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What does the area under a Pressure volume diagram equal?

I read in my textbook it equals 'external' work done, but why is this?

First of all, what exactly is external work?

Can you get it external work by the simple formula $W= F\cdot s$?

Also, why does the area under a $pV$ diagram equal 'external' work?

What is the logic behind that?

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I think it depends on the case if it equals external or internal work. However, I still don't know what the difference is between external and internal work and why the area under a pV-diagram equals work.. –  user14445 Oct 31 '12 at 15:58
    
Just a hint for learning to interpret this kind of thing yourself: whenever you see "the area under the curve" you should be thinking of an integral. Then when you see "work" you should think "Ah...$W = \int \vec{F} \cdot d\vec{l}$!". Then you put those two thoughts together and see where it takes you. –  dmckee Oct 31 '12 at 16:21
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I haven't actually had integrals at school yet. –  user14445 Oct 31 '12 at 16:23
    
Well, that certainly makes my advice less than useful in the mean time, but when you get there it is worthwhile to remember. –  dmckee Oct 31 '12 at 16:25
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1 Answer

up vote 3 down vote accepted

The area under the curve in the pV-diagram is the integral $$ \int p \;\mathrm dV = \int \frac FA A\;\mathrm ds=\int F \;\mathrm ds \equiv W $$ by definition of pressure as force per area and (infinitesimal) volume as area times distance.

This is the mechanical work done by the system on the environment in case of expansion or by the environment on the system in case of compression, which differ by sign. It is called external to emphasize the interaction with the environment.

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I haven't gotten integration, so I'm afraid this is quite meaningless to me at this juncture.. –  user14445 Oct 31 '12 at 16:24
    
@user14445: the integral $\int p\,\mathrm dV$ is how you compute the area under the curve - you can't use simple multiplication $p\cdot V$ as the value of $p=p(V)$ changes as you move along the $V$-axis; if you don't want to read up on (Riemann) integration, just think of integrals $\int y\,\mathrm dx$ as products $y\cdot x$ and you'll get the right intuition for the physics... –  Christoph Oct 31 '12 at 16:37
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