Entropy change for reversible and irreversible processes

In summary: It's very confusing when it comes to Second Law of Thermodynamics. I have seen in some online sources that this law applies to a closed system, while other sources/textbooks say it applies to an isolated system. Who's right?The law applies to an isolated system, which is a system that is not in contact with any other system.
  • #1
vcsharp2003
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TL;DR Summary
Change in entropy for reversible and irreversible processes
I came across the following statement from the book Physics for Engineering and Science (Schaum's Outline Series).
CamScanner 01-25-2023 16.58.jpg

I cannot seem to find a satisfactory answer to the questions.
  1. Is the statement in above screenshot talking about entropy change the statement of Second Law of Thermodynamics or is it a conclusion based on definition of entropy?
  2. If the statement of entropy change is the Second Law of Thermodynamics, then the entropy change truth needs to be accepted without any questions and we don't need to bother about the reasons why entropy changes occur in the way described. But, I still wonder how can entropy change for an irreversible process increase as per the statement in above screenshot, when this change depends only on the end points of the irreversible process that when joined by a reversible process will yield a 0 change in entropy as per the same statement for the system and it's surroundings.
 
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  • #2
vcsharp2003 said:
TL;DR Summary: Change in entropy for reversible and irreversible processes

I came across the following statement from the book Physics for Engineering and Science (Schaum's Outline Series).
View attachment 321092
I cannot seem to find a satisfactory answer to the questions.
  1. Is the statement in above screenshot talking about entropy change the statement of Second Law of Thermodynamics or is it a conclusion based on definition of entropy?
  2. If the statement of entropy change is the Second Law of Thermodynamics, then the entropy change truth needs to be accepted without any questions and we don't need to bother about the reasons why entropy changes occur in the way described. But, I still wonder how can entropy change for an irreversible process increase as per the statement in above screenshot, when this change depends only on the end points of the irreversible process that when joined by a reversible process will yield a 0 change in entropy as per the same statement for the system and it's surroundings.
The statement is messed up, and I disagree with the last sentence in the statement all-together.

When looking at the change in entropy for an irreversible process, you need to separate the system from the surroundings and subject each of them to new reversible paths separately. The reversible path for the system will be different from the reversible path for the surroundings.

See my Physics Forums Insights Articles
https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
and
https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
 
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  • #3
Chestermiller said:
The statement is messed up, and I disagree with the last sentence in the statement all-together.

When looking at the change in entropy for an irreversible process, you need to separate the system from the surroundings and subject each of them to new reversible paths separately. The reversible path for the system will be different from the reversible path for the surroundings.

See my Physics Forums Insights Articles
https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
and
https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
As per the first article provided by you, it seems that the entropy change for any real process is less than or equal to the entropy change for a reversible process joining the end points of the real process. Is that what the article is saying? And furthermore, the entropy change is only for the thermodynamic system undergoing the real process and excludes it's surroundings.
 
  • #4
vcsharp2003 said:
As per the first article provided by you, it seems that the entropy change for any real process is less than or equal to the entropy change for a reversible process joining the end points of the real process. Is that what the article is saying?
Basically, yes.

vcsharp2003 said:
And furthermore, the entropy change is only for the thermodynamic system undergoing the real process and excludes it's surroundings.
The surroundings experiences a change in entropy too.. But that has to be calculated as ii the surroundings were a separate system.

It is important to recognize that entropy is a physical property of the material(s) comprising the system, defined at thermodynamic equilibrium. In this sense, it is analogous to internal energy U and enthalpy H. None of these depend directly on the manner (path) by which the material arrived at that state.

There are only two ways that the entropy of a system can change:

1. By heat transfer through the boundary between the system and its surroundings, at the boundary temperature ##T_B##, such that the entropy transferred to the system is the integral of ##dQ/T_B##. This mechanism is present in both reversible- and irreversible processes.

2. By entropy generation within the system ##\sigma##. This mechanism is present only in irreversible processes.

So the entropy change a system for an irreversible process is $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$
 
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  • #5
Chestermiller said:
The surroundings experiences a change in entropy too.. But that has to be calculated as ii the surroundings were a separate system.
It's very confusing when it comes to Second Law of Thermodynamics. I have seen in some online sources that this law applies to a closed system, while other sources/textbooks say it applies to an isolated system. Who's correct is a big question mark.

But if we include system alongwith it's surroundings, we end up with an isolated system across whose boundary neither matter nor energy is exchanged. So, the system + surroundings is also another system. On the other hand, a closed system allows energy exchanges across it's boundary but no matter exchange. So, the fact that some books/online sources say this law applies to only closed systems while some say it applies to only isolated systems gets very, very confusing for a student studying this law for the first time.

I thought there would be one and only one statement for Second Law of Thermodynamics since a law must be uniform. It's like different sources are giving different interpretations for this law unlike the Newton's Laws of Motion which are so beautifully stated that there is zero chance of any confusion.
 
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  • #6
vcsharp2003 said:
I thought there would be one and only one statement for Second Law of Thermodynamics since a law must be uniform. It's like different sources are giving different interpretations for this law unlike the Newton's Laws of Motion which are so beautifully stated that there is zero chance of any confusion.
The 2nd law is not a "law" in the common sense of the word in physics. It is fundamentally a statement about probabilities, a "statistical law" if you want. Small systems can (at least temporarily) break the 2nd law.
 
