Expert: Daniel Mazur - 9/16/2009

Question
QUESTION: Hi Daniel,

I just can not get my head around this idea of reversible processes.

This is what I know.
A quasi static process is one which has the state functions specified at all points of time.
And this will be possible only if we change things infinitely slowly.
So if a process is quasi static and doesn't have any dissipative forces in play, it fulfils the conditions for being a reversible process.

Also the following definitions are given on wikipedia:

REVERSIBLE PROCESS

"An alternative definition of a reversible process is a process that, after it has taken place, can be reversed and causes no change in either the system or its surroundings"

IRREVERSIBLE PROCESS

"In an irreversible process, finite changes are made; therefore the system is not at equilibrium throughout the process. At the same point in an irreversible cycle, the system will be in the same state, but the surroundings are permanently changed after each cycle."

Now suppose we have an adiabatic cylinder filled with an ideal gas and a light frictionless piston attached to it.
If Suppose I compress the piston all of a sudden, then initially the pressure and tempertaure will be different at different points in the gas. I did some work on the gas that led to this change in state of the gas.
After this, if i stop pushing it further and leave the gas to itself, due to the increased pressure, the gas will tend to expand back to its initial state where the internal pressure equals the external pressure.
So when the gas expands back, it does the same amount of work back on me (does it or does it not??). So the final result that we have is the gas coming back to its initial state.
How is this process not reversible?? We didn't carry out the process infinetely slowly but still it comes  back to it's initial state. And what "permanent" changes to the surroundings, as the wikipedia definition says, has the gas caused in this case that it wouldn't have if we had carried out the process infinetely slowly???


Thanks,
Shikhin

ANSWER: Dear Shikhin,

Firstly, when you suggest your non-equilibrium process ("the pressure and temperature will be different at different points in the gas"), your global state variables have undefined values. Instead, you have defined them locally and only hope (!) that they still make sense, which they might or might not.

Secondly, you say "compress the piston all of a sudden", which means "very fast", but you do not say "Fast - compared to what?" Usain Bolt may be fast, a thrown cricket ball will be faster, yet a bullet will be faster still etc. Speed is a relative quantity and here it is critical in deciding, whether the state variables (previous paragraph) make sense or not. State variables are defined only for large number of gas molecules via a robust (Poisson) distribution of their microscopic properties (e.g. position and momentum of the molecules). This means that you need to consider finitely (not infinitesimally) large minimal volumes and finite (not infinitesimal) times over which you evaluate the (non-)equilibrium of your system.

In real systems when your state variables become undefined, you cannot PREDICT, what will happen next. However, it MAY happen that after equilibrium is reached again, the state variables will have precisely those values that they would have had, had they reached the final state by the infinitely slow quasi-static adiabatic process. This MAY happen, if only by chance. Now, what are the conditions under which this MUST happen? Ideal gas and constant number of molecules, no heat exchanged with the surroundings.... that's an ADIABATIC CONTAINER!

And here we are: By using the construct of adiabatic container you have removed the absolute necessity for slowness and quasi-static character of the process and effectively cheated the laws of REAL physical thermodynamic systems. In this case you can have a fast reversible process - but only in an abstract theory. Wikipedia speaks about permanent changes to the surroundings, which proves that they did not use the adiabatic container in their consideration.

Cheers,
Daniel



---------- FOLLOW-UP ----------

QUESTION: Hi Daniel,

I think I understood most of it.

Here's what I understood. Please correct me if I'm wrong.

If we have a diathermal container instead of an adiabatic one and if we compress the piston quickly enough so that our global variables are not defined, the temperature and pressure will be different at diffeerent points of the container and thus the heat lost to the surroundings will vary from point to point.
Now, if we compress it to a particular volume and then start with the reverse process of pulling the piston back, again quickly enough, so that the global variables aren't defined, heat will be drawn in from the surroundings, again in different amounts at different points of the container. And thus the net heat given and withdrawn from the surroundings in the two processes might not be equal thus casuing a permanent change to the surroundings.
Is that what you meant?? Am I missing anything here??

Also, suppose if we compress the gas at a fast enough(so that state variables are undefined) but constant speed and then expand it at this very speed, will the net heat given to the surroundings in the compression process be equal to the heat withdrawn in the expansion process??


