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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 ?

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    $\begingroup$ 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. $\endgroup$
    – hyportnex
    Commented Dec 8, 2016 at 15:27
  • $\begingroup$ 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? $\endgroup$
    – Ruben Verresen
    Commented Dec 8, 2016 at 21:24
  • $\begingroup$ @RubenVerresen The examples I discussed are precisely not quasistatic. $\endgroup$
    – Dimitri
    Commented Dec 13, 2016 at 10:10
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    $\begingroup$ @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. $\endgroup$
    – Dimitri
    Commented Dec 13, 2016 at 10:12
  • $\begingroup$ 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. $\endgroup$
    – G. Bergeron
    Commented Dec 20, 2016 at 11:22

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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.

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    $\begingroup$ 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. $\endgroup$
    – Dimitri
    Commented Dec 13, 2016 at 10:17
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    $\begingroup$ @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, ... $\endgroup$
    – march
    Commented Dec 14, 2016 at 16:56
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    $\begingroup$ 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. $\endgroup$
    – march
    Commented Dec 14, 2016 at 16:56
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    $\begingroup$ 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 $\endgroup$
    – joshphysics
    Commented Jul 1, 2017 at 16:23
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    $\begingroup$ (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?" $\endgroup$
    – joshphysics
    Commented Jul 1, 2017 at 16:24
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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.

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Squeezing toothpaste out of a tube.

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Reversible: The change is slow enough so that the system is in a sequence of equilibrium states, in a sense strong enough to ensure entropy never increases at any time because no equilibrium recovery is needed. (I recommend reading a good book of statistical physics to understand why)

Quasistatic: The change is slow enough so that the system is in a sequence of equilibrium states, in the weak sense that the macro-variables are mostly defined (temperature, pressure… are uniform almost everywhere in the system) and that formulas about equilibrium states can be used along the path (such as $dU=TdS-PdV$, $PV=Nk_B T$... ). In this case, equilibrium “catches up” with macroscopic changes but there is still an underlying entropy creation because of equilibrium recovery.

In some cases, you may describe a progressive process as a continuous path in the space of equilibrium states, but it may not necessarily be a true succession of equilibrium states. It is slow enough to draw a continuous path, but it hides a fine-grained disequilibrium. An example is vibrating a piston at supersonic speed to progressively increase the temperature of a gas. This will look like a continuous path of equilibrium states $(V,T)$ but it hides a fine-grained disequilibrium. Another good example is a succession of small free expansions. These transformation are quasistatic but they are not reversible.

Elementary thermodynamics uses an ideal gas with a piston as an example most of the time, examples are a bit "extreme" or "artificial" because equilibrium happens very fast and using a piston is a rather smooth action. You need violent actions to drive the system out of equilibrium. When friction or viscosity is involved, even an (apparently) very slow action is irreversible. Heat exchange at finite temperature difference is another good example of irreversible quasistatic process but possibly confusing because the requirement of being quasistatic "the system is at equilibrium at all stages" is arguably not met.

Both definitions start by “the change is slow enough so that the system is in a sequence of equilibrium states” and this is very confusing. In the literature (most often), “quasistatic” means what I just defined except (sometimes) when the text focuses on explaining what reversibility/irreversibility is and here, writers may use the word quasistatic to mean slow, which in this context means reversible.

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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Dimitri
    Commented Dec 13, 2016 at 10:14
  • $\begingroup$ @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). $\endgroup$
    – march
    Commented Dec 14, 2016 at 17:04
  • $\begingroup$ I was meant that reversible process has zero entropy change. I don't know why it is not true. $\endgroup$
    – user115350
    Commented Dec 14, 2016 at 17:31
  • $\begingroup$ 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. $\endgroup$
    – user115350
    Commented Dec 18, 2016 at 17:19

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