How can I interpret $P(t) = \frac{1}{Q(t)} \frac{dW(t)}{dt}$ physically?

Question

Generally speaking, one can calculate the work, $W$ (energy) required to compress a volume, $V$ of gas by integrating the pressure-volume 'loop' in a phase-space as

$$W=\oint PdV$$ where $P$ is the pressure.

or alternatively if $W$ and $V$ are expressed as functions of time

$$W(t)=\int_0^t P(t)Q(t)dt$$

where $Q$ is the flow rate giving rise to the volume

From the alternate expression I can solve for pressure to get

$$P(t) = \frac{1}{Q(t)} \frac{d}{dt}W(t)$$

This says that pressure is the time rate of change of work (energy) divided by flow rate. But I'm having a hard time trying to visualize/interpret this meaning.

Can someone please help me interpret this expression?

Answer 1

By the way, it's customary to use primes or some other indicator inside the integral to be slightly less confusing. I.e. $$W(t) = \int_{t_0}^{t} P(t')Q(t')dt'$$

Be careful about how you think about Q(t). You've described it as a "flow rate", but really, you are saying $Q(t) = \frac{dV}{dt}$, or the rate of change of the volume with time.

Now, the term $$P(t)=\frac{1}{Q(t)} \frac{dW(t)}{dt} = \frac{dt}{dV} \frac{dW(t)}{dt}=\frac{dW(t)}{dV}$$ This is telling you that pressure is a measure of how much work you do by changing the volume of your system at a given time. Your original expression was just a slightly odd way of expressing the same thing.

Systems at high pressure have a lot of energy to release, so changing your volume results in a lot of work being done, i.e. $dW/dV$ is large. Taking the opposite example, for a system in vacuum (P(t) = 0) changing the volume does not result in any change in energy or any work being done, hence W(t) = 0, again corresponding to what you'd expect physically.