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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.

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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.

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  • $\begingroup$ Your explanation for work makes sense and I see where ๐‘ฮ”๐‘‰ comes from. I think I am confused about the difference between ๐›ฟ๐‘ค (the differential of ๐‘ค) and ๐‘ค. I always thought work was defined as ๐‘ค = ๐‘ฮ”๐‘‰. So is ๐›ฟ๐‘ค = ๐‘ฮ”๐‘‰ an equivalent statement? $\endgroup$
    – LamGyro
    Oct 6, 2020 at 1:35
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    $\begingroup$ @JamesBond $\delta \ w$ implies an infinitesimal amount of work you do on the system. We consider an infinitesimal work when we can not make fast changes in the system, so that the equilibrium conditions are always satisfied. These are called quasi-static processes. I would recommend to focus more on intuition in solving doubts like these. The sign conventions, symbols in the book can sometimes be quite confusing, and even misleading ($\delta$ is mostly used for a change, and this change in work makes no sense obviously) $\endgroup$
    – Physiker
    Oct 6, 2020 at 3:32
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    $\begingroup$ @Sarthak Girdhar "๐›ฟ 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'. $\endgroup$ Oct 6, 2020 at 14:19

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