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Apparently trying to plot a $PV$ diagram for an adiabatic free expansion of ideal gas is undefined. I'm taking issue as to why this is the case. From what I understand using my diagram attached as my aid, free expansion is equivalent to removing the solid partition on the left, so that no work is done but the volume increases since the particles aren't given any more energy, they just take longer to hit an obstacle in $+y$. But, I would think that since the volume increases, pressure would as well at the exact same degree, so the $PV$ diagram would be an incremental increase of $V$ and decrease of $P$, having some $y=-x$ behavior. But I've been told that is not the case. Just wondering what about my reasoning is incorrect and why it's undefined in this case.

enter image description here

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Points on a $PV$ diagram represent equilibrium states. During the transition, the gas is not in equilibrium. This is easy to see when you consider that the pressure will be different in different parts of the gas ... high near where the partition was, low at the wall away from the partition.

What value would you use for $P$ when plotting your points?

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    $\begingroup$ So, if I'm understanding you correctly, are you saying that, until late enough $t$, the pressure and volume are both indeterminate, because the pressure depends on where it is in the box (as the pressure at the top will be less at least at the beginning) and the volume will as well, we have no way of knowing how to join the points at $t=0$ and sufficient $t$ where there is finally an equilibrium state. $\endgroup$ – sangstar Oct 10 '17 at 8:25
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    $\begingroup$ Yes. This is basically correct. Between the initial and final equilibrium states of the system, the pressure (and temperature, and density) will not be spatially uniform within the container. There will be pressure waves traveling through the gas, and viscous stresses will be present within the gas to dissipate mechanical energy. This can all be analyzed in detail by applying computational fluid dynamics (CFD), but it is not needed if you are only interested in the final equilibrium state. $\endgroup$ – Chet Miller Oct 10 '17 at 12:13

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