# Given an irreversible process from A to B, is there a reversible process of the same type from A to B?

I'm an undergraduate student, and I'm facing problems with entropy for the first time.

The word "adiathermal" is used here, as suggested in the comments, in order to describe a process occurring without any flow of heat and matter. "Adiabatic" was the word used at first. (Use the word you prefer, the meaning is what matters)

I know that if I consider an irreversible adiathermal process from A to B, the change in entropy $\Delta S_{AB}$ is more than $0$, and that if I consider a reversible adiathermal process starting from A, its path will not pass through B (and its difference of entropy will be $\Delta S_{AP} = 0$, where $P$ is its ending point).

My question is: when I have an irreversible $x$ process (where $x =$ isothermal, isobaric, etc...) from A to B, is there any reversible $x$ process going from state A to state B? I.e. Irreversible isothermal process from A to B, is there a reversible isothermal process from A to B?

As explained before, I have found a counterexample for $x =$ adiathermal, so if I want to evaluate $\Delta S_{AB}$ I need to find another reversible path from A to B (i.e. rev. isobaric + rev. isothermal).

I think such $x$ process does not exist, otherwise we would just use Clausius integral ($\int_{A}^{B} \frac{\delta Q}{T}$) without noticing any difference between a reversible and irreversible process. Furthermore, I think there should be some deeper concept that I'm missing here.

• @diracula thanks, I'll correct the question. The terminology aspect is due to the fact my book and my lessons are not in English, so my translation was just wrong! – moonknight Feb 4 '18 at 11:43
• @diracula Never heard of this distinction. What is your source? – valerio Feb 4 '18 at 12:05
• @valerio92 I did not mean to suggest this is the common usage, nor that the question should be altered, only that it is something the asker may want to look out for when reading. It occurs in for example Blundell and Blundell's Concepts in Thermal Physics. I'll delete my comment as it is causing confusion. – diracula Feb 4 '18 at 12:47
• No problem, any suggestion is accepted. I have edited the answer only to make my question more understandable. – moonknight Feb 4 '18 at 13:02

## 2 Answers

Disclaimer: I don't agree with diracula's remark that a generic process with no heat exchange should be called "adiathermal", and only a reversible "adiathermal" process is called adiabatic. To me, adiabatic = no heat exchange, like in all thermodynamic texts that I know of. Therefore, I will just use the term "adiabatic".

My question is: when I have an irreversible $x$ process (where $x$= isothermal, isobaric, etc...) from A to B, is there any reversible $x$ process going from state A to state B?

The answer is yes, and it is fairly easy to see this.

Let $x$ be the thermodynamic variable that you want to keep constant. Now let's work in the $xy$ plane, where $y$ is any other thermodynamic variable describing the system.

If the state points $A$ and $B$ are connected by an irreversible iso-$x$ proces, then $x(A)=x(B)$: in other words, they lay on an horizontal line in the $xy$ plane.

Is there a reversible iso-$x$ process connecting $A$ and $B$? Sure. Just draw a continuous segment connecting $A$ and $B$. This segment is horizontal in the $xy$ plane, i.e. is an iso-$x$ process. That is the reversible process you are looking for.

I know that if I consider an irreversible adiabatic process from A to B, the change in entropy $\Delta S_{AB}$ is more than $0$ and that if I consider a reversible adiabatic process starting from A, its path will not pass through B (and its difference of entropy will be $\Delta S_{AP} = 0$, where $P$ is its ending point)

The case of an adiabatic process is different: "adiabatic" means that $\delta Q=0$, but doesn't correspond in general to any iso-$x$ process, unless we restrict ourselves to reversible processes: in this case, $\delta Q = T dS$ and therefore adiabatic=iso-$S$.

Your analysis for an adiabatic process is right. For an irreversible adiabatic process between A and B, as a consequence of Clausius' inequality we have

$$\Delta S_{AB} > \int_A^B \frac{\delta Q} T = 0 \longrightarrow \Delta S_{AB} > 0$$

while for a reversible process we would have

$$\Delta S_{AB} = \int_A^B \frac{\delta Q} T = 0 \longrightarrow \Delta S_{AB} = 0$$

Since entropy is a state function, $\Delta S_{AB}$ is independent on the path connecting A and B. Therefore, this means that we cannot find both an irreversible and a reversible adiabatic process connecting the same points A and B.

• Thanks for the answer. About the second quote: I meant $\Delta S_{AP}$, I have corrected my question. I am not confusing the entropy change of the system with the entropy change of sys+surroundings, I only stated that entropy change of the system is $0$ for every reversible adiabatic process, whereas given an irreversible adiabatic process from A to B, there is no rev. adiabatic process going from A to B so I have to use a combination of different rev. processes in order to go from A to B (and then I can evaluate $\Delta S_{AB}$ because entropy is a state function as you said). – moonknight Feb 4 '18 at 12:35
• about the first quote: so the answer is "yes" for iso-x processes and "no" for adiabatic processes? – moonknight Feb 4 '18 at 13:04
• @moonknight Yes, this is correct. I updated the answer. – valerio Feb 4 '18 at 13:22

I have a different perspective than @valerio92. My perspective is that it depends on (a) how you define an irreversible isobaric- or isothermal process and (b) whether you are including just the system, or the system plus surroundings. Let's assume that the answer to (b) is that you are including just the system. Now, let's consider (a).

If you define an irreversible isobaric process as one in which the initial pressure is $p_0$, and the external force per unit area applied to the system is also $p_0$ for the entire process path, while you vary the temperature at the system boundary in some arbitrary way T(t) between $T_0$ and $T_1$, then the entropy change for this path will be the same as for a reversible path between the same two end (equilibrium) states. However, during the irreversible path, there will be entropy generated within the system which is transferred through the boundary to the surroundings during the process, over and above the entropy that is exchanged with the surroundings during the reversible path.

A second definition of an irreversible isobaric process is one in which the initial pressure of the system is $p_0$, but, at time zero, you suddenly change the external force per unit area applied to the system to $p_1$, and hold it constant at this value for all subsequent times; in this scenario, we also assume that the temperature of the boundary is held at $T_0$ for both the reversible and irreversible cases. In this case, the reversible process between the two end states would, of course, be quite different, but the entropy change would be the same. Here again, entropy would be generated within the system for the irreversible change, and this entropy would be transferred through the boundary to the surroundings during the process, over and above the entropy that is exchanged with the surroundings during the reversible path.

The same types of considerations could be applied to what we call isothermal processes. So, the real question is, what do you define as an isothermal or isobaric process?