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Some time ago, I asked for a definition of thermodynamic reversibility without mentioning entropy, and a user came up with the following formal definition of a reversible process:

Definition:

Reversible process means that given the outside controllable mechanical, electrical, magnetic, chemical, etc., macroscopic parameters $\hat x_1,\hat x_1,\hat x_2,...,\hat x_n$ of the surroundings and its temperature $\hat T$ at which heat exchange can also take place any and all internal thermodynamic properties (parameters), say $z$, of the system at any time instant during the process can be written as a function of said external parameters: $z(t) = f(\hat T(t), \hat x_1(t),\hat x_2(t),...,\hat x_n(t))$. Notice the function depends only on the instantaneous values and not on the time rates of the external parameters. The $t$ in the function is just a process index by which the various consecutive stages of the thermodynamic process is marked, i.e., time."

Using this definition, can you come up with an example of irreversible process and explain why it is irreversible, again, using the provided definition or explain why this definition is wrong?

Qmechanic
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armoredchihuahua
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    to confirm my understanding, would heat capacity be an example of "an internal thermodynamic property." Could it be an example of a $z$? – AXensen Mar 30 '23 at 22:36
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    I think that technically yes. As i understood, $z$ can be any thermodynamic property be it a state variable (like pressure, volume, temperature, etc.) and functions of those variables. – armoredchihuahua Mar 30 '23 at 22:41
  • @AndrewChristensen, yes $z$ could be any thermodynamic property of interest and that includes heat capacity. – hyportnex Mar 31 '23 at 11:19

3 Answers3

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The most obvious example is magnetic hysteresis. The current value of magnetization $M=M(H_{ext})$ does not just depend on the current bias field but it also depends on the history of how the sample has been biased. What makes this example so interesting is that actual bias history can be quasi-static, the time rate of bias can be arbitrarily low, and it is still irreversible. This is very much unlike the usual example, say a quick piston movement against a gas in a cylinder, that creates all kinds of internal disequilibrium, and the time-dependent relaxation is irreversible.

hyportnex
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    And we met again. Would it be too much to ask for a more elementary example, maybe involving gases and friction? – armoredchihuahua Mar 30 '23 at 22:46
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    Read the last sentence again, it is about a gas and a fast moving piston. The position of the piston does not determine the pressure of the gas unless it is moved very slowly in or out where slowly is measured relative to the internal relaxation rate. Another famous example would be stress-strain hysteresis. – hyportnex Mar 30 '23 at 22:51
  • Ok, but consider a slow moving piston, but with kinetic friction this time, the process would be quasistatic and it would not have hysteresis correct? But it would still be irreversible, why is that? – armoredchihuahua Mar 30 '23 at 22:56
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    It depends on what you count as your "system". If the friction between the piston and the cylinder wall is counted as part of your system then yes, the wall dissipation contributes to the irreversibility of your system, But if you model your gas as $p(t), V(t), T(t)$ then for the gas it does not matter how sticky the piston is as long as it moves slowly so that at any instant the gas parameters $p,V,T$ are well defined and uniform. – hyportnex Mar 30 '23 at 23:03
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    @armoredchihuahua you have an example with the piston here (see the figure at the end of the answer) https://physics.stackexchange.com/a/747464/226902 . In the case of the piston you can have a rate-dependent hysteresis (the slower the transformation the smaller is the hysteresis effect). – Quillo Mar 31 '23 at 08:38
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I think there are some cases that are not covered by this definition of reversibility and are, therefore, irreversible. The way I interpret this definition is that the state is a unique function of the control parameters. One way to violate it is by finding a state that is not uniquely determined by its parameters (one example is already given by @hyportnex), but the other way is by finding a situation where the external parameters cannot be controlled (or different at the beginning and the end of the process.)

For example:

  • Gas expansion into vacuum If we were able to put the gas back into the container that it came from, so that its volume is the same as it initially was, we conceivably could reverse the process - that is, it falls under the definition... but we cannot reverse it.
  • Irreversible chemical reactions - e.g., one can have a mixture of oxygen and hydrogen in a container, which exists for a very long time as is. It may spontaneously explode and transform into water (provided that the container is strong enough to withstand the explosion - a situation often discussed in the context of combustion.) The external parameters remain the same, but the situation has drastically changed.

