# Is there a quasistatic process that is not reversible?

I have seen several questions and good answers on the link between reversible and quasistatic processes, such as here or here. However, these questions only adress one side of the problem : a reversible process is necessarily quasistatic.

I am interested in the other side of the equivalence : is there a process that is quasistatic, yet not reversible ? It looks to me that an irreversible process cannot be made perfectly quasistatic.

The wikipedia article about quasistatic processes takes as an example the very slow compression of a gas with friction. As the compression occurs very slowly, the transformation is quasistatic, and the friction makes it irreversible. I am not convinced by this example : if you press on the piston with a vanishingly small force you will have to reach the threshold of the Coulomb law for solid friction before moving the piston anyway. It makes the process non-quasi-static, however small the Coulomb threshold might be.

Another example I've heard of is the reaction between a strong acid and a strong base. It is always an irreversible process, and you could make it quasistatic by adding very small drops of base into acid at a time. But by trying to do that, you would inevitably reach a limit to the size of the drop imposed by surface tension.

Even if "reversible" and "quasistatic" mean very different things, is it true to consider that in practice, a reversible process and a quasistatic process is essentially the same thing ?

• All macroscopic processes in a ferromagnet are essentially irreversible the result being that no matter how slowly the process proceeds it is irreversible. The irreversibility manifests itself in the $B/H$ curve hysteresis whose size is independent of the process speed. Quasistatic is not the same as reversible: all reversible processes are quasistatic but not all quasistatic processes are reversible. – hyportnex Dec 8 '16 at 15:27
• First you discuss examples that are in practice clearly quasi-static (yet not reversible), and then you ask whether it is true that in practice quasi-static = reversible? – Ruben Verresen Dec 8 '16 at 21:24
• @RubenVerresen The examples I discussed are precisely not quasistatic. – Dimitri Dec 13 '16 at 10:10
• @hyportnex Thanks for the ferromagnet example, I had not thought about that. Could you elaborate a bit more in an awnser ? I know quasistatic doesn't mean the same thing as reversible, I was asking for a concrete example of a quasistatic irreversible phenomenon. – Dimitri Dec 13 '16 at 10:12
• With the precision you ask, nothing is quasi-static. At equilibrium, your system is at a local minimum which you try to move around by affecting the parameters of the system. If done infinitely slowly, you get a perfect quasi-static process. In any real process, irreversibility shows itself as the excitations the driving of the system causes are dissipated through the anharmonicity of the potential and transferred to the thermally populated states. – G. Bergeron Dec 20 '16 at 11:22

## 6 Answers

Most quasi-static processes are irreversible. The issue comes down to the following: the term quasi-static applies to the description of a single system undergoing a process, whereas the term irreversible applies to the description of the process as a whole, which often involves multiple interacting systems.

• In order to use the term quasi-static, one has to have a certain system in mind. A system undergoes a quasi-static process when it is made to go through a sequence of equilibrium states.

• A process is irreversible if either (a) the system undergoes a non-quasi-static process, (b) the system undergoes a quasi-static process but is exchanging energy with another system that is undergoing a non-quasi-static process, or (c) two systems are exchanging energy irreversibly, usually via heat flow across a finite temperature difference.

One can imagine a (admittedly idealized, as most of basic thermodynamics in physics) process in which two systems undergo quasi-static processes while exchanging energy via heat due to a finite temperature difference between them. The irreversibility comes about due heating due to the temperature difference between them rather than due to irreversibilities inside each system.

