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The standard proof of $\Delta S_{u}\geq0$ is the following:

If I perform first a reversible process and then an irreversible process the result is opposite: $\Delta S_{u}\leq0$

What am I doing wrong?

I doubt that when I perform my calculation I'm not allowed to invert the extreme of integration (clearly adding before also a minus) of the irreversible process, as I did...

$\int_{1}^{2} \frac{dQ}{T} rev + \int_{2}^{1} \frac{dQ}{T}irrev \leq0 $
$\int_{1}^{2} \frac{dQ}{T}rev - \int_{1}^{2} \frac{dQ}{T} irrev\leq0 $
$\int_{1}^{2} \frac{dQ}{T}rev \leq \int_{1}^{2} \frac{dQ}{T} irrev $
$\Delta S_{u}\leq0$

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  • $\begingroup$ Just for clarity: it is understood that we integrate along a line here (not just along a real axis). Check line integral $\endgroup$
    – Roger V.
    Commented Jun 10, 2022 at 12:25
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    $\begingroup$ Just to be clear, the processes shown in your figure cannot be both adiabatic. It is impossible to connect the same two equilibrium states with a reversible and irreversible adiabatic process. $\endgroup$
    – Bob D
    Commented Jun 10, 2022 at 12:28
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    $\begingroup$ What is the source of your diagram and equations? $\endgroup$
    – Bob D
    Commented Jun 10, 2022 at 15:37
  • $\begingroup$ Unless you provide the source for your diagram and equations, I will vote to close. $\endgroup$
    – Bob D
    Commented Jun 10, 2022 at 16:56

3 Answers 3

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You are misunderstanding the derivation of non-decrease of entropy of an "isolated system" when its subsystem undergoes a cyclic process. The wrong assumption here is that the "red" path in the state space is irreversible no matter what due to its position and shape in space of states.

On the contrary, irreversibility is not property of the path in the space of states, but it is property of the real process. The same path can be done both reversibly and irreversibly, in both directions; it is the details of the process which determine which one it is.

Let's start with stating the assumptions. The "isolated system" is made of two parts: the system of interest S which exchanges heat with its environment, and that environment E, and it is assumed that the whole supersystem S+E does not interact with other systems, hence "isolated". Classical thermodynamics says nothing about "universe"; that system is too big and no heat exchange and volume work experiments can be done on it, so we can't replace "isolated system" with "universe" here. The entropy non-decrease theorem is about systems which can be controlled, either isolated systems, or adiabatic systems (can do work but not exchange heat).

If the system S undergoes a cyclic process, it may or may not follow a well defined path in the space of thermodynamic states. If it follows a closed well-defined path in the space of states at all, such as that in your pictures above, and part of the cyclic process is irreversible, this is due to other circumstances outside the system, not encoded in the path itself. For example, such circumstance can be the fact that heat exchange happens over finite temperature difference between the system S and its environment E.

The Clausius inequality for the system S undergoing cyclic process $\gamma$ with reservoir of single temperature $T_r$ states:

$$ \oint_\gamma \frac{dQ}{T_r} \leq 0. $$

It is important to realize system's temperature $T$ may differ from $T_r$.

So, let's calculate change of entropy of the supersystem S+E in one cycle:

$$ \Delta S_{S+E} =\Delta S_S + \Delta S_E. $$ Since the system S returns to the original state, change of its entropy in one cycle is zero: $\Delta S_S = 0$. Thus we are interested in changes of entropy of the environment now.

Now we make another assumption: that changes in $S_E$ obey

$$ dS_E = -\frac{dQ}{T_r} $$ which is reasonable if the environment state changes quasistatically, so that it is in an equilibrium state at all times during the cycle. Integrating over the cycle, we obtain $$ \Delta S_E = \oint_\gamma -\frac{dQ}{T_r}. $$ Applying the Clausius inequality now, we obtain

$$ \Delta S_E \geq 0. $$

So there is the result, the non-decrease of entropy of our special isolated system. If the cyclic process in the subsystem is reversible, then equality applies.