Related:
Difference between Reversible and Irreversible processes in Physics vs. Chemistry
Can we call rusting of iron a combustion reaction?

hyportnex
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Roger V.
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  • But, the word "controllable" is included in the definition: "given the outside controllable [...] macroscopic parameters". Doesn't that cover both scenarios? – Filip Milovanović Mar 31 '23 at 21:08
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    According to this definition a reversible process is one in which the internal thermodynamic parameters are functions of the external control parameters at all time instants. In both irreversible examples you have given some internal parameters are explicitly uncontrolled, and thus are not covered by the above definition of reversibility. – hyportnex Mar 31 '23 at 21:52
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    @hyportnex the way I understand the question it is to give examples of processes that are irreversible according to the definition *or* demonstrate that the definition does not always apply. Is control of the parameters the necessary condition for reversibility? – Roger V. Apr 01 '23 at 02:00
  • @FilipMilovanović indeed, they violate a different clause of the definition. I tried to communicate this idea in the first paragraph of the answer, but perhaps not clearly enough. I am open to suggestions for more precise wording. – Roger V. Apr 01 '23 at 03:34
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    the first sentence wording was just ambiguous. – hyportnex Apr 01 '23 at 11:39
  • Ah, I see. I interpreted the original wording to mean that there are cases of reversibility not covered by the definition (i.e. the definition is too narrow or incomplete). – Filip Milovanović Apr 02 '23 at 13:59
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If I'm reading the definition right, I'm not sure there is any "reversible process" according to it. The problem is this part of the definition:

"the function depends only on the instantaneous values and not on the time rates of the external parameters"

In practice, physical systems always have some amount of inertia, such that sufficiently rapid changes in their external environment are only reflected in their internal state after some delay. Indeed, if we had a parameter to which a system reacted truly instantaneously, with no delay at all, it's hard to see how such a parameter could be described as "external" to the system.

But you were asking for an example of an "irreversible process" according to the definition you quoted. As I argue above, I think almost any process should qualify as irreversible according to it, but the obvious example that comes to mind reading it would be heat exchange.

Specifically, let our system consist of a 3-dimensional block of solid material in a heat bath at temperature $\hat T(t)$, and let the one of the internal thermodynamic properties $z(t)$ be the temperature at the center of the block, which I'll call $T_c(t)$.

At thermal equilibrium, of course, $T_c(t) = \hat T(t)$, at least assuming that no heat is generated within the block. But if we change $\hat T(t)$ sufficiently fast (or, technically, at all!), then $T_c(t)$ will lag behind $\hat T(t)$, since it will take time for heat to diffuse in or out of the block and for a new thermal equilibrium to be reached.

Ilmari Karonen
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  • Since armoredchihuahua is referring to an answer I gave to his question I would like to comment here that this definition is no more realistic for any real process than the usual definition of reversibility. In fact, it is the same in physical meaning, it only adds a practical formulation by default. That is, if you characterize a system by $p(t)V(t)=RT(t)$ and $dU(t)=C_v(T(t))dT(t)$ for all time instants $t$ then you already have restricted that to a reversible transformation. – hyportnex Mar 31 '23 at 11:27
  • "if we had a parameter to which a system reacted truly instantaneously" - I think hyportnex is not saying that the physical system itself, when looked at in all the gory detail, reacts truly instantaneously, but that it can be, given certain assumptions, modeled (described by) a function that treats it as such with respect to these parameters. – Filip Milovanović Mar 31 '23 at 20:41
  • Indeed, no real process is truly reversible for the reasons cited in the answer. – Roger V. Apr 01 '23 at 03:36
  • $FilipMilovanovic Exactly. I was just trying to formulate mathematically what the usual verbiage of defining reversibility might mean in a thermodynamic equation, such as, infinitely slow, quasistatic, a process that is always in equilibrium, infinitesimal steps that can spontaneously go either way globally, etc. Maybe the vague verbiage is more general but I have not yet seen such an example of it. – hyportnex Apr 01 '23 at 11:53