Adiabatic expansion of a piston in a cylinder filled with ideal gas

[Hiero]

Homework Statement

A cylinder of cross section A is divided into two chambers 1 and 2 by means of a frictionless piston. The piston as well as the walls of the cylinder are heat insulating, and the chambers initially have equal length L. Both chambers are filled with 1 mol of helium gas, with pressures $P_0$ and $2P_0$ respectively. The piston is then allowed to slide freely, whereupon the gas in chamber 1 pushed the piston a distance “a” to equalize the pressure. Find a.

Relevant Equations

$PV^\gamma =$ constant

I was puzzling over how to solve this and finally peeked at the solution. They used the relevant equation above.

I disagree with this though. The problem specifically says “the piston is allowed to slide freely!” This means that we don’t let it happen slowly. So then we are not in quasi-static equilibrium throughout the expansion. Thus the pressure is not well defined at each moment, and so the relevant equation does not apply! (It is derived from dU = -PdV which seems to me to rely on the pressure being well defined and hence the process being reversible, unlike in this problem.)

Do you agree with me that the relevant equation only applies for reversible (quasi-equilibrium) adiabatic processes, and not irreversible adiabatic processes?

(And if so, how would you solve this problem as it stands?)

 

[Isaac0427]

Do you agree with me that the relevant equation only applies for reversible (quasi-equilibrium) adiabatic processes, and not irreversible adiabatic processes?

Yes. However, thermodynamics becomes far, far more difficult for non-quazi-static processes. You are correct that the equation will not give a 100% correct answer when applied to the real world...unfortunately, no other equation in classical physics does either. Physics is full of approximations, and the quazi-static approximation made in this problem turns out to be incredibly useful in thermodynamics.

It may also be helpful to note that the concept of a quazi-static process in general is an approximation. A true quazi-static process would take an infinite amount of time. So, I would consider the answer given to be correct, as much as any answer in classical physics is correct.

The problem specifically says “the piston is allowed to slide freely!” This means that we don’t let it happen slowly

I also think you are confusing "freely moving" with "fast." Just because there is no friction does not mean there is not a force requirement to get the piston moving. A "freely moving" object in deep space could move quite slow, depending on its mass and the force applied to it.

 

[Hiero]

I also think you are confusing "freely moving" with "fast." Just because there is no friction does not mean there is not a force requirement to get the piston moving. A "freely moving" object in deep space could move quite slow, depending on its mass and the force applied to it.

But the problem already stated earlier in the question that it’s frictionless! Why wouldn’t I interpret the additional statement of “freely” as “unconstrained”? I guess your point is that if the piston is very heavy then it will occur slowly? (I was indeed imagining a light piston.) They should really say something like that then because that really confused me.

Yes. However, thermodynamics becomes far, far more difficult for non-quazi-static processes. You are correct that the equation will not give a 100% correct answer when applied to the real world...unfortunately, no other equation in classical physics does either. Physics is full of approximations, and the quazi-static approximation made in this problem turns out to be incredibly useful in thermodynamics.

It may also be helpful to note that the concept of a quazi-static process in general is an approximation. A true quazi-static process would take an infinite amount of time. So, I would consider the answer given to be correct, as much as any answer in classical physics is correct.

At least I can take solace in the fact that I couldn’t solve the problem (as I interpreted it) because it was unsolvable. It’s frustrating that they couldn’t just explain the problem better. I don’t think I like this author.

Thanks for your comments. Take care.

 

[Chestermiller]

Homework Statement:: A cylinder of cross section A is divided into two chambers 1 and 2 by means of a frictionless piston. The piston as well as the walls of the cylinder are heat insulating, and the chambers initially have equal length L. Both chambers are filled with 1 mol of helium gas, with pressures $P_0$ and $2P_0$ respectively. The piston is then allowed to slide freely, whereupon the gas in chamber 1 pushed the piston a distance “a” to equalize the pressure. Find a.
Relevant Equations:: $PV^\gamma =$ constant

I was puzzling over how to solve this and finally peeked at the solution. They used the relevant equation above.

I disagree with this though. The problem specifically says “the piston is allowed to slide freely!” This means that we don’t let it happen slowly. So then we are not in quasi-static equilibrium throughout the expansion. Thus the pressure is not well defined at each moment, and so the relevant equation does not apply! (It is derived from dU = -PdV which seems to me to rely on the pressure being well defined and hence the process being reversible, unlike in this problem.)

Do you agree with me that the relevant equation only applies for reversible (quasi-equilibrium) adiabatic processes, and not irreversible adiabatic processes?

(And if so, how would you solve this problem as it stands?)

