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I have heard that all reversible processes are quasi-static but I have not yet found a strong reason for the fact that no non-quasistatic process is reversible.

I have seen on some other answers that it has to do with this $dS=dq/T$ formula for entropy not being valid for non-quasistatic processes but I don't get that completely if that's true

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    $\begingroup$ Explain how you know all reversible processes are quasistatic $\endgroup$
    – Bob D
    May 21, 2021 at 9:39
  • $\begingroup$ thanks for this comment it actually made me rethink it. Frankly I am pretty new to thermodynamics being a high school student and I don't get the correlation between reversibility and the process being quasistatic or not. But till now from what I have searched on this topic I got to know that reversible processes are quasistatic but again I cant find a strong reason for that. $\endgroup$ May 21, 2021 at 9:56
  • $\begingroup$ OK, I'll post an answer to try and explain. $\endgroup$
    – Bob D
    May 21, 2021 at 10:37

2 Answers 2

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I have heard that all reversible processes are quasi-static

That is correct. It's because a reversible process is one where the system is in equilibrium with its surroundings all during the process. That means the process has to be carried out very slowly, which is what quasi-static means. In reality process can only approach being reversible. Thats because natural processes occur as a consequence of disequilibrium.

Consider a heat transfer process. Heat is energy transfer due to temperature difference. You probably already know that spontaneous heat transfer always occurs from high temperature to low temperature. Never in the reverse direction. That makes all real heat transfer processes irreversible. In order to move heat from cold to hot you need to do work. That's how heat pumps and air conditioners work.

But we can make a heat transfer process approach being reversible by making the temperature difference infinitesimal (i.e. approach zero). The smaller the temperature difference, the slower the process proceeds (becomes quasi-static) approaching a reversible process. To put it another way, if the temperature difference is tiny, then at each step in the process it only takes a tiny amount of work to reverse it.

but I have not yet found a strong reason for the fact that no non-quasistatic process is reversible.

For the same reason a process has to be quasi-static to be reversible, a non quasi-static process can not be reversible. In the heat transfer example above, the greater the temperature difference the greater the rate of heat transfer and the more irreversible it becomes (the more work required by the surroundings to reverse the heat transfer).

I have seen on some other answers that it has to do with this $dS=dq/T$ formula for entropy not being valid for non-quasistatic processes

That is not correct. To understand why you need to learn that entropy is a state function. In other words, the difference in entropy between two equilibrium states does not depend on the process connecting the states. The formula you gave is the definition of a differential change in entropy. But it should read

$$dS=\frac{\delta q_{rev}}{T}$$

where $\delta q_{rev}$ means a reversible transfer of heat.

Even if the actual process is irreversible, you will learn that you can assume any convenient reversible process connecting the states and calculate the difference in entropy using equation defining entropy.

One word of caution. Although all reversible process are quasi-static, not all quasi-static processes are reversible. An example is a quasi-static process involving mechanical friction. All processes involving friction are irreversible.

Hope this helps.

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  • $\begingroup$ Thanks a lot for the answer. And really brilliant explanation man I was stuck on this bit for a long time without realizing how intuitive it was $\endgroup$ May 21, 2021 at 15:23
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    $\begingroup$ @KalpeshBhatnagar Your'e welcome, glad it helped. $\endgroup$
    – Bob D
    May 21, 2021 at 21:49
  • $\begingroup$ @BobD I have one confusion, You said that if temperature difference is large then heat transfers from hot body to cold body spontaneously and we can't reverse it by doing infinitesimal changes , is this the reason that it is irreversible as it cannot be reversed by inducing some infinitesimal small changes $\endgroup$ Jun 24, 2021 at 16:36
  • $\begingroup$ @lalittolani You can reverse the process by doing infinitesimal changes to bring the system to its original state but the surroundings will not return to its original state because the initial process was irreversible generating entropy. A reversible process is one in which both the system and surroundings are returned to their original state. $\endgroup$
    – Bob D
    Jun 25, 2021 at 1:30
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In general, none of the statements "reversible${}\Rightarrow{}$quasi-static" or "quasi-static${}\Rightarrow{}$reversible" is true.

