How does reversible =>no entropy change in the universe?

In summary: We can determine the change in entropy of the universe and the change in entropy of the surroundings in a precise manner for the Joule expansion. Here's how. We now focus on the surroundings and treat it as a second separate "system" that can experience a possible change during the Joule expansion that occurred in the primary system. Treading the...In summary, the Attempt at a Solution states that a reversible process in the universe does not imply the change of entropy in the universe is zero. If the overall entropy increases, what would the entropy do in the time-reversed process? Is that possible?
  • #1
Clara Chung
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Homework Statement


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2.Relevant equations[/B]

The Attempt at a Solution


How does a reversible process in the universe imply the entropy doesn't increase? I understand that the change of entropy in a closed reversible cycle is 0 in the system, but I don't get why a not closed reversible process in a system can imply zero change in entropy in the universe.
 

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  • #2
If the overall entropy increases, what would the entropy do in the time-reversed process? Is that possible?
 
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  • #3
mfb said:
If the overall entropy increases, what would the entropy do in the time-reversed process? Is that possible?
Thank you. The entropy change will be smaller than zero which is not possible. There is a similar post that I posted with more entropy confusions. https://www.physicsforums.com/threads/very-confused-about-entropy.963359/#post-6112379
I hope you can take a look.
 
  • #4
By a Joule expansion, do you mean that, initially you have an insulated volume divided in half, with gas in one chamber and vacuum in the other. You remove the barrier between the chambers and allow the system to re-equilibrate?
 
  • #5
Clara Chung said:

Homework Statement


View attachment 236661

2.Relevant equations[/B]

The Attempt at a Solution


How does a reversible process in the universe imply the entropy doesn't increase? I understand that the change of entropy in a closed reversible cycle is 0 in the system, but I don't get why a not closed reversible process in a system can imply zero change in entropy in the universe.
Clara,
You are confused over these results because they are incorrect, at least for the surroundings and for the universe. After you read my PF insights article on how to determine the entropy change for an irreversible process, we will have a basis for discussing why they are incorrect, and what the correct answers are.
 
  • #6
Chestermiller said:
Clara,
You are confused over these results because they are incorrect, at least for the surroundings and for the universe. After you read my PF insights article on how to determine the entropy change for an irreversible process, we will have a basis for discussing why they are incorrect, and what the correct answers are.

But this thread is not about an irreversible processs, it is about a reversible process, so Clara has not included the complete example. In the part of the example given, equation (14.30) and the equation preceding it are for a reversible isothermal expansion. In the part of the example not included, expressions are given for an irreversible Joule expansion.
 
  • #7
George Jones said:
But this thread is not about an irreversible processs, it is about a reversible process, so Clara has not included the complete example. In the part of the example given, equation (14.30) and the equation preceding it are for a reversible isothermal expansion. In the part of the example not included, expressions are given for an irreversible Joule expansion.
I have no idea what you mean. Joule expansion is an irreversible (spontaneous) process.
 
  • #8
Chestermiller said:
I have no idea what you mean. Joule expansion is an irreversible (spontaneous) process.
For completeness, I give below the complete example and the section from which the example is taken. In this particular thread, Clara is asking only about the solid line process of Fig. 14.5.

Joule 1.jpg
Joule 2.jpg
Joule 3.jpg
 

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  • #9
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I am sorry for not posting the whole section. This is the "above" section and I think the problem is discussing about the devised reversible process from two equilibrium states...(not the joule expansion.)
If the process is irreversible, the entropy change of the gas is still the same as the reversible process. However, in the universe the change of entropy is an upper bound of the integral of q'(t)/T, so the change of entropy must be larger than 0.
 

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  • #10
Clara Chung said:
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I am sorry for not posting the whole section. This is the "above" section and I think the problem is discussing about the devised reversible process from two equilibrium states...(not the joule expansion.)
You are correct that it is discussing the devised reversible process. However, the devised reversible process has the same two end states as the irreversible Joule expansion. And, as I said in Step 2 of my Insights article, you need to totally forget about the actual process path when determining the change in entropy between the end states. It is no longer of consequence. The only thing that matters is the two end states.
If the process is irreversible, the entropy change of the gas is still the same as the reversible process.
Yes.
However, in the universe the change of entropy is an upper bound of the integral of q'(t)/T, so the change of entropy must be larger than 0.
We can determine the change in entropy of the universe and the change in entropy of the surroundings in a precise manner for the Joule expansion. Here's how. We now focus on the surroundings and treat it as a second separate "system" that can experience a possible change during the Joule expansion that occurred in the primary system. Treading the surroundings as a second system, please describe the change in the thermodynamic state of the surroundings between its initial and final states, when the gas in the closed insulated primary system experiences an irreversible Joule expansion.
 
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  • #11
Chestermiller said:
We can determine the change in entropy of the universe and the change in entropy of the surroundings in a precise manner for the Joule expansion. Here's how. We now focus on the surroundings and treat it as a second "system" that can experience a possible change during the Joule expansion that occurred in the primary system. Treading the surroundings as a second system, please describe the change in the thermodynamic state of the surroundings between its initial and final states, when the gas in the closed insulated primary system experiences an irreversible Joule expansion.
There are no heat change and no change in internal energy of the second system. The original volume of the second system is V0, the final volume of it is Vf=V0 + V.
So, the initial pressure is P0 and final pressure Pf, the final pressure is given by the ideal gas law.
Pf = P0 * (V0 / V0 + V)
Now I forget about the irreversible and define a reversible process from the initial state to the final state which the macroscopic properties are well defined in the process. The step is an isothermal expansion from V0 to Vf, the entropy change in this reversible process will be nRln(Vf/V0). Also, after the isothermal expansion, the pressure matches the pressure of the final state in the irreversible process.
And the change of entropy of the universe will be the change in entropy of the second system plus the change in entropy of the first system...
?
 
