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

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.

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.
• @Sarthak Girdhar "𝛿 is mostly used for a change, and this change in work makes no sense obviously" I agree, My use of $\delta w$ was an uneasy compromise. I wanted a symbol for a small quantity (of work) but not for an increment of work, as if work were some function of state. That's why I kept off $dw$, but I agree that $\delta w$ is not ideal notation. Zemansky in his wonderful 1950s (?) thermodynamics textbook used a special symbol, $dW$ in which the up-stroke of the d had a horizontal bar through it to mean a small amount of work but not an increment of a larger quantity, 'work'. Commented Oct 6, 2020 at 14:19