Yes. I totally agree with you. See my posts in the following thread, in which Andrew Mason and I argued for a while on how to go about getting the most accurate approximation to the solution for this kind of problem: https://www.physicsforums.com/threa...n-an-adiabatic-container.981529/#post-6271576

 

[Isaac0427]

I guess your point is that if the piston is very heavy then it will occur slowly? (I was indeed imagining a light piston.) They should really say something like that then because that really confused me.

Yes-- say it is a very heavy piston, with the container on its side. The process would be very slow, but close to quazi-static. I think that the problem should have specified that it's a quazi-static process, but you could probably assume that given that it is likely all you have talked about in the course.

At least I can take solace in the fact that I couldn’t solve the problem (as I interpreted it) because it was unsolvable. It’s frustrating that they couldn’t just explain the problem better. I don’t think I like this author.

Unsolvable exactly. However that is the story with most of physics. I am sure there are some very sophisticated, probably computer-based approximations for irreversible processes. I think that is far beyond the scope of your course-- @Chestermiller seems like he knows more about that than I do. Also, out of curiosity, what book are you using?

 

[Chestermiller]

Yes-- say it is a very heavy piston, with the container on its side. The process would be very slow, but close to quazi-static. I think that the problem should have specified that it's a quazi-static process, but you could probably assume that given that it is likely all you have talked about in the course.

Even if the piston is massive and the deformations are approximately quasi static, we still wouldn't be able to satisfy the constraint that the total internal energy change is zero. This is because the kinetic energy of the massive piston will eventually be dissipated and irreversibly converted to internal energy. So the pressures on the two sides of the piston will not match, and the difference in work will be finite, and equal to the kinetic energy of the piston. How this irreversibility affects the final states in the two chambers is unknown. Try doing it this way, and see that the change in U is not zero because of the unaccounted for affect of the massive piston.

 

[Isaac0427]

Even if the piston is massive and the deformations are approximately quasi static, we still wouldn't be able to satisfy the constraint that the total internal energy change is zero. This is because the kinetic energy of the massive piston will eventually be dissipated and irreversibly converted to internal energy. So the pressures on the two sides of the piston will not match, and the difference in work will be finite, and equal to the kinetic energy of the piston. How this irreversibility affects the final states in the two chambers is unknown. Try doing it this way, and see that the change in U is not zero because of the unaccounted for affect of the massive piston.

I think that’s far too deep for an introductory thermodynamics course. At the level the OP is asking, the system is approximately adiabatic and quazi-static.

 

[Chestermiller]

I think that’s far too deep for an introductory thermodynamics course. At the level the OP is asking, the system is approximately adiabatic and quazi-static.

But, if you treat it that way and you match the final pressures, you won't be able to get the changes in internal energy to sum to zero.

 

[Isaac0427]

But, if you treat it that way and you match the final pressures, you won't be able to get the changes in internal energy to sum to zero.

How do you figure that?

 

[Hiero]

I think that’s far too deep for an introductory thermodynamics course. At the level the OP is asking, the system is approximately adiabatic and quazi-static.

I’m not even in a thermo course... not sure why you assume that... it’s self study. And just because the problem is trivial doesn’t mean I want a trivial discussion. Or else I wouldn’t have come here pointing out the flawed assumption of reversibility and asking how to solve it in the irreversible case.

but you could probably assume that given that it is likely all you have talked about in the course.

Problems should be self contained and clear and not depend on the context of a course.

Also, out of curiosity, what book are you

It is introduction to statistical physics by Huang, but there have been multiple reasons that I dislike the authors presentation and problems so I’ve already deleted it. I think I’ll try the book by Chandler next.

 

[Hiero]

Even if the piston is massive and the deformations are approximately quasi static, we still wouldn't be able to satisfy the constraint that the total internal energy change is zero. This is because the kinetic energy of the massive piston will eventually be dissipated and irreversibly converted to internal energy. So the pressures on the two sides of the piston will not match, and the difference in work will be finite, and equal to the kinetic energy of the piston.

I’m a bit unclear on what you mean. If the piston is massive enough to move slowly enough to let the chambers equilibrate as it moves, and if it is truly frictionless, then I imagine the piston would oscillate back and forth endlessly? Is your point that to avoid this oscillation we must have a small amount of friction which then transfers some internal energy from the gas to the piston+cylinder?

 

[Chestermiller]

I’m a bit unclear on what you mean. If the piston is massive enough to move slowly enough to let the chambers equilibrate as it moves, and if it is truly frictionless, then I imagine the piston would oscillate back and forth endlessly? Is your point that to avoid this oscillation we must have a small amount of friction which then transfers some internal energy from the gas to the piston+cylinder?

