Understanding Entropy and the 2nd Law of Thermodynamics - comments

[Chestermiller]

Chestermiller submitted a new PF Insights post

Understanding Entropy and the 2nd Law of Thermodynamics

Continue reading the Original PF Insights Post.

 

[Greg Bernhardt]

Awesome first entry chestermiller!

 

[Jimster41]

Thanks Chester.
Yes. I really did find that clear.

Which is not to say I understood it...

What is special about the reversible paths? Are all the other paths, the non-reversible ones, the same, or do some integrate to different values <= DeltaS than others?

 

[anorlunda]

Kudos chestermiller. That was clear, and understandable, The historical perspective really helped.I look forward to the day when chestermiller makes it similarly easy to understand why this second law implies that "the entopy of the universe tends to a maximum". And how it relates to the kind of information debated in the Hawking/Susskind "black hole wars."

 

[Chestermiller]

Thanks Jimster.

Thanks Chester.
Yes. I really did find that clear.

Which is not to say I understood it...

What is special about the reversible paths?

Reversible paths minimize the dissipation of mechanical energy to thermal energy, and maximize the ability of temperature differences to be converted into mechanical energy. In reversible paths, the pressure exerted by the surroundings at the interface with the system is only slightly higher of lower than the pressure throughout the system, and the temperature at the interface with the surroundings is only slightly higher or lower than the temperature throughout the system. This situation is maintained over the entire path from the initial to the final equilibrium state of the system.

For irreversible paths, the dissipation of mechanical energy to thermal energy is the result of viscous dissipation. The same thing happens if you compress a combination of a spring and (viscous) damper connected in parallel. If you compress the combination very rapidly from an initial length to a final length, you generate lots of heat in the damper (since the force carried by the damper is proportional to the velocity difference across the damper). On the other hand, if you compress the combination very slowly, the force carried by the damper is much less, and you generate much less heat. The amount of work you need to do in the latter case to bring about the compression is also much less. This is a very close analogy to what happens when you cause a gas in a cylinder to compress.

Are all the other paths, the non-reversible ones, the same, or do some integrate to different values <= DeltaS than others?

They integrate to different values < ΔS. The equal sign does not apply to irreversible paths. They are all less.

Chet

 

[Jimster41]

Thanks Chet, I hope it's okay if I keep asking you questions. It really is my favorite way to learn, and I can get enough of the second law, and I'm sure I will learn something - not the least of which will be precision of terms.

In the the case of a gas in a cylinder with a piston (aka "the damper") why does the amount of dissipation vary with the amount of force per unit time? what does the time rate of force have to who with the efficiency of conversion to mechanical energy? Why does the difference at any given time between the system and surroundings, dictate the reversibility, as opposed to say the amount of energy transferred altogether?

 

[Chestermiller]

Thanks Chet, I hope it's okay if I keep asking you questions. It really is my favorite way to learn, and I can get enough of the second law, and I'm sure I will learn something - not the least of which will be precision of terms.

In the the case of a gas in a cylinder with a piston (or damper) why does the amount of dissipation vary with the amount of force per unit time? what does the time rate of force have to who with the efficiency of conversion to mechanical energy? Why does the difference at any given time between the system and surroundings, dictate the reversibility, as opposed to say the amount of energy transferred altogether?

Hi Jimster. You ask great questions.

Why don't you introduce this in a separate thread, and we'll work through it together? First we'll consider the spring/damper system to get an idea of how a difference between a rapid deformation and a very slow deformation (between the same initial and final states) plays out in terms of mechanical energy dissipated in the damper and work done. The idea is for you to get a feel for how this works.

Chet

 

[Jimster41]

Thanks Chet. Your explanation of the "Clausius Inequality" and you answer on the difference between reversible and non-reversible paths were helpful and lucid, and It means a lot to know they are at least sensible questions.

I don't suppose @techmologist and I could get your help reading an old Galvin Crooks paper from 1999 on the "generalized fluctuation theorem"? We've got a thread going in the cosmology forum. PeterDonis has been helping us (humoring us more like it). It's under @techmologists question "why are there heat engines?" It's pretty rambly at this point so I would be more than happy to restart it focusing it back on Crooks' paper and handful of equations, and drill in with your guidance.

 

[Chestermiller]

Thanks Chet. Your explanation of the "Clausius Inequality" and you answer on the difference between reversible and non-reversible paths were helpful and lucid, and It means a lot to know they are at least sensible questions.

I don't suppose @techmologist and I could get your help reading an old Galvin Crooks paper from 1999 on the "generalized fluctuation theorem"? We've got a thread going in the cosmology forum. PeterDonis has been helping us (humoring us more like it). It's under @techmologists question "why are there heat engines?" It's pretty rambly at this point so I would be more than happy to restart it focusing it back on Crooks' paper and handful of equations, and drill in with your guidance.