• A counterexample to the second implication are systems with internal state variables, which cannot be made non-dissipative, no matter how slowed-down they are. See the discussion and mathematical analysis in Astarita § 2.5.

• A counterexample to the first implication is a system of spins in a crystal lattice. It is possible to reversibly bring the system form an equilibrium state to another with opposite temperature by reversing the external magnetic field as fast as possible – and therefore not through a quasi-static process. In fact it is key here that the process be not-quasi-static, but as fast as possible, because a slow change of the external magnetic field would lead to an irreversible process with dissipation. For more details see the discussion in Buchdahl, Lecture 20.

The point is that for some systems a fast change can actually prevent the onset of dissipative phenomena, and so the process needs to be fast if we want it to be reversible. Adiabatic processes often also need to be fast (as a curious historical fact, Truesdell & Bharatha, Preface p. xii, remark that "In introducing what we today call an 'adiabatic process', Laplace called it 'a sudden compression', in which he was followed by Carnot).

In fact, clearly non-quasi-static phenomena such as explosions can in some circumstances be described by reversible processes! This is possible if the explosion involves many shock waves, as explained by Oppenheim, chap. 1 p. 63:

If there is more than one shock, the losses in available energy are diminished, so that in the limit, with an infinite number of shocks, they become negligible, and the process acquires the character of a thermodynamically optimal, i.e., reversible, change of state. The study of explosion processes reveals that, indeed, they are associated not with one but with a multitude of shocks.

For explosions see also the mathematical analysis by Dunwoody: Explosion and implosion in a mixture of chemically reacting ideal gases, where again reversible-process equations are used.

A caveat about reversible and quasi-static associations is given by Ericksen (§ 1.2):

Some tend to associate nearly reversible processes with those taking place very slowly – the "quasi-static" processes. This probably stems, at least in part, from experience with classical theories of heat conduction, viscosity, and so on. However, a ball made of silly putty behaves almost reversibly when bounced rapidly and various other high polymers have similar predilections. So, it seems prudent to be open-minded in considering what may be reversible processes for particular systems.

He later discusses (§ 3.1) the case of bars subjected to dead loads, for which we can have reversible processes under sudden jumps in elongation. He concludes (p. 46) that "the sudden jump provides an example of a process that is reversible but not reasonably considered to be quasi-static".


• But there's an important question that underlies our discussion: what do we actually mean by "quasi-static"? We need to specify a time scale, otherwise the term is undefined. For example, a geological process (say, tectonic motion) can be considered as quasi-static – or even completely static – on time scales of minutes or days; but it is not quasi-static on time scales of millions of years.

Whether a process is reversible or not, within any tolerance needed, is an experimental question. We can measure any relevant quantities, say pressure $p$ and exchanged heat $q$, under the process, and compare them with those, $p^*$ and $q^*$, determined by the equations for a reversible process. We may find for example that at all times $$\biggl\lvert\frac{p - p^*}{p^*}\biggr\rvert < 0.001 \ , \quad \biggl\lvert\frac{q - q^*}{q^*}\biggr\rvert < 0.001 $$ and conclude that the process is reversible, if relative discrepancies of $0.1\%$ or less are negligible in our concrete application.

But suppose that someone tells us "if you want the process to be reversible, you must make sure that it is quasi-static". Alright, but how much is "quasi-static"? is it OK if the piston moves with a speed of 1 cm/s? or is that too much? How about 1 mm/s? – In fact we may find that for some kind of fluid 1 cm/s is absolutely acceptable for the process to be reversible, whereas for another kind of fluid that speed would lead (at the same temperature) to an irreversible process.

