However if the graph is a closed shape, where the pressure and volume starts at a point and then comes back to that same point, how is there work done? The system simply starts at a certain pressure and volume, and ends up in the same place.

Answer 1

The work done by the system is not equal to the product of pressure and volume (P times V), but rather the integral of pressure with respect to volume (W = ∫ PdV).

In a PV diagram, the area under the curve does represent the work done by the system, however you have to break the curves into appropriate segments and then calculate the work under each segment. When the curve on the PV diagram forms a closed shape, it means the system has gone through a complete cycle and returned to its initial state. In this case, the work done is represented by the area enclosed by the cycle, not the area under the curve. This area can be positive, negative, or zero, depending on the direction and shape of the cycle. The system still does work throughout the cycle, but the net work done is not zero because the system returns to its initial state.

Answer 2

Break the cycle into segments. Suppose we have a rectangle-shaped path in $pV$ space, and that we traverse the path clockwise. The top segment has lots of area underneath it--large amounts of work--and it's positive since the pressure pushes outward, in the same direction as the change in volume. The bottom segment has less area underneath it--and that's negative since the pressure opposes the volume change. The sides have no are underneath them--no change in volume, no work. Add up those four contributions and you end up with a net positive amount of work, even though we go back to the same place. It's the path that dictates the work, not just the start/end states.

Answer 3

I understand it mathematically (P times V = Work done).

Mathematically, work is not P times V, but is

$$W=\int_1^2PdV$$

The integral is the area under the path from state 1 to state 2.

However if the graph is a closed shape, where the pressure and volume starts at a point and then comes back to that same point, how is there work done?

Again, work is not simply the product of the pressure and volume. It's the area under the pressure volume curve.

You are describing a cycle, i.e., a series of processes that bring you back to the same initial state. The net work done over the cycle is the area enclosed by the processes. The net work done by the system equals the areas under the curves for the expansion processes (work done by the system) minus the areas under the curves for the compression processes (work done on the system). See the figures below for the example of a Carnot heat engine cycle.

In addition to the pressure and volume returning to their initial state, so does the internal energy, $U$, return to its initial state, i.e., $\Delta U=0$ for the cycle. From the first law:

$$\Delta U =Q-W=0$$ $$W=Q$$

Meaning the net work done by the system during the cycle equals the net heat added to the system during the cycle, for conservation of energy over the cycle.

Hope this helps.