How does Pressure times Volume = Work Done?

Question

I understand the equation mathematically-- $p$ times (change in) $V$ = Work done. However, I am confused about it conceptually. Work is (in a non-calculus context) constant force applied over a set distance. But, while the volume of, say, an ideal gas increases, pressure decreases--in other words, while the volume increases, the pressure is not constant. So then, what value of P would be used in the equation P times (change in) V?

Note: I presume that the concept of a Quasi-static process explains this question.

Answer 1

Work is (in a non-calculus context) constant force applied over a set distance.

Ok, so you only want to work with constant forces. That's fine.

But, while the volume of, say, an ideal gas increases, pressure decreases--in other words, while the volume increases, the pressure is not constant.

Not necessarily. This is assuming constant temperature. You can have isobaric expansion of an ideal gas where you also increase the temperature (add heat) to keep the pressure constant. This is the situation you want.

So then, what value of P would be used in the equation P times (change in) V?

The constant pressure described above.

Note: I presume that the concept of a Quasi-static process explains this question.

It certainly applies, but it's not the explanation.

Answer 2

Note: I presume that the concept of a Quasi-static process explains this question.

Somehow, yes, in the sense quasi-static means a sequence of infinitesimal actions during which each variable can be considered a "constant". It's all about summing a sequence of infinitesimal values.

You can imagine that the pressure $P$ is constant during a short (infinitesimal) time. This results in a change of volume we write $dV$ and the work done by the system during this infinitesimal time is $\delta W=PdV$. (note: this in only valid for reversible work, but all courses start by describing reversible work only, so you don't have to bother at this stage).

Of course, at the end of this infinitesimal change, the pressure may have changed and became $P+dP$. So the question could be: "why do we use $P$ in the formula rather than $P+dP$ or something in between?". This reasoning is just infinitesimal calculus. The answer is "it does not matter, use the simplest formula possible". Compare the two calculations:

$$(P+dP)dV=PdV+dPdV$$

$dPdV$ is a second order infinitesimal and we always eliminate them when a first order infinitesimal is already relevant. It's not even an approximation, it is mathematically irrelevant.

What I explained in not thermodynamics. You will find it absolutely anywhere physics does some infinitesimal reasoning. For example, the area between two circles or radius $r$ and $r+dr$ is $2\pi rdr$ : we multiply the length of the smallest circle by the distance $dr$. Why not the largest circle? For the same reason: it does not matter because the difference is a second order infinitesimal. See how it finally works to calculate the area of the disk: $$\int_0^R 2\pi rdr=2\pi (R^2/2-0)=\pi R^2$$

It is just about getting used to infinitesimal reasoning, and how we can use the simplification "during an infinitesimal change, we can consider this or that is constant".

Now, during a non infinitesimal process, the pressure truly varies (in general) and the formula $W=P\Delta V$ is very false. Instead we need to sum all the infinitesimal values:

$$W=\int \delta W=\int PdV$$

Answer 3

To simplify the issue let's look at a simple piston in a cylinder

Pressure = Force/area

When a piston moves the swept volume = Area* displacement

We know that work = Force * displacement

From these we can derive that Pressure * displacement * area = work done