1
$\begingroup$

I have read that a reversible process is one in which $\Delta S_{\text{universe}}=\Delta S_{\text{system}}+\Delta S_{\text{surroundings}}=0$ and an irreversible process being one in which $\Delta S_{\text{universe}}>0$

However, what actually makes a process either reversible or irreversible? How does the change in entropy of the system bring about an opposite change in entropy of the surroundings in the reversible case? What is an intuitive explanation for reversibility?

$\endgroup$

2 Answers 2

2
$\begingroup$

However, what actually makes a process either reversible or irreversible?

A reversible process is one which can be taken from its initial state to another state, and then back again without any change to either the system or surroundings. A process that is not reversible is irreversible.

In reality, there are no reversible processes. All real processes are irreversible due to various effects that generate entropy including, but not limited to, processes involving various forms of friction, heat transfer from a hot object to a cold object (energy transfer through a finite temperature difference), unrestrained expansion of fluids, the mixing of different substances, the combustion of substances, etc.

One would not expect any of the above processes to naturally reverse, or "undo" themselves. Some form of intervention is needed which would alter the state of the source of that intervention, preventing either the system or surroundings from returning to their original state. Consequently, each generates entropy in either the system or surroundings.

That said, it is possible to approach, in theory, a reversible process as discussed below for energy transfer by heat.

How does the change in entropy of the system bring about an opposite change in entropy of the surroundings in the reversible case?

What you are referring to is entropy transfer, which, for a closed system, can only occur by means of reversible heat transfer which itself is an idealization, as discussed below. (Note that work cannot transfer entropy. Irreversible work generates entropy)

To illustrate, let the system be a hot thermal reservoir of constant temperature $T_{H}$ and the surroundings be a cold thermal reservoir at constant temperature $T_{C}$, where $T_{H}\gt T_C$. Let heat $Q$ flows from the system to the surroundings. From the definition of entropy, we have

$$\Delta S_{sys} =-\frac{Q}{T_{H}}$$

$$\Delta S_{surr} =+\frac{Q}{T_{L}}$$

$$\Delta S_{TOT}=\Delta S_{sys}+\Delta S_{surr}=-\frac{Q}{T_{H}}+\frac{Q}{T_{L}}$$

Since $T_{H}\gt T_{C}$ we have $\Delta S_{TOT}\gt 0$ and the process is irreversible.

In order for the process to be reversible, i.e., for $\Delta S_{TOT}=0$ it is necessary for $T_{H}=T_{C}$. However, if $T_{H}=T_{C}$ there can be no energy transfer by heat since, by definition, heat requires a temperature difference. This illustrates the fact that all real heat transfer processes are irreversible and can only approach being reversible as $\Delta T$ approaches zero. That is, when $\Delta T$ becomes infinitesimal.

What is an intuitive explanation for reversibility?

Keeping in mind that reversibility is an idealization and not a reality, in order for a process to approach being reversible it is necessary to eliminate or minimize, a far as possible and practicable, all potential sources of entropy. That necessitates (1) carrying out processes infinitesimally slowly (quasi-statically) so that the system and surroundings are always in approximate equilibrium with one another (since disequilibrium generates entropy) and (2) eliminating sources of friction.

Hope this helps.

$\endgroup$
6
  • $\begingroup$ @Chemomechanics. Yes, i was only thinking about a closed system. See edit. Thanks. $\endgroup$
    – Bob D
    Sep 6, 2023 at 17:33
  • $\begingroup$ Thank you for your response. Is it therefore fair to say that a reversible process is one that is able to return to its initial states without changing the total entropy? You mentioned that, in reality, nothing is reversible, but we approach reversibility by conducting processes infinitesimally slowly. Is this why, in a "reversible" heat engine, the isothermal expansion and compression must be conducted so slowly that the change in temperature in the engine is always approximately equal to the heat reservoir? As such, the heat flowed into the engine during expansion will equal that flowed out? $\endgroup$ Sep 6, 2023 at 20:07
  • $\begingroup$ Not only is it fair to say that, it is the requirement for the total change in entropy to be zero. $\endgroup$
    – Bob D
    Sep 6, 2023 at 20:10
  • $\begingroup$ Could I say that, in the reversible case, since the net entropy change of the universe is 0, the universe is neutral as to in which direction the process proceeds. Thus, a process can go in both the forward and reverse direction. However, in the irreversible process, the universe gains energy, so if the process were to reverse by itself, the universe will now have to lose entropy, which is not possible according to the second law. $\endgroup$ Sep 7, 2023 at 5:40
  • $\begingroup$ @Cold_Spaghetti “However, in the irreversible process, the universe gains energy, so if the process were to reverse by itself, the universe will now have to lose entropy”. First of all I assume you meant “gains entropy”, and not “gains energy”. Secondly, for the irreversible process if the process is reversed to return the system to its original state, the surroundings transfers entropy to the system but not all the entropy that was generated in the system. That generated entropy remains in the surroundings so that the total entropy of the universe is greater than zero. $\endgroup$
    – Bob D
    Sep 7, 2023 at 12:24
0
$\begingroup$

You should not think of thermal interaction as "heat" or "work" exchange. Instead think of thermal interaction as exchanging entropy at some temperature. This is in complete analogy with chemical interaction that is exchanging a chemical species at some chemical potential, or electric charge at some electric potential, or volume at some pressure, or surface area at some surface tension. In a thermal interaction the quantity transported is entropy. In other words, it is the agent of the interaction in the same sense that electric charge is the agent of electric interaction, etc. We may call these agents of interaction energetic quantities.

In your question regarding the thermodynamic "universe" the entropy is exchanged between the "system" and its "environment". In an electric interaction the electric charge is absolutely conserved, whatever moves form here the same amount goes there, it cannot be created and it cannot be destroyed. The same with the total mass, the same with total volume, these are all conserved quantities. But not all energetic quantities are conserved. It is not the same with surface area because area is not conserved, it can shrink and it can increase, similarly the quantity of electric dipoles is not conserved quantities, dipoles can be created and destroyed. But unlike these non-conserved examples of energetic quantities, while entropy is also not conserved it cannot be destroyed, it may stay the same or it maybe created.

By definition, a process in which the total entropy stays the same is called reversible, the one in which increases is called irreversible. Now usually, the concepts of reversibility and irreversibility are introduced differently but you are already asking about their relationship to entropy, so I thought I would give the definition in the reversed order from the usual that starts with reversibility/irreversibility and then defines entropy from heat concepts.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.