Thanks,
Shikhin

ANSWER: Hi Shikhin,

once state variables are not defined, energy (heat) transfer to and from walls can be described only on a microscopic level. Nothing like averaging will work any longer to allow you to speak about the gas as a single entity (for example). You must move to the description using ~10^23 equations of motion with twice as many initial conditions (even more for gas molecules that have rotational and vibrational degrees of freedom). A small deviation in the initial conditions can cause almost arbitrarily large error in the whole calculation. You will need to know the kinetics (trajectory) and dynamics (energy balance) for every gas molecule from the initial equilibrium to the final equilibrium re-gained. Need I say the word "impossible"?

Putting aside the fact that no-one can manage that number of simultaneous equations of motion, it is principally impossible to know the exact positions and momenta of all the gas molecules. In fact, we can't measure even a single molecule's state of motion to a good enough precision to be any better use in calculation than random numbers. You may still ask why the the process will be still irreversible... Well, compression and expansion are at high speeds NOT symmetrical processes. No process is physically reproducible backwards, when you have that many particles to take care for. Time-reversal symmetry is a purely theoretical topic. Reversibility is tightly related to entropy and entropy stems from the exact microscopic arrangement (in space and in energy) of all molecules. In a non-equilibrium process without any state variables it is a pure chance, how far from an equilibrium-like distribution of molecules the real one will be, what will be the rate of energy exchange between the gas and container walls at any point in time, and therefore, to what kind of equilibrium the system will get in the end.

A reasonably slow compression (so that state variables stay defined) will result in a polytropic process somewhere between an adiabatic and isothermal one. If the container acts as a thermal bath and you compress the piston slowly, the process will be isothermal. The faster you compress and depress the piston, the more adiabatic-ish the process will be. In some non-equilibrium processes you can define state variables locally. The closer this will be to the global equilibrium, the smaller error you will make by still using the state equations. But an error will always be there and no analytic way to predict its magnitude.

Does this sound like a solid reason to you?

Daniel

---------- FOLLOW-UP ----------

QUESTION: Hi Daniel,

That is precisely what even i wanted to say in my last follow up, that when the state variables aren't defined, every molecule will have different conditions and equations associated to it and thus heat lost or gained will be different by different molecules of the gas.

So basically doesn't that mean that the heat lost to the surroundings in the compression process of the gas isn't equal to the heat gained by it during the exapnsion process?

Also, I still don't feel quite right about the second question.
If the speeds of compression and of expansion are exactly equal (fast enough so that the process is non-equilibrium and state variables are not defined), every molecule is affected, if not exactly, in nearly the same way when it is expanded as it was affected when it was compressed. So why aren't the two processes of expansion and compression almost symmetrical when the speed of expansion and compression is the same, with the result that the heat gained and lost to the surroundings equal each other in the two processes? and thus the non euilibrium process behaves like an equilibrium one with the initial and final states being the same??


Thanks,
Shikhin

Answer
Dear Shikhin,

the asymmetry that troubles you is difficult to prove without the proper (complicated) maths. Nevertheless, let's try an illustration. When you are compressing fast, you create a shock wave in front of the piston, where locally a high density (and hence pressure) of molecules will be. For that brief time you can imagine that heat exchange of the gas with the piston is very much increased, because pressure corresponds - microscopically - to higher momentum transfer rate.

Forced expansion, on the other hand, means that you drag much lower local density and pressure (empty space is the extreme) behind your piston and the rate of molecule impacts is momentarily decreased.

When you compare the two, the ideal gas will in compression be allowed to give more of its rising internal energy TO the piston than how much it is allowed to withdraw FROM the piston during expansion. All this is true, when the container (including the piston) is diathermic, which is to say that it will immediately obey the thermal state of the gas inside. I think diathermic container is defined by infinite thermal conduction and zero thermal capacity. If this is so, here comes your asymmetry.

A real-life container is not quite diathermic, so it acts always a bit like a thermal bath - a reservoir of temperature. The effect of this can be to a large extent included in the non-equilibrium thermodynamic equations of the problem. But it does get complicated and the properties of the gas during(!) the expansion or compression are still more or less nondescript.

Cheers,
Daniel