Clara Chung
I never really understand the concept of entropy through classical thermodynamics. Here are a few questions.
1. The change in entropy dS in an isolated system is always >=0, but how does it imply the system tends to a state with maximum entropy? How to know that there exist a maximum?
2. Why is the change of entropy 0 in equilibrium?
3. What is the correlation between zero change of entropy and a reversible process?

Gold Member
1. I think, if your system is an ideal gas that is expanding to a vacuum of infinite volume, there is no maximum entropy.

2. No thermodynamical state functions can change in equilibrium. There's the conceptual problem of why doesn't the entropy of a gas increase indefinitely because of its mixing with itself, but this is resolved by QM and the concept of indistinguishable particles.

Clara Chung
Clara Chung
1. I think, if your system is an ideal gas that is expanding to a vacuum of infinite volume, there is no maximum entropy.

2. No thermodynamical state functions can change in equilibrium. There's the conceptual problem of why doesn't the entropy of a gas increase indefinitely because of its mixing with itself, but this is resolved by QM and the concept of indistinguishable particles.

Thank you. Also, how do you prove that a reversible process implies zero change of entropy in an isolated system?

Gold Member
A reversible process can happen to either direction, so there can be no entropy change. Otherwise one of the directions would have a physically impossible ##\Delta S < 0##.

Clara Chung
jartsa
1. I think, if your system is an ideal gas that is expanding to a vacuum of infinite volume, there is no maximum entropy.

By using large smooth walls we can bounce expanded and still expanding gas back towards its origin. This can be described as reversing the process.

A small wall would reverse the direction of a part of the gas moving into a certain direction, this is of course known as mixing.

Clara Chung
Hi I have one more question. If I have a process going on in an isolated system, the isolated system has no entropy change during the process, does it imply my process is reversible

Clara Chung
Also, if the entropy change is 0 in equilibrium, why do some quasistatic process has entropy change (according to wikipedia)? Isn't a quasistatic process always in equilibrium?

Gold Member
Hi I have one more question. If I have a process going on in an isolated system, the isolated system has no entropy change during the process, does it imply my process is reversible

An example of a reversible process could be something like this: You have ##1 dm^3## of gas at ##20^{o}C##, enclosed in a rigid container that can exchange heat with surroundings but can not change shape or volume. The surroundings are initially at ##20^{o}C##, too. Then you heat up the surroundings to ##30^{o}C## so slowly compared to the heat conductivity of the rigid container that the gas in the container is practically at the same temperature as the surroundings during the whole process. The gas in the container is both in internal equilibrium, as well as in equilibrium with the surroundings at all times.

If you have an isolated system, you can't have a reversible process as far as I know. If the system is initially at internal equilibrium, and can't be affected by the surroundings, then it will just remain in the same equilibrium state indefinitely.

Clara Chung
Clara Chung
An example of a reversible process could be something like this: You have ##1 dm^3## of gas at ##20^{o}C##, enclosed in a rigid container that can exchange heat with surroundings but can not change shape or volume. The surroundings are initially at ##20^{o}C##, too. Then you heat up the surroundings to ##30^{o}C## so slowly compared to the heat conductivity of the rigid container that the gas in the container is practically at the same temperature as the surroundings during the whole process. The gas in the container is both in internal equilibrium, as well as in equilibrium with the surroundings at all times.

If you have an isolated system, you can't have a reversible process as far as I know. If the system is initially at internal equilibrium, and can't be affected by the surroundings, then it will just remain in the same equilibrium state indefinitely.

For an isolated system, I mean the universe, if there is a system which is not isolated with its surroundings, the process ongoing in the not isolated system does not increase the entropy of the universe. Can I be sure that such process is reversible?

jartsa
An example of a reversible process could be something like this: You have ##1 dm^3## of gas at ##20^{o}C##, enclosed in a rigid container that can exchange heat with surroundings but can not change shape or volume. The surroundings are initially at ##20^{o}C##, too. Then you heat up the surroundings to ##30^{o}C## so slowly compared to the heat conductivity of the rigid container that the gas in the container is practically at the same temperature as the surroundings during the whole process. The gas in the container is both in internal equilibrium, as well as in equilibrium with the surroundings at all times.

If you have an isolated system, you can't have a reversible process as far as I know. If the system is initially at internal equilibrium, and can't be affected by the surroundings, then it will just remain in the same equilibrium state indefinitely.

The gas gained entropy. If some warm object gave heat energy to the gas, said warm object might have lost the same amount of entropy as the gas gained.

If the gained entropy was equal to the lost entropy then the process was reversible.

If the process was not very slow, then we know that the gained entropy was larger that the lost entropy, and the process was irreversible.

Clara Chung
Gold Member
For an isolated system, I mean the universe, if there is a system which is not isolated with its surroundings, the process ongoing in the not isolated system does not increase the entropy of the universe. Can I be sure that such process is reversible?