I'll take a look and see if I can contribute. There are lots of pages and lots of posts, so it may take me a while to come up to speed. No guarantees.

Chet

 

[INFO-MAN]

Hello Chestermiller.

"There have been nearly as many formulations of the second law as there have been discussions of it."
~P. W. BridgmanEntropy and the Second Law of Thermodynamics is not exactly an intuitive concept. While I think your article is basically a good one, it is obviously somewhat limited in scope, and my only critique is that you did not cover some of the most important aspects of entropy.

I agree that most people have a very hard time grasping entropy and the second law of thermodynamics. But I am not sure I understand why your article keeps referring to reversible processes and adiabatic idealizations. In natural systems, the entropy production rate of every process is always positive (ΔS > 0) or zero (ΔS = 0). But only idealized adiabatic (perfectly insulated) and isentropic (frictionless, non-viscous, pressure-volume work only) processes actually have an entropy production rate of zero. Heat is produced, but not entropy. In nature, this ideal can only be an approximation, because it requires an infinite amount of time and no dissipation.

You hardly mention irreversible processes. An irreversible process degrades the performance of a thermodynamic system, and results in entropy production. Thus, irreversible processes have an entropy production rate greater than zero (ΔS > 0), and that is really what the second law is all about (beyond the second law analysis of machines or devices). Every naturally occurring process, whether adiabatic or not, is irreversible (ΔS > 0), since friction and viscosity are always present.

Here is my favorite example of an irreversible thermodynamic process, the Entropy Rate Balance Equation for Control Volumes:

And here are are a couple of other important things you did not mention about entropy:

1) Entropy is a measure of molecular disorder in a system. According to Kelvin, a pure substance at absolute zero temperature is in perfect order, and its entropy is zero. This is the less commonly known Third Law of Thermodynamics.

2) "A system will select the path or assemblage of paths out of available paths that minimizes the potential or maximizes the entropy at the fastest rate given the constraints." This is known as the Law of Maximum Entropy Production. "The Law of Maximum Entropy Production thus has deep implications for evolutionary theory, culture theory, macroeconomics, human globalization, and more generally the time-dependent development of the Earth as a ecological planetary system as a whole." http://www.lawofmaximumentropyproduction.com/

And apparently, I just got another trophy since this is my first post!

 

[Jimster41]

No problem Chet, don't feel obliged. Might be just as well if you were to do something you thought would be most helpful, rather than follow us down a rabbit hole. This is the Crooks paper

http://arxiv.org/abs/cond-mat/9901352

The Entropy Production Fluctuation Theorem and the Nonequilibrium Work Relation for Free Energy Differences
Gavin E. Crooks
(Submitted on 29 Jan 1999 (v1), last revised 29 Jul 1999 (this version, v4))
There are only a very few known relations in statistical dynamics that are valid for systems driven arbitrarily far-from-equilibrium. One of these is the fluctuation theorem, which places conditions on the entropy production probability distribution of nonequilibrium systems. Another recently discovered far-from-equilibrium expression relates nonequilibrium measurements of the work done on a system to equilibrium free energy differences. In this paper, we derive a generalized version of the fluctuation theorem for stochastic, microscopically reversible dynamics. Invoking this generalized theorem provides a succinct proof of the nonequilibrium work relation.
I'm interested in the fluctuation theorem, just understanding it really, it seems to underpin the second law? What I liked about Crooks formulation is that I thought I could see more how "entropy" path selection and work are related. But I have little confidence I understand it.

 

[Chestermiller]

Hello Chestermiller.

Entropy and the Second Law of Thermodynamics is not exactly an intuitive concept. While I think your article is basically a good one, it is obviously somewhat limited in scope, and my only critique is that you did not cover some of the most important aspects of entropy.

Thanks INFO_MAN. It's nice to be appreciated.

Yes. You are correct. I deliberately limited the scope. Possibly you misconstrued my objective. It was definitely not to write a treatise on entropy and the 2nd law. I was merely trying to give beginning thermodynamics students who are struggling with the basic concepts the minimum understanding they need just to do their homework. As someone relatively new to Physics Forums, you may not be aware of the kinds of questions we get from novices. Typical of a recurring question is: How come the entropy change is not zero for an irreversible adiabatic process if the change in entropy is equal to the integral of dq/T and dq = 0? Homework problems frequently involve irreversible adiabatic expansion or compression of an ideal gas in a cylinder with a piston. Students are often asked to determine the final equilibrium state of the system, and the change in entropy. You can see where, if they were asking questions like the the previous one, how they would have trouble doing a homework problem like this.

My original introduction to the tutorial was somewhat longer than in the present version, and spelled out the objectives more clearly. However, the guidelines that Physics Forums set a goal of about 400 words