Why Is Understanding Entropy So Challenging?

In summary, the concept of entropy in classical thermodynamics can be difficult to understand. There are several questions about the relationship between entropy and isolated systems, equilibrium, and reversible processes. One explanation is that in an ideal gas expanding into a vacuum, there is no maximum entropy. In equilibrium, the change in entropy is 0 because no thermodynamic state functions can change. However, the issue of mixing in gases is resolved by quantum mechanics and the concept of indistinguishable particles. A reversible process occurs when there is no change in entropy, as one direction would result in a physically impossible negative change. This can be demonstrated by using large smooth walls to bounce expanded gas back towards its origin. A small wall would only reverse the direction of a part
  • #1
Clara Chung
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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?
 
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  • #2
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.
 
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  • #3
hilbert2 said:
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?
 
  • #4
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##.
 
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  • #5
hilbert2 said:
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.
 
  • #6
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
 
  • #7
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?
 
  • #8
Clara Chung said:
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.
 
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  • #9
hilbert2 said:
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?
 
  • #10
hilbert2 said:
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.
 
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  • #11
Clara Chung said:
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.
 
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  • #12
Grinkle said:
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...?
 
  • #13
Clara Chung said:
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...
 
  • #14
Clara Chung said:
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.
 
  • #15
Clara Chung said:
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.
 
  • #16
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.
 
  • #17
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
Chestermiller said:
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.
 
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  • #18
Clara Chung said:
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.
 
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  • #19
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
 
  • #20
Chestermiller said:
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
 

1. What is entropy?

Entropy is a measure of the disorder or randomness in a system. In scientific terms, it is the amount of energy in a system that is unavailable for work.

2. How does entropy relate to thermodynamics?

In thermodynamics, entropy is a measure of the energy dispersal in a system. It is closely related to the 2nd law of thermodynamics, which states that the total entropy of a closed system will always increase over time.

3. Why does entropy tend to increase over time?

This is due to the 2nd law of thermodynamics, which states that in any isolated system, the total entropy will tend to increase over time. This is because energy tends to spread out and become more evenly distributed, leading to an increase in disorder or randomness.

4. Can entropy be reversed?

In some cases, yes. For example, in a refrigerator, the entropy of the system (the food inside) decreases while the entropy of the surroundings (the room) increases. However, in closed systems, the 2nd law of thermodynamics states that entropy will always increase over time, and it cannot be reversed.

5. How is entropy used in other fields besides thermodynamics?

Entropy is a fundamental concept in many fields of science, including physics, chemistry, biology, and information theory. It is used to describe and quantify the randomness or disorder in a system and can be applied to a wide range of phenomena, from chemical reactions to the behavior of living organisms to the transmission of data in communication systems.

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