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Topic: Why does entropy of surroundings not depend on reversible q?  (Read 4716 times)

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Offline RobNk

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Why does entropy of surroundings not depend on reversible q?
« on: November 05, 2010, 02:26:44 PM »
Hi,

I'm trying to understand the "asymmetry" in the basic entropy equations.  The dS for a system is dqrev/T.  So if the change from state A to B is irreversible, you don't use the actual q, but instead use the q from a reversible path from A->B.  Okay.  But why is it not the same for the surroundings?   My book says you use the actual dq (reversible or irreversible), to calc dSsurr

With the 1st law (dU = dq + dw), it seemed like it you could call either half of the universe the "system" and things were the same.  I don't see why the surroundings are different from the system here when it comes to entropy.

Unfortunately, I'm so confused I can't even ask a clear question.  But if anyone wants to take at clarifying this for me, thanks...

Rob

Offline Juan R.

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Re: Why does entropy of surroundings not depend on reversible q?
« Reply #1 on: November 07, 2010, 09:49:15 AM »
According to 21st century thermodynamics, the total change of entropy is

dS = diS + deS

diS denotes the changes of entropy due to processes inside the system (e.g. chemical reactions).

deS denotes the changes of entropy due to processes of flow with the exterior of the system (e.g. a heat flow with surroundings).

The second law of thermodynamics states that

diS >= 0

The equality for reversible processes and the inequality for irreversible processes. The flow deS is undefined, it can be positive, negative or zero. Using the second law, the total variation of S can be rewritten like

dS >= deS

where

deS = dU/T + pdV/T + dmatterS

Where the last term accounts for interchanges of matter with the surrounds (recall entropy is an extensive quantity). For a system with constant matter (dmatterS = 0) and

dS >= dU/T + pdV/T = dQ/T

There is no reason to call to that dQ reversible or irreversible (the definition of irreversibility follows from diS and the second law), except in some bad textbooks that confound people with 19th century formalisms and non-rigorous treatments.
« Last Edit: November 07, 2010, 10:09:22 AM by Juan R. »
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Offline RobNk

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Re: Why does entropy of surroundings not depend on reversible q?
« Reply #2 on: November 11, 2010, 10:32:06 PM »
Thanks for the reply.

You mention textbooks... is there one you recommend?  I'm auditing a university class that uses Atkins, 9th ed.  I'm suspicious of books that cost almost $200 and have a new version every few years.  I found McQuarrie and Simon's, Physical Chemistry: A Molecular Approach and I think I like better.  My class started with thermodynamics, which didn't feel right to me... but them I'm not sure I can understand quantum mechanics...

Rob

Offline Juan R.

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Re: Why does entropy of surroundings not depend on reversible q?
« Reply #3 on: November 13, 2010, 07:41:49 AM »
All textbooks in physical chemistry that I know present the outdated formulation of thermodynamics of the 19th century and, even doing only that, many of them show an absolute lack of rigor. For instance, I have found pure nonsense about the second law of thermodynamics in the Physical Chemistry by Levine. Atkins is something better but not anything that I would recommend.

McQuarrie and Simon have a specific textbook, their molecular thermodynamics, but is also outdated and lacks rigor, in despite that they give a molecular approach.

In their benefit, one would remark that none of those authors is an expert in thermodynamics, therefore they are writing about something beyond their scope.

For an introduction to our modern understanding of macroscopic thermodynamics, I would recommend Modern Thermodynamics: From Heat Engines to Dissipative Structures by Dilip Kondepudi and Ilya Prigogine.

This textbook is rather rigorous and up to date, covers both equilibrium and nonequilibrium thermodynamics, correct many mistakes found in other textbooks [see correction note below], there is interesting links with relativity and quantum field theory that put thermodynamics in context (using a particle-antiparticle pair creation and anhilation results from quantum field theory, Kondepudi and Prigogine propose a novel absolute scale for the chemical potential), apart from giving the old Gibbs-Duhem stability theory (that you find in many textbooks) in the chapter 12, they give the modern and more general stability theory in the chapter 14, etc.

Note that many that I wrote is found in their textbook. For instance, the modern expression dS = diS + deS that I wrote above is just (3.4.5) in Kondepudi and Prigogine.

Maintain in mind that Prigogine won the Nobel Prize in Chemistry for his extension of the old thermodynamics ;D.

[Some corrections to other textbooks]

E.g. in pages 39-40 Kondepudi and Prigogine explain why the other textbooks, based in old formulations and mathematical idealizations, are forced to say you that DQ is an "imperfect differential" (other authors use the delta notation δQ or a "d" with a stroke) and write expressions as

dU = DQ + DW.

However, the term dQ is completely correct and rigorous in the modern formulation (which is not built over idealizations of the physical and chemical transformations as in the old formulation), doing that one can write the rigorous and general expression (for a closed system)

dU = dQ + dW

They also correctly note in pages 89-90 that something like

dS = dQ/T

that you find in many other textbooks as the "definition of entropy" is not a definition neither is valid for an open system.

Consider an open system that loses one half of its matter in ten minutes at constant T. If the expression dS = dQ/T was valid for the open system interchanging matter, you would be forced to say that nonzero dS (entropy is an extensive quantity) implies nonzero dQ. But dQ is exactly zero in this case, what happen is that the variation of entropy due to the loss of mass is given by the flow of matter. This is the kind of mistakes that one finds in those outdated and unrigorous textbooks.

In chapter 15 Kondepudi and Prigogine give the general expression for the flow of entropy JS as

JS = (Jq/T) + Sumk skJk

where Jq is the heat flow, sk the partial molar entropy of component k and Jk the diffusion flow of component k (moving with velocity vk)

Here you can see that JS = Jq/T is only valid when there is not interchange of matter. And you can see that for an open system without interchange of heat, the variation of entropy is given by the flow of matter

JS = Sumk skJk
« Last Edit: November 13, 2010, 08:15:09 AM by Juan R. »
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Offline RobNk

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Re: Why does entropy of surroundings not depend on reversible q?
« Reply #4 on: November 18, 2010, 11:22:45 AM »
Thanks.  I can't understand all of what you said, but I hopefully I can come back to it later.  I'm waiting for "Modern Thermodynamics" from my library.

Rob

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