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I am posting about this because it seems to be a big issue and misconception in the thermodynamic literature. My issue is about adiabatic processes. As I see it there are two intrinsically different definitions of adiabatic processes:

  1. Processes for which $\delta Q_\mathrm{irr}=Td_iS+Td_eS=0$ ($Td_iS$ is the irreversible heat produced and $Td_eS$ the heat due to heat transfer). This means that in these processes there is no heat generation whatsoever. This also means that any adiabatic process is isentropic. Actually, I think this definition is wrong, because every irreversible process will produce entropy $Td_iS$ which cannot be compensated, because the system is thermally isolated ($Td_eS=0$), so that $\delta Q_\mathrm{irr}>0$.
  2. (I think the right definition) Processes for which $\delta Q_\mathrm{rev}=Td_eS=0$. This means that no heat transfer is allowed into the system, but still irreversible processes can generate heat.

The second one should be in principle correct, as an adiabatic, irreversible expansion of a gas can heat it up due to entropy production. To name an example for definition 2, I could name the expansion of the universe which is adiabatic in the sense of no heat transfer (no environment). Still, the entropy is increasing, since $Td_eS\neq 0$ (while it is assumed to be 0 in definition 1).

However, large parts of the literature work with the first definition also.

One example for the use of the first definition is https://chemistry.stackexchange.com/questions/16260/derivation-of-the-relation-between-temperature-and-pressure-for-an-irreversible or https://chemistry.stackexchange.com/questions/38127/reversible-and-irreversible-adiabatic-expansion Here, the authors claim that they derived the expression for the volume change of an irreversible, adiabatic process and starts with the equation (I have seen exactly the same equation in Lecture notes of my Thermodynamics class and other books):

$\mathrm{d}U=-p_\mathrm{ex}\mathrm{d}V$

However, according to the first law:

$\mathrm{d}U=\delta Q_\mathrm{irr}+\delta W_\mathrm{irr}=\delta Q_\mathrm{irr}-p_\mathrm{ex}\mathrm{d}V$

So to reproduce the equation above $\delta Q_\mathrm{irr}=0$ which means that the use of the above equation assumes definition 1, which makes no sense in my opinion.

In my opinion, for any process we should have:

$\mathrm{d}U=\delta Q_\mathrm{rev}+\delta W_\mathrm{rev}=0-pdV=-pdV$

which means that no matter the reversibility, a specific volume change will always induce the same change in the internal energy.

I would appreciate any opinion on this issue.

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  • $\begingroup$ Another example: from a practical standpoint, an internal combustion engine can be considered adiabatic. The burning fuel inside the cylinder does PV work so very quickly that there isn't time for significant heat transfer to the environment. For example, an engine operating at 3600 rpm (a reasonable speed) has it's pistons moving through one complete cycle in 1/60 if a second. The power stroke of a 4 cycle engine occurs only on the downstroke, so all the PV work of that stroke is done in 1/120 of a second (0.00833 seconds). $\endgroup$ Mar 8, 2021 at 17:18

3 Answers 3

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So I think there are two things going on here. The first thing is that, yes, some people use the term adiabatic to mean any process with no heat transfer, whilst others use it to mean specifically a reversible process in which there is no heat transfer. The use of the same term for both concepts is sometimes inconvenient and something to watch out for, but there is not a great deal we can do at this point. It should be obvious when put in these terms, however, that both concepts are well defined and important.

The second issue is what is going on in your definition 1. Now from the Clausius inequality we have that $$ dS \ge \frac{dQ}{T} $$ If we add an extra term to turn it into an equality we get \begin{align} dS &= \frac{dQ}{T} + dS_{irr}\\ dS_{irr} &\ge 0 \end{align} so the extra (positive) term goes on the other side of the equation to what you have in your definition 1. That is in an irreversible process we transfer less heat into the system and do more work on it to compensate. Turning the minus signs around this means that in irreversible processes the system gives out more heat and does less work, i.e. irriversible processes are less efficient.