Friction is not the only dissipation mechanism that would allow the piston oscillation to be damped. The gas (even in the ideal gas limit) is viscous (which is a major effect in irreversible processes), and this would dissipate the kinetic energy of the piston, converting it to internal energy.

 

[Hiero]

Friction is not the only dissipation mechanism that would allow the piston oscillation to be damped. The gas (even in the ideal gas limit) is viscous (which is a major effect in irreversible processes), and this would dissipate the kinetic energy of the piston, converting it to internal energy.

Great point. However doesn’t the kinetic energy of the piston come from the internal energy of the gas in the first place? The high pressure chamber does more positive work on the piston than the low pressure chamber does negative work. So isn’t the energy just going from the gas to the piston and then back to the gas?

Edit:
I suppose the piston would probably heat up a bit as well though, which would take energy from the gas.

 

[Isaac0427]

I’m not even in a thermo course... not sure why you assume that... it’s self study. And just because the problem is trivial doesn’t mean I want a trivial discussion. Or else I wouldn’t have come here pointing out the flawed assumption of reversibility and asking how to solve it in the irreversible case.Problems should be self contained and clear and not depend on the context of a course.It is introduction to statistical physics by Huang, but there have been multiple reasons that I dislike the authors presentation and problems so I’ve already deleted it. I think I’ll try the book by Chandler next.

That book is written for an introductory thermodynamics/thermal physics course. I also disagree that a problem should be self-contained and independent of the context of the course. Introductory physics textbooks don’t include the phrase “in the nonrelativistic approximation” in every kinematics problem.

If the book you are using is much like the one I used at all (Thermal Physics by Daniel Schroeder— I would highly recommend it), everything in the book assumed a quazi-static approximation. I agree that it’s worth-while asking about the case without that approximation, but it’s a standard assumption that you’ll find in all introductory textbooks that I know of.

 

[etotheipi]

I don't think the question is a good one. Evidently the given solution has assumed that the process is reversible ($PV^{\gamma}$ is only conserved in a reversible adiabatic process), but that is definitely not the case in this problem. The piston will not be in equilibrium apart from in the very final rest state. We would need to take into account things like viscous stresses in the gas and get into the fluid dynamics in order to get a reasonable answer!

 

[Hiero]

it’s a standard assumption that you’ll find in all introductory textbooks that I know of.

I know you’ve seen the basic example of the “FREE” expansion of a gas which is not quasi static. Why would I not assume that “the piston moves FREELY” would mean something similar? It’s not difficult to say “we slowly move the piston until the pressures equalize” and then I would just imagine someone’s hand holding the piston instead of expanding FREELY...

I also disagree that a problem should be self-contained and independent of the context of the course. Introductory physics textbooks don’t include the phrase “in the nonrelativistic approximation” in every kinematics problem.

It is not the same because if you solve such problems relativistically you’ll get basically the same answer (which can be shown by truncating the Taylor series expaned in powers of the velocity). It does not fundamentally change the dynamics of the problem like in this example.

At any rate, my friend, we are quite off topic. Let's not get carried away with silly things.

 

[Chestermiller]

Great point. However doesn’t the kinetic energy of the piston come from the internal energy of the gas in the first place? The high pressure chamber does more positive work on the piston than the low pressure chamber does negative work. So isn’t the energy just going from the gas to the piston and then back to the gas?

My point is that, in order for the change in internal energy of the two chambers to sum to zero, the kinetic energy of the piston has to be converted back to internal energy. The ideal adiabatic reversible changes of the two gases would otherwise result in a change in total internal energy of the gases inside this rigid container if the final pressures are required to match. Something has got to give. The thing that has to give in reality is the assumption that the gases can undergo adiabatic reversible changes (with the final pressures matching), even with a massive piston.

Edit:
I suppose the piston would probably heat up a bit as well though, which would take energy from the gas.

Not if the piston had negligible heat capacity, which is certainly conceptually possible. Allowing the piston to have finite heat capacity would, of course, complicate the analysis somewhat.

 

[Hiero]

Thanks for your insight @Chestermiller you are truly a master of thermodynamics!

 

[Chestermiller]

Thanks for your insight @Chestermiller you are truly a master of thermodynamics!

Your recognition that the processes would not be adiabatic reversible speaks very highly of you also, even as one not as experienced yet. I can recognize raw talent when I see it.

 

[Hiero]

Your recognition that the processes would not be adiabatic reversible speaks very highly of you also, even as one not as experienced yet. I can recognize raw talent when I see it.

Aww you’re too kind I’ve been down recently but you’ve just brightened my day :)

You sir are a treasure for everyone here.

Take care