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I'm studying thermochemistry and I don't get this. What's the difference between $w = Fd$ and $w = -p \Delta V$?

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3 Answers 3

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Aside from a sign convention artifact, they're the same thing.

Pressure is force per unit area:

$$ P = {F\over A} $$

If a surface with a given area is moved through a displacement $\Delta x$, then the volume of displacement is the product of $A$ and $\Delta x$:

$$ \Delta V = A \Delta x $$

So, the $P\Delta V$ work formula can be rewritten:

$$ w = -P\Delta V = -P\left(A \Delta x\right) = - \left({F \over A}\right)\left(A \Delta x\right) = - F \Delta x $$

There are a lot of assumptions embedded in the above, and you have to be a lot more careful than I was about defining positive and negative quantities correctly, but on a high level that's all there is to it.

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Here's an example to illustrate the difference.

At a pressure of one atmosphere, a musket (gun) is fired due to the expansion of exploding gunpowder. The musket has a 1 meter barrel, with 10mm inner diameter, and 100 gram musket ball (bullet). The musket ball exits the barrel at a velocity of 100 meters per second. How much work is done by the expanding gunpowder products by the time the musket ball leaves the barrel?

There are many types of work: gravitation/acceleration, electromagnetic, etc.
PV (pressure-volume) work is just one type of work.

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The first equation $w = F\,\mathrm dx$ is the definition of how we define work, which in itself is the energy expanded to move an object of some mass $F$ some distance $\mathrm dx$.

While $w=-P \Delta V$ is an expression that tells us work that has to be done by a gas $w$ to expand into a volume. The '$-$' is there because we define work done by a system as expanded energy

Remember that work is esentially energy so its ok to compare

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  • $\begingroup$ Probably mean "expended" and not "expanded". $\endgroup$
    – Buck Thorn
    Commented Mar 26 at 8:02

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