Back to your question, what if we make the super-system retrace the irreversible cyclic process in the opposite direction? In thermodynamics, we believe we can't; it would violate the Clausius inequality and 2nd law of thermodynamics. Later with developments in statistical physics and molecular theory of matter we realized that such thing could be possible, but is extremely improbable to happen with macroscopic systems.

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Frankly, I can't make sense of the equations at the end of your post and how you reached the conclusion that $\Delta S_{universe}\le0$.

Consider the following:

  1. For any combination of processes, in any order, that takes you from equilibrium state 1 to 2 and returns you to the original equilibrium state 1, it is always true that $\Delta S_{system}=0$. That's because entropy is a state property having a unique value for each equilibrium state.

  2. If all processes going from state 1 to state 2 and back again are reversible, then in addition we have $\Delta S_{surroundings}=0$. It then follows that

$$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surroundings}=0$$

This is because, for a reversible cycle, entropy transferred to/from the system equals the entropy transferred from/to the surroundings so the equality applies to the Clausius theorem.

  1. If any process in the cycle between state 1 and 2 is irreversible, entropy is generated in the system. In order to return the system to its original state the system needs to rid itself of the entropy generated. This means more entropy needs to be transferred from the system to the surroundings than from the surroundings to the system, for a net increase in entropy of the surroundings. So we have, for an irreversible ccyle,

$$\Delta S_{system}=0$$ $$\Delta S_{surroundings}\gt 0$$ and $$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surroundings}\gt0$$

This is the crux of the Clausius theorem for a cycle where the inequality applies. It says if the cycle is irreversible, more entropy is transferred from the system to the surroundings than from the surroundings to the system.

As a final comment, with regard to the statement in the diagram "For adiabatic processes.." It is impossible to connect the same two equilibrium states with a reversible and irreversible adiabatic process.

Hope this helps.

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If I invert the processes and perform first a reversible process and then an irreversible process the result is opposite: $\Delta S_u\leq 0$.

This is pretty much the point: if we could reverse the process (invert the process in your language) it wouldn't be irreversible, that is the entropy would not change: $S_1=S_2$. The statement is essentially: for processes that can go only in one direction (called irreversible) the entropy is increasing.

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  • $\begingroup$ Sorry, I was not clear in the question. I edited the question. By invert, I simply meant that we perform first a reversible process and than an irreversible one. $\endgroup$
    – DieMann
    Commented Jun 10, 2022 at 12:16
  • $\begingroup$ @DieMann this means that you go from 1 to 2 by the reversible path and from 2 to 1 by the irreversible one. But in this case the irreversible process is reversed. (If you do not change the direction of the processes, it doesn't matter whether you start at 1 or 2, i.e., which of the two processes is actually performed first.) $\endgroup$
    – Roger V.
    Commented Jun 10, 2022 at 12:21
  • $\begingroup$ Could you be more clear by what you mean when you say: the irreversible process is reversed? I don't reverse any process. I simply try to prove the principle by first performing a reversible process from 1 to 2 and an irreversible process from 2 to 1. If I then apply the same method used in my post, I invert the extreme of integration of the irreversible process and add a minus. Then, doing some basic algebra I obtain dSu <=0. I doubt that the problem is inverting the extreme of integration to the irreversible process is not allowed, but why? $\endgroup$
    – DieMann
    Commented Jun 10, 2022 at 12:31
  • $\begingroup$ But the irreversible process runs from 1 to 2. If you run it from 2 to 1, you are running it in reverse direction. I am not sure about your argument about swapping limits - please add your equations to the question. $\endgroup$
    – Roger V.
    Commented Jun 10, 2022 at 12:34
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    $\begingroup$ @DieMann the inequality holds for the direction of the physical process - mathematically, you can run it in the opposite direction, but then the inequality should be reversed: $\leq \longrightarrow \geq$ $\endgroup$
    – Roger V.
    Commented Jun 10, 2022 at 13:00

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