I couldn’t quite follow your question. That said, an isolated system requires a boundary, so the universe does not qualify as an isolated system unless you define it as bounded - that leaves you with needing to define what happens at the boundary you defined; does it transfer heat or not for instance. Such bounded universe definitions would be arbitrary as there is no evidence that our universe is indeed bounded.

Clara Chung
Clara Chung
I couldn’t quite follow your question. That said, an isolated system requires a boundary, so the universe does not qualify as an isolated system unless you define it as bounded - that leaves you with needing to define what happens at the boundary you defined; does it transfer heat or not for instance. Such bounded universe definitions would be arbitrary as there is no evidence that our universe is indeed bounded.
Sorry I couldn't understand.. Why must an isolated system be bounded? If I defined the universe (everywhere) as a system... There is nothing else that can exchange heat with the universe...?

Clara Chung
Also, if the entropy change is 0 in equilibrium, why do some quasistatic process has entropy change (according to wikipedia)? Isn't a quasistatic process always in equilibrium?
This is the only question left...

Gold Member
Sorry I couldn't understand.. Why must an isolated system be bounded? If I defined the universe (everywhere) as a system... There is nothing else that can exchange heat with the universe...?

Entropy is only defined for a bounded system. I might argue that bounded is part of the definition of isolated, but that might be a semantic rathole. I should have advised in my first post to use the term bounded instead of isolated - isolated might mean different things to different people, bounded is relatively unambiguous.

Mentor
Hi I have one more question. If I have a process going on in an isolated system, the isolated system has no entropy change during the process, does it imply my process is reversible
Yes.

Mentor
Do you have any idea how to determine the change in entropy for an irreversible process that takes place in a closed system (no mass flow into or out of the system)? If so, please describe your game plan. Please use a specific example to illustrate this.

Clara Chung
Take the Joule expansion as an example, the closed system is the two boxes connected with a valve. All gas molecules are contained in one box. When the valve is opened, the entropy can be calculated as follow,
dU=0
ds= p/T dV
Δs=nRln2
Do you have any idea how to determine the change in entropy for an irreversible process that takes place in a closed system (no mass flow into or out of the system)? If so, please describe your game plan. Please use a specific example to illustrate this.

StellarCore
Mentor
Take the Joule expansion as an example, the closed system is the two boxes connected with a valve. All gas molecules are contained in one box. When the valve is opened, the entropy can be calculated as follow,
dU=0
ds= p/T dV
Δs=nRln2
Clara,

The reason that you are having so much trouble with entropy is because your thermodynamics book SUCKS (big time).

Over the past few years, I have written two short Physics Forums Insights Articles to help confused students like you. The first article explains how the entropy evolved historically, primarily by the work of Clausius, and how entropy is a state function.

https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

The second article presents a cookbook recipe (game plan) for determining the entropy change of a system as a result of any irreversible process. For any problem that you are going to encounter, the general procedure will always be exactly the same:

https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Please read these over and get back to me with any questions you have. After that, we can work on applying the approach in the 2nd article to the specific problems you are trying to solve (like your Joule-Thompson problem). You will not regret assimilating the information in these articles.

Buzz Bloom, StellarCore, Clara Chung and 1 other person
Gold Member
2022 Award
Although I like your Insight articles very much (I just voted 5 stars for each), that's the way I never understood entropy. Of course the math of axiomatic phenomenological thermodynamics is not too complicated, you just have to remember the names of all these potentials for different changes of states and use the appropriate Legendre transformations to switch between them, but that for never has led to much understanding of the physics behind it. Maybe, it's because I've never played with toy steam engines in my childhood (simply because they were not in fassion when I was a kid; I rather got an electronics kit to play with, which got me interested in physics to begin with).

For me the revelation was a lecture by a professor who taught thermodynamics as statistical physics, where you start from the fundamental microscopic theory to derive macroscopic behavior, and he used the modern approach to entropy due to information theory a la Shannon, Jaynes, and von Neumann. I can recommend a very nice book by a colleague of mine:

https://www.amazon.com/dp/B076DGST77/?tag=pfamazon01-20

Clara Chung
Clara,

The reason that you are having so much trouble with entropy is because your thermodynamics book SUCKS (big time).

Over the past few years, I have written two short Physics Forums Insights Articles to help confused students like you. The first article explains how the entropy evolved historically, primarily by the work of Clausius, and how entropy is a state function.

https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

The second article presents a cookbook recipe (game plan) for determining the entropy change of a system as a result of any irreversible process. For any problem that you are going to encounter, the general procedure will always be exactly the same:

https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Please read these over and get back to me with any questions you have. After that, we can work on applying the approach in the 2nd article to the specific problems you are trying to solve (like your Joule-Thompson problem). You will not regret assimilating the information in these articles.
Thank you so much. I think I get the definitions of reversible and irreversible process that my professors never explained... Also, the way of devising a reversible process