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Adiabatic Free Expansion Question

by brainpushups
Tags: adiabatic, expansion, free
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brainpushups
#1
May9-12, 09:34 AM
P: 35
A typical free expansion scenario described in texts is that of a gas confined to a thermally insulated container with a removable membrane. When the membrane is removed the gas expands into vacuum. I understand that no work can be done because the gas does not exert a force. However, I am confused by the statement given at this source:

http://www.etomica.org/app/modules/s...ckground2.html

Which says that because the expansion is free, the change in volume is zero. Obviously this must be the case because this is the only way for the area under the path on the PV curve to equal zero. Can anyone articulate this idea? It is not obvious to me why we can say the change in volume during free expansion is zero.
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nonequilibrium
#2
May9-12, 09:45 AM
P: 1,417
Where exactly on that page do they say that they take the change in volume to be zero?
brainpushups
#3
May9-12, 09:56 AM
P: 35
Right below the first diagram. It actually says that the "the volume change of the system is zero," not the gas. Is that what matters? That the container has a constant volume? Seems a little ad hoc.

nonequilibrium
#4
May9-12, 10:20 AM
P: 1,417
Adiabatic Free Expansion Question

Ah now I see it. Yes, I agree it's a very odd statement. It seems wrong. But anyway, the rest of their argument doesn't change if you leave that bit out, so I suppose it shouldn't be a source of worry (although I understand it was a source of confusion).
brainpushups
#5
May9-12, 10:42 AM
P: 35
I don't have a problem with the website, I was just using it since it gets to the heart of my question: If the work done by a system is the area under the PV diagram curve and a gas undergoes an adiabatic free expansion how can the area under the curve be zero when there is clearly a change in volume and pressure of the gas?

Clearly no work is done and no heat is added so the internal energy must stay the same. For an ideal gas this means the temperature stays the same.

In my book this free expansion scenario (1) is pictured next to a scenario where the gas expands and does work on a piston by absorbing energy by heat from its surroundings (2). The process is done slowly so the temperature remains constant.

These two scenarios are pictured together to illustrate the path dependence of heat as a form of energy transfer. In both (1) and (2) the gas starts at the same initial temperature, pressure, and volume and each end up at the same pressure temperature and volume. The states of the gases are the same and the beginning and end, but the paths they took to get to the final states differ.

Back to my question: Can you explain how the volume change for an ideal gas undergoing a free expansion can be zero? Where is the subtlety that I am missing?
brainpushups
#6
May9-12, 11:30 AM
P: 35
What do you think about this:

The state changes discontinuously from the initial to the final so the integral is ill defined and cannot be evaluated.

or perhaps it is better to think about the system as a whole:

The average pressure remains constant and the total volume does not change therefore the work done is zero. Therefore the state doesn't actually change so there is no curve on the PV diagram.

Thoughts?
nonequilibrium
#7
May9-12, 11:48 AM
P: 1,417
The idea is pretty straightforward actually: PdV does not always equal the work done by the system. Maybe you've seen the derivation of PdV = dW by using the idea of a lot of particles pushing against a wall that is moving outward (and then using the fundamental notion of Fdx = dW). However, that derivation assumes pressure is homogeneous in the system (otherwise, what does P mean?), and for this to be true, the process needs to happen very slowly (also called quasistatically). But you can see that free expansion is not quasistatical, it's quite sudden. As soon as you remove that inner membrane, molecules start to diffuse into the right half and you can intuitively imagine that the "pressure of the system" to be an ill-defined concept, or that it at least can only be defined locally.

In short: PdV does not equal the work done by the system in irreversible transitions.
brainpushups
#8
May9-12, 12:04 PM
P: 35
Got it. Thanks.

The explanation by Count Iblis in this thread

http://www.physicsforums.com/showthread.php?t=225555

explains it quite nicely.
Dickfore
#9
May9-12, 12:21 PM
P: 3,014
You could say that P = 0 in the volume where the gas expands freely. That is why W = 0.
Ken G
#10
May10-12, 03:17 AM
PF Gold
P: 3,135
Actually, I really don't like the language that says we must through out PdV in a free expansion. In fact PdV still has a very important and useful physical meaning even in a free expansion, we simply need to treat it locally rather than globally. Each tiny volume in the gas (treated as a fluid) still has a perfectly meaningful pressure P and volume V, and as each of those volumes change by dV, they do indeed do work PdV on their surroundings. It's just that their "surroundings" are more of the same gas, not walls or pistons. So the work the gas does is done on itself, and it shows up as lost thermal energy, in favor of gained bulk flow kinetic energy.

Thus one can say an adiabatic free expansion converts the internal kinetic energy of the gas into bulk-flow kinetic energy, and what mediates that conversion is good old-fashioned PdV work, simply interpreted locally instead of globally. So tossing out the concept completely seems ill-advised, though mr. vodka mentioned that the meaning becomes local rather than global. That's fine though-- P has a perfectly useful local meaning, it is simply (nonrelativistically) 1/3 of the local kinetic energy density in a frame locally moving with the bulk speed of the gas (so the gas appears locally stationary in the frame where P is being reckoned). All you need is for the gas motions to continue to appear isotropic in that frame, but I believe that generally does continue to hold quite well (as physical idealizations go).


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