You see how this imprecise situation can lead to circular definitions: "if the process is irreversible, then it means it isn't quasi-static" – but then we are actually defining "quasi-static" in terms of "reversible"! Any statement of the kind "reversible${}\Rightarrow{}$quasi-static" or "quasi-static${}\Rightarrow{}$reversible" then becomes not a matter of experimental verification, but of pure semantics. At this point we can simply get rid of "quasi-static" terminology since it doesn't bring any new physics to the table. This circularity is admitted for example by Callen in discussing irreversible gas expansion (Problem 4.2-3 p. 99):

The fact that $dS > 0$ whereas $dQ = 0$ is inconsistent with the presumptive applicability of the relation $dQ = T\,dS$ to all quasi-static processes. We define (by somewhat circular logic!) the continuous free expansion process as being "essentially irreversible" and non-quasi-static.

A similar criticism can be read in Astarita, § 2.9, p. 62, where he also provides a mathematical quantification of quasi-static, similar to the one given above for reversibility:

Often this point is circumvented by bringing in another difficult concept, that of a quasi-static transformation, which proceeds "through a sequence of equilibrium states." Quasi-static is an impressive word, but the only meaning which can be attached to it is the less impressive word "slow" – and how can one speak of slowness without implying the concept of time? How slow is slow enough? If one chooses to develop a thermodynamic theory (rather than a thermostatic one), the answer is easy. For instance, in the case of a system where the state is $V, T, \dot{V}$ [the latter is the rate of change of $V$], one needs to assume that [the non-equilibrium pressure] $p(V,T,\dot{V})$ is a Taylor-series expandable at $\dot{V} = 0$ to obtain [that $$ p = p^* + \frac{\partial p}{\partial \dot{V}}\biggl\lvert_{\dot{V}=0}\dot{V} + \mathrm{O}(\dot{V}^2) \ , $$ where $p^* = p(V,T,0)$ is the pressure at equilibrium]. One then reaches the conclusion that if the condition $$\dot{V} \ll \frac{p^*}{\partial p/\partial \dot{V}\lvert_{\dot{V}=0}} $$ is satisfied, then indeed the difference between $p$ and $p^*$ is negligibly small as compared to $p^*$, and thus the process can be regarded as a quasi-static one.

Criticisms against the fuzzy notion of "quasi-static" have appeared in many other works. Truesdell & Bharatha (Preface p. xii), make the historical remark that "the 'quasi-static process' was barely mentioned for the first time in 1853 and was altogether foreign to the early work [in thermodynamics]". See also the mathematical analysis by Serrin: On the elementary thermodynamics of quasi-static systems and other remarks.


• I also want to point out that "quasi-static" in some works has specific meanings somewhat unrelated to the discussion above. For example that the rate of increase of the total kinetic energy $K$ of the system is negligible, so that the law of energy balance, which in its full generality is $$ \frac{\mathrm{d}(U+K)}{\mathrm{d}t} = Q + W$$ (that is, the rate of increase of internal energy $U$ and kinetic energy is equal to the heat rate $Q$ and work rate $W$ provided to the system) can be approximated by $$ \frac{\mathrm{d}U}{\mathrm{d}t} = Q + W \ .$$ Or that similar inertial terms in the motion of the system are negligible. See for example the book by Day, chap. 2.

But note that such definitions of "quasi-static" have, again, no a-priori relation with reversibility.


• Finally, the equation $\mathrm{d}S = Q/T$ is only valid for a process that is:

  • reversible (by definition),
  • closed (no exchange of mass),
  • with a homogeneous surface temperature,
  • without bulk heating (such as instead happens in a microwave oven).

Under the last three conditions we have in general that $\mathrm{d}S \ge Q/T$; when the equality sign is satisfied, then the process is defined as reversible. See Astarita, § 1.5, or Müller & Müller, for the different forms of the second law under different circumstances. This equation may be valid in quasi-static and non-quasi-static processes, as explained above.


References

I recommend you do your own reading and eventually reach your own conclusions about your question.

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  • $\begingroup$ While you are technically correct in your first statement, as you go on to explain, it is worth saying that in conventional thermodynamics (which was always based around classical one-phase systems as I understand), the implication "reversible ⇒ quasi-static" is indeed true. $\endgroup$
    – Noldorin
    Nov 20, 2021 at 18:03

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