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  • $\begingroup$ If this equation is right, then $dU=dW+dQ=-p_{ex} dV+T(dS-dS_{irr})$ for an irreversible process? And for an reversible process: $dU=dW+dQ=-pdV+TdS$? Then for an adiabatic process dQ=0 independent if the process is reversible or not, so that for a reversible process: $dU=-pdV$ and for an irreversible process $dU=-p_{ex}dV$? That would explain my confusion I think. $\endgroup$
    – Guiste
    Mar 10, 2021 at 2:07
  • $\begingroup$ So we could define $dQ_{irr}=T(dS-dS_{irr})$ and $dQ_{rev}=TdS$, so it seems that my definition was just switched before. $\endgroup$
    – Guiste
    Mar 10, 2021 at 2:09
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    $\begingroup$ What you have written here is not exactly wrong, but isn't really the right way to think about it either. You are trying to think of the extra entropy generated as a type of heat, but this isn't really true. Irreversible processes generate entropy in addition to what is transferred into the system as heat. The $dQ_{rev}$ you define here is not "the reversible part of the heat". It is "the heat transfer on a reversible path with the same initial and final states". $\endgroup$ Mar 10, 2021 at 11:25
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    $\begingroup$ It is also important to note that $dU = TdS - pdV$ is true for both reversible and irreversible processes. This is because all the quantities involved are functions of state. We can consider a reversible process with the same initial and final states as the process we are interested in. The above equation holds on this reversible path, but since all the quanities involved depend only on the initial and final states it must also hold on the path we are interested in. $\endgroup$ Mar 10, 2021 at 11:28
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    $\begingroup$ What this tells you is that, in your notation, $dQ_{irr} = - dW_{irr}$. The irreversible process generates entropy, so for a reversible process to have the same final state more heat has to be transferred in, so to to have the same final energy more work has to be transferred out. $\endgroup$ Mar 10, 2021 at 11:31
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It is much simpler than any of this. The variations of U, P, V, and S are irrelevant. An adiabatic process is one in which the heat transferred between the surroundings is zero at every instant during the process. This applies irrespective of whether the process is reversible or irreversible.

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  • $\begingroup$ Thanks, I tried to combine my understanding and put it into the answer of By Symmetry, I would appreciate if you could check it also. $\endgroup$
    – Guiste
    Mar 10, 2021 at 2:24
  • $\begingroup$ Well, the first paragraph, I agree with. The 2nd paragraph, I don't fully follow. But I will say that, in the equation for irreversible entropy change, the T should be the temperature at the boundary TB through which the heat dQ is flowing (not the system temperature, whatever that would be for an irreversible process); and the dSirrev would usually be referred to as the entropy generated within the system in the irreversible process. $\endgroup$ Mar 10, 2021 at 3:01
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I would like to point out a connection between the comment by @David White and the answer by @Chet Miller which, in my opinion, are both good and that compliment one another.

As Chet said, an adiabatic process is one in which the heat transferred between the system and surroundings is zero irrespective of whether the process is reversible or irreversible. An adiabatic process that is reversible is one where the system and surroundings are thermally insulated from one another and where the process is carried out very slowly (quasi-statically) without friction.

@David White commented that an adiabatic process can be one that happens "so very quickly that there isn't time for significant heat transfer to the environment". That is an example of a process in which the boundary between the system and the surroundings (environment) is not necessarily thermally insulated but where the process is carried out so quickly there is no time for heat transfer to occur. A process carried out very quickly is not quasi-static and is thus irreversible. That's an example of an adiabatic process that is irreversible per Chet's answer.

Hope this helps.

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  • $\begingroup$ Thanks for the comment, I put a comment to the answer of By Symmetry which refers to this also. $\endgroup$
    – Guiste
    Mar 10, 2021 at 2:24

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