• Thanks for the clarification between single and multiple systems. However I disagree with your example, if there is a finite temperature difference then the process is not quasistatic because there is no thermal equilibrium at all times. – Dimitri Dec 13 '16 at 10:17
• @Dimitri. I don't think I agree with your assessment. Imagine an idealized situation where the two systems in contact have very large specific heats and large thermal conductivities so that the temperature necessarily changes very slowly, allowing each system to equilibrate very quickly when it gains a little bit of thermal energy. This is really what we mean by quasi-static, that the system's behavior approaches that of being in equilibrium at all times. In this way, each system is undergoing a quasi-static process. However, since the two systems are at different temperatures, ... – march Dec 14 '16 at 16:56
• the combined system is not undergoing a quasi-static process, because it's not in thermal equilibrium, as you said. This is exactly what I meant by the difference between the two terms: you get to choose the system you're talking about when you use the term quasi-static, but you don't when you use the term reversible. – march Dec 14 '16 at 16:56
• I'm concerned that your definition of quasistatic might be misinterpreted the way it's stated. Consider, for example, the adiabatic free expansion of an ideal gas into a sequence of extremely small extra volumes. If we let the gas come to equilibrium before allowing it to expand into the next small compartment, we might be tempted to call this process quasistatic based on your definition, but I personally think that would violate the spirit of the term. Shouldn't one add something like "as the process becomes slower/more incremental, the successive equilibrium states become more dense on – joshphysics Jul 1 '17 at 16:23
• (contd.) some reasonably smooth curve in the thermodynamic state space of the system, and this curve can be used to correctly compute any thermodynamic quantity one chooses for the process by integrating an appropriate differential form, e.g. heat or work, along it?" – joshphysics Jul 1 '17 at 16:24

In your question you mentioned two examples -- (1) slowly moving something that has friction, and (2) gradually mixing two chemicals that react spontaneously ($\Delta G\gg0$).

Then you said neither of these count as quasi-static because of (1) stiction, and (2) minimum droplet size due to surface tension.

I see your objections as pointless nitpicking. First, with slight creativity, we can get around these objections. (1) Instead of friction between two solids, call it viscous drag of a solid in a liquid. (2) Put the acid and base on two sides of a barrier with extraordinarily small pores in it, such that one molecule passes through every minute. OK, you'll say, but that's still one molecule at a time, not truly infinitesimal. That brings us to the second point, which is that you can do this kind of nitpicking with any so-called quasi-static process. Take an ideal Carnot engine. It's ideal! It has perfectly-insulating walls and perfectly-frictionless pistons and infinite reservoirs with infinitesimally slow heat transfer. None of these things are physically possible!

The whole notion of "quasi-static" is an ideal which is conceptually useful even if it is kinda inconsistent with practical realities in many (perhaps all) cases.

What we mean by "quasi-static" is really: Start with fast change, and make it slower and slower, and see what the limit is as rate goes to zero. If a Carnot engine has the same efficiency at one cycle per minute, per week, and per century, we can safely extrapolate that an ideally-quasi-static Carnot engine, with one cycle per eternity, would have the same efficiency. The latter may not be physically possible for various reasons, but that's OK, we don't need to actually imagine building it.

Likewise, if mixing chemicals together over the course of one hour releases the same amount of heat (within 0.0001%) as over the course of one month, we can say that both mixing processes are essentially quasi-static, and nobody really cares whether or not it's physically possible to mix them together smoothly over the course of 500 millenia.

Squeezing toothpaste out of a tube.

There are several definitions for quasi-static. They are not equivalent. There is some confusion about it in the literature, notably on Wikipedia's page.

First, before you ask if quasi-static implies reversible, you must make clear you are talking about the same system. March's answer explains this. I'll focus on the question once this confusion is cleared: you are talking about the same system.

As far as I know, the word reversible does not make sense for a non thermally isolated system. Hence you must choose a thermally isolated system, in other words, an adiabatic process. If a sub-system is not thermally isolated, then consider the larger system. Now we're ready to get at the core of the question.

