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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 𝑝Δ𝑉 comes from. I think I am confused about the difference between 𝛿𝑀 (the differential of 𝑀) and 𝑀. I always thought work was defined as 𝑀 = 𝑝Δ𝑉. So is 𝛿𝑀 = 𝑝Δ𝑉 an equivalent statement? $\endgroup$
    – LamGyro
    Commented Oct 6, 2020 at 1:35
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    $\begingroup$ @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'. $\endgroup$ Commented Oct 6, 2020 at 14:19

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