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My chemistry professor recently showed this in a presentation explaining thermodynamics. In particular, he used it as a demonstration that global entropy rises when starting from the assumption that the work produced by an irreversible expansion is always less than the work produced by a reversible one.

I translated this from Italian

What I find dumbfounding is why he would claim that in an irreversible process, the entropy of the surrounding environment, ${\Delta}S_{amb}$, is the same as it would be if the process was reversible.

Wouldn't it instead be equal to $$\frac{q}{T}=\frac{-w}{T},$$ since it's giving $q$ to the system, since that's all that it needs to do its work?

Also, wouldn't the entropy change of the system be equal to $$\frac{0}{T},$$ since it's accepting $q$ heat and losing $w = -q$ of it as work?

The total entropy comes out negative! That can't be right! I noticed the formulas he is using don't take into account volume, which also affects entropy. Is that the mistake?

I have checked on a textbook (Atkins-Jones-Laverman) and they don't take this case into consideration when dealing with the same topic: they talk about the reversible transformation (the first in the image), and the completely irreversible one (I cut it out but it was the third, it was the same as the second but with $q = 0$), while ignoring the case for a generic irreversible transformation.

The textbook, though, provides a formula for isothermal expansions with non-zero work in a different section, and a formula for entropy change of isothermal processes in the same section. I tried to work it out based on those but I couldn't get very far.

Does anyone know if what my professor is saying in these slides makes any sense? If not, what is the correct process to come to the same conclusion (the entropy of the universe increases) from the same starting assumptions (isothermal expansions do less work/consume less heat when irreversible)?

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    $\begingroup$ What he means is that if the irreversible and reversible processes have the same endpoints (ie they start and end at similar initial and final conditions of pressure, temperature, etc) then the entropy of those two processes is the same. This is of course true because entropy is a function of state $\endgroup$ Nov 16, 2023 at 19:14
  • $\begingroup$ I always find that, if I am having trouble understanding generalized concepts like this, it helps to dream up specific problems and solve them to better see how it plays out in simpler cases. Would you be willing to dream up and solve a simple specific problem and maybe even solve it, or would you like me to dream one up and help you solve it? The reasoning is, if you can't apply it to a simple case, you'll never be able to do it for more general complicated cases. $\endgroup$ Nov 16, 2023 at 20:56
  • $\begingroup$ "What I find dumbfounding is why he would claim that in an irreversible process, the entropy of the surrounding environment, ${\Delta}S_{amb}$, is the same as it would be if the process was reversible." Maybe i'm missing something, but where does he claim that? Isn't he claiming that the change in entropy of the environment is less negative for the irreversible process? $\endgroup$
    – Bob D
    Nov 17, 2023 at 15:08
  • $\begingroup$ Also, I can't read the details of the diagrams, but is this example for an ideal gas where the initial an final temperatures are the same? $\endgroup$
    – Bob D
    Nov 17, 2023 at 15:22
  • $\begingroup$ Here is a cookbook Primer with worked examples on how to determine the entropy changes for irreversible processes: physicsforums.com/insights/grandpa-chets-entropy-recipe $\endgroup$ Nov 17, 2023 at 20:45

2 Answers 2

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To determine the change in entropy of a combination of system and surroundings that has suffered a mutual irreversible process between an initial thermodynamic equilibrium state and a final thermodynamic equilibrium state, you first find a new set of surroundings that can apply an alternate reversible process to the original system between the same two end states and evaluate the integral of dQ/T for that alternate process. Then you find a new system that can apply a second alternate reversible process to the original surroundings between its same two end states and evaluate the integral of -dQ/T for that second alternate process. Then you add the entropy changes of the system and surroundings for the two individual alternate reversible processes. This gives you the entropy change for the combination of original system and original surroundings in the irreversible process. So the alternative reversible processes for the system and surroundings don't have to match one another.

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  • $\begingroup$ Thank you so much! The last sentence is what I originally didn't understand (because it wasn't at all explained to us!). It's not intuitive that you have to construct mismatched reversible processes to get the correct result... $\endgroup$
    – trerri
    Nov 18, 2023 at 12:13
  • $\begingroup$ Yes. The literature doesn’t present this well at all. $\endgroup$ Nov 18, 2023 at 12:30
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Does anyone know if what my professor is saying in these slides makes any sense?

The following is an example to explain the slides. See the PV diagrams below.

The red and blue curves are two alternative processes between the initial and final equilibrium states. The red path is for the reversible isothermal process (your professor's bottom slide), and the blue path is an example of an irreversible process between the same two states (your professor's top slide).

Both processes are carried out in the same constant temperature surroundings. The pressure in the graphs is the external pressure. It is also the equilibrium pressure for the gas in the reversible process. For the irreversible process the pressure and temperature of the gas is only the same as the surroundings at the boundary. Within the gas pressure and temperature vary spatially.

The reversible process is carried out very slowly (quasi-statically) such that the gas is always in thermal and mechanical equilibrium with the surroundings with $PV$=constant. Note that heat is added to the system all along the process path.

The irreversible process is carried out as follows:

  1. The external pressure is suddenly (instead of gradually) reduced to equal the final external. pressure of the reversible process. Note that unlike the reversible process, there is no time for heat to transfer to the system during this rapid pressure drop.*

  2. The gas is then allowed to expand rapidly (irreversibly) against constant external pressure, until the gas re-establishes thermal and mechanical equilibrium at state 2.

Observations:

  1. Since for an ideal gas the change in internal energy between two equilibrium states depends only on the change in temperature between the two states, the change in internal energy for both processes is zero. Thus for both processes, per the first law, $q=w$, where $q$ is the energy transferred to the gas by the surroundings in the form of heat, and $w$ is energy transferred to the surroundings in the form of work. t

  2. Since the work done by the gas is the area under the process paths, less work is done for the irreversible process than the reversible process. It follows then that less energy is transferred into the gas in the form of heat from the surroundings for the irreversible process.

  3. Since $q$ is less for the irreversible process, the change in entropy of the surroundings for the irreversible process is less negative (more positive) than the reversible process. Since the change in entropy of the gas is the same for both process, for the irreversible process $\Delta S_{univ}\gt 0$ whereas for the reversible process $\Delta S_{univ} = 0$

Hope this helps.

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