  • #12
Clara Chung said:
There are no heat change and no change in internal energy of the second system. The original volume of the second system is V0, the final volume of it is Vf=V0 + V.
No. There is no change in the surroundings (this second separate system) whatsoever. It doesn't even know that anything happened in the rigid, insulated container. So, at the end of the irreversible change in the container where the Joule expansion took place, the final state of the surroundings is exactly the same as it was in its initial state. What does this tell you about the entropy change for the surroundings?
 
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  • #13
Chestermiller said:
No. There is no change in the surroundings (this second separate system) whatsoever. It doesn't even know that anything happened in the rigid, insulated container. So, at the end of the irreversible change in the container where the Joule expansion took place, the final state of the surroundings is exactly the same as it was in its initial state. What does this tell you about the entropy change for the surroundings?
Ah...right... That will be equivalent with no process have ever happened, dQrev will be zero, so the entropy change of the surroundings is zero.
 
  • #14
Clara Chung said:
Ah...right... That will be equivalent with no process have ever happened, dQrev will be zero, so the entropy change of the surroundings is zero.
Right. So, now, what is the entropy change of the universe (system + surroundings)?
 
  • #15
So what you see here is that, in determining the entropy change of a system, you can separate various parts of the system, and devise separate reversible paths for each of them. And, considering the system and surroundings as. part of a larger overall isolated system, you can separate the system from the surroundings, and devise separate reversible paths for each of them.
 
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  • #16
Chestermiller said:
So what you see here is that, in determining the entropy change of a system, you can separate various parts of the system, and devise separate reversible paths for each of them. And, considering the system and surroundings as. part of a larger overall isolated system, you can separate the system from the surroundings, and devise separate reversible paths for each of them.
Thank you. The entropy change in the universe will be the entropy change in the primary system. I think I got the recipe too. Thank you
 
  • #17
Clara Chung said:
Thank you. The entropy change in the universe will be the entropy change in the primary system. I think I got the recipe too. Thank you
You can also now see how the answer in your book for this problem is incorrect. You have shown that, as expected, the increase in entropy of the universe for the irreversible Joule expansion is greater than zero.

Now what you need to do is get some practice applying this methodology (to lock it in). Maybe an interesting practice problem might be the one we discussed in the other thread, which also features an initial temperature difference between the two chambers. If you decide to go ahead with this, I will specify the problem statement in a precise manner.
 
  • #18
Chestermiller said:
You can also now see how the answer in your book for this problem is incorrect. You have shown that, as expected, the increase in entropy of the universe for the irreversible Joule expansion is greater than zero.

Maybe I am missing something, but how is this different from the answer given in the example from the book? The book's answer also gives that the increase in entropy of the universe for the irreversible Joule expansion is greater than zero.
 
  • #19
George Jones said:
Maybe I am missing something, but how is this different from the answer given in the example from the book? The book's answer also gives that the increase in entropy of the universe for the irreversible Joule expansion is greater than zero.
That's not what the image in post #1 seems to show. What am I missing?
 
  • #20
Chestermiller said:
That's not what the image in post #1 seems to show. What am I missing?

The image in post #1 is a subset of the images that I gave in post #8. The example starts on the page 145 and continues on to page 146. The three equations labeled (14.30) at the bottom of page 145 are for a reversible isothermal process. The three equations (14.31) at the top of page 146 are for the Joule expansion.

Sorry about the quality of the images; I had to zoom my browser to read them.
 
  • #21
George Jones said:
The image in post #1 is a subset of the images that I gave in post #8. The example starts on the page 145 and continues on to page 146. The three equations labeled (14.30) at the bottom of page 145 are for a reversible isothermal process. The three equations (14.31) at the top of page 146 are for the Joule expansion.

Sorry about the quality of the images; I had to zoom my browser to read them.
Aah. You can see how the OP (and I) got confused.

I think that, as a result of our efforts, her understanding has greatly improved.
 

1. What does it mean for a process to be reversible?

A reversible process is one in which the system and its surroundings can be returned to their original state after the process is completed. This means that there is no net change in the system or surroundings, and the process can be reversed without any energy losses.

2. How does a reversible process affect entropy in the universe?

In a reversible process, there is no net change in entropy in the universe. This is because the system and surroundings are returned to their original state, so there is no overall increase or decrease in entropy. However, there may be local changes in entropy within the system or surroundings.

3. Why does a reversible process have no entropy change in the universe?

This is because a reversible process involves an idealized, frictionless system where all energy transfers are perfectly efficient. In reality, all processes involve some amount of energy loss due to factors like friction, which results in an increase in entropy. But in a reversible process, these losses are eliminated, resulting in no net change in entropy in the universe.

4. Can all processes in the universe be reversible?

No, not all processes in the universe can be reversible. In fact, most natural processes are irreversible, meaning that there is a net increase in entropy in the universe. This is due to the presence of factors like friction and heat transfer, which result in energy losses and an increase in entropy.

5. How does the concept of reversibility relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy in a closed system will always increase over time. In a reversible process, there is no net change in entropy, so it does not violate the second law. However, irreversible processes result in an overall increase in entropy and are in line with the second law.

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