How to derive that $delta w = - PdV$?

Question

I am not understanding how to derive this particular expression, which relates the inexact differential of work to the exact differential of volume,
$$delta w = -PdV $$
My attempt:
Reversible work can be defined as:
$$w=-int P dV  $$
First, I integrate both sides with respect to volume,
$$frac{d}{dV}(w)=-frac{d}{dV}(int P dV) $$
$$ frac{dw}{dV}=-P $$
Since the differential of work is inexact:
$$ delta w=-PdV $$
Mathematically, I am unsure about my first step. Nonetheless, this was my approach.

Philip Wood · Accepted Answer

It seems to me that you've gone round in a circle. What's wrong with this simple argument?
Suppose that the fluid exerts a force $F_n$ on a small area $A$ of the container wall, in a direction normal to that area. If that area moves outwards by a small distance $Delta x$ normal to $A$ then the work done by the fluid on $A$ will be
$$delta w=F_n Delta x=frac{F_n}{A} times A Delta x = p Delta V.$$
We are not in any way assuming that $delta w$ is a differential of some function of state, so there is no suspicion that $delta w$ is an exact differential.

How to derive that $delta w = - PdV$?

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