Among all the definitions for quasi-static we find:

• the equilibrium is sufficient at each infinitesimal step for the macroscopic variables (such as temperature) to be defined. Not clear: for each sub-system, for the global system?
• the motion is very slow.
• the change is slow enough so that the system is at equilibrium at each infinitesimal step

I'll focus the latest definition, which I'll call "truly quasi-static". With this definition, quasi-static is equivalent to reversible. Let's focus on the statistical mechanics foundations (with classical mechanics as a background). In statistical mechanics, a "truly quasi-static" process can be defined as:

"The process is equivalent to a progressive change in the system's Hamiltonian that is:

• Adiabatic : the change does not depend on the unknown micro-state (position in the phase space).
• slow and smooth enough so that the system has time to go through its full energy orbit (in the phase space) for each infinitesimal change in the Hamiltonian. In other words the system can be considered at equilibrium at each infinitesimal step."

This is officially the definition of "adiabatic reversible". When you write $$dU = -PdV$$ ($$V$$ is any variable the Hamiltonian depends on and $$P$$ is the generalized force), you mean this kind of process. Even though this defines "reversible", it is interesting because it does so thanks to an intuitive concept of quasi staticity (slow change) instead of the intuitive concept of reversibility (the reversed process leads to the initial state). The two definitions are equivalent. This constitutes a theorem.

Usual examples :

• moving the piston of an insulated gas chamber (much slower than the speed of sound).
• moving a piston of a gas chamber in thermal contact with another gas (slow enough to allow thermal equilibrium at each step). In this example, the system under consideration is the union of the two gases.
• Counter-example : the irreversible Joule (free) expansion

This definition excludes heat transfer, since in the case of heat, the Hamiltonian varies in a way that depends on the micro-state (during each collision with a mollecule from the other system for example). It excludes friction. If the motion happens to be very slow but the pores are very small, then the Hamiltonian varies abruptly. It also excludes this interesting case mentioned by Huang : “a gas that freely expands into successive infinitesimal volume elements”. Indeed, if the potential wall is smooth, this cannot be a free expansion but a usual reversible expansion.

Now, consider this definition of quasistatic : "The system is at equilibrium during the process (so that temperatue and other variables are defined almost everywhere) except possibly in small places where some irreversibility happens".

This allows friction and viscosity. With this definition, you can say that quasistatic does not imply reversible.

Quasi-static: process sufficiently slow so that the system is practically in equilibrium at each instant. Reversible: when the process temporal inverse is realizable in the practical etc.

Vide Kerson Huang, Statistical Physics, pg 4, Reversible implies Quasistatic Quasistatic does not imply reversibility.

And more... Quasistatic and without friction does NOT imply reversibility.

An example of quasi-static and without dissipation process but NOT REVERSIBLE: A sequence of successive infinitesimal free expansions with long time intervals between steps.

A reversible process is that its entropy change is zero. This could be $$\triangle S=\frac{\delta Q}T$$ So, for an isothermal process, if there is no heat exchange, then it is reversible.

A quasi-static process is when the system is almost at equilibrium state at any moment. If there is no heat transfer, it can be a quasi-static process. But with heat transfer, it can also be a quasi-static process. And due to heat exchange, it become irreversible. For example, you can very very slowly heat up gas in a volume by increasing its ambient (reservoir) temperature by $1.0\times10^{-9}$ degC/s.

• A reversible process can have entropy change, this is precisely what you wrote with $\Delta S \neq 0$. And it is not true that an isothermal adiabatic process is necessarily reversible. – Dimitri Dec 13 '16 at 10:14
• @Dimitri. I agree with you here, except that probably what user115350 meant here is that the total change in entropy of the universe is zero. Sure, the systems in contact with each other during the process will likely have entropy changes, but for a reversible process, the net change in entropy of all the systems exchanging energy will necessarily be zero (pretty much be definition of reversible). – march Dec 14 '16 at 17:04
• I was meant that reversible process has zero entropy change. I don't know why it is not true. – user115350 Dec 14 '16 at 17:31
• Sorry for my negligence; I took a sufficient condition in my answer though it is not necessary. Here I can make an example of a quasi-static process but not reversible: when we very slowly bend a steel bar till it yields, the process is quasi-static but irreversible. Breaking bond between molecules can generate a small amount of heat but also making it irreversible. – user115350 Dec 18 '16 at 17:19