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  • #7
DrClaude said:
Small systems can (at least temporarily) break the 2nd law.
I see. I need to figure this out.

Do you know of an example where this law is violated?
 
  • #10
vcsharp2003 said:
That's a surprising revelation for something that's called a law in Physics. I guess knowledge is always evolving.
It is a question of domain of application. For any sufficiently big system (which in practice is actually quite small) the 2nd law applies. Not much different from Newton's laws that only apply if speed is low compared to c.
 
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  • #11
vcsharp2003 said:
It's very confusing when it comes to Second Law of Thermodynamics. I have seen in some online sources that this law applies to a closed system, while other sources/textbooks say it applies to an isolated system. Who's correct is a big question mark.

But if we include system alongwith it's surroundings, we end up with an isolated system across whose boundary neither matter nor energy is exchanged. So, the system + surroundings is also another system. On the other hand, a closed system allows energy exchanges across it's boundary but no matter exchange. So, the fact that some books/online sources say this law applies to only closed systems while some say it applies to only isolated systems gets very, very confusing for a student studying this law for the first time.
While it is being taught to you in a lecture it is kinda like that makes sense and understandable.
And then they give you a problem to solve and you are, wait, what, let's go over this again.
 
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  • #12
DrClaude said:
It is a question of domain of application. For any sufficiently big system (which in practice is actually quite small) the 2nd law applies. Not much different from Newton's laws that only apply if speed is low compared to c.
Yes that makes a lot of sense. It's like saying that a law in Physics can have limitations but still is true in many situations.
 
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  • #13
vcsharp2003 said:
It's very confusing when it comes to Second Law of Thermodynamics. I have seen in some online sources that this law applies to a closed system, while other sources/textbooks say it applies to an isolated system. Who's correct is a big question mark.
The form of the equation I wrote applies to a closed system: $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$
When we consider a system and surroundings, we can regard "the system" as System A,, and its surroundings as a separate system, System B. System B is the surroundings for System A, and system A is the surroundings for System B. So we have, $$\Delta S_A=\int{\frac{dQ_A}{T_{BA}}}+\sigma_A$$and$$\Delta S_B=\int{\frac{dQ_B}{T_{BB}}}+\sigma_B$$But, since Systems A and B are a complementary system and surroundings, $$Q_B=-Q_A$$and$$T_{BB}=T_{BA}$$we have $$\Delta S_A+\Delta S_B=\sigma_A+\sigma _B$$So for an isolated system consisting of System A (system) and System B (surroundings), the entropy change is positive.
vcsharp2003 said:
But if we include system alongwith it's surroundings, we end up with an isolated system across whose boundary neither matter nor energy is exchanged. So, the system + surroundings is also another system. On the other hand, a closed system allows energy exchanges across it's boundary but no matter exchange. So, the fact that some books/online sources say this law applies to only closed systems while some say it applies to only isolated systems gets very, very confusing for a student studying this law for the first time.
So there are two versions of the law, depending on whether we are talking about a closed system or an isolated system. Big deal!!
vcsharp2003 said:
I thought there would be one and only one statement for Second Law of Thermodynamics since a law must be uniform. It's like different sources are giving different interpretations for this law unlike the Newton's Laws of Motion which are so beautifully stated that there is zero chance of any confusion.
The version I always work with is the closed system version, which is easier to use in solving problems.
 
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FAQ: Entropy change for reversible and irreversible processes

What is the difference between entropy change in reversible and irreversible processes?

In a reversible process, the entropy change of the system and the surroundings is zero when considered together, as the process can be reversed without any net change in the universe. For an irreversible process, the entropy change of the universe is positive, as the process increases the total entropy due to the production of entropy within the system or surroundings.

How is entropy change calculated for a reversible process?

For a reversible process, the entropy change (ΔS) can be calculated using the formula ΔS = Q_rev / T, where Q_rev is the heat exchanged reversibly and T is the absolute temperature at which the exchange occurs. This relationship holds because the process is carried out in infinitesimally small steps, maintaining equilibrium.

Why is the entropy change greater in irreversible processes compared to reversible processes?

In irreversible processes, additional entropy is generated due to factors like friction, unrestrained expansion, or rapid mixing, which drive the system away from equilibrium. This generates entropy within the system or surroundings, leading to a greater overall increase in entropy compared to a reversible process, where no such additional entropy is produced.

Can the entropy change of a system be negative in an irreversible process?

Yes, the entropy change of a system can be negative in an irreversible process, but the total entropy change of the universe (system plus surroundings) will always be positive. This is due to the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time. The negative entropy change in the system is offset by a larger positive entropy change in the surroundings.

What is the significance of entropy change in understanding the direction of a process?

Entropy change is crucial in determining the spontaneity and direction of a process. A process will naturally proceed in the direction that increases the total entropy of the universe. For reversible processes, the entropy change is zero, indicating a delicate balance. For irreversible processes, the positive entropy change indicates a spontaneous process. This concept helps in predicting whether a given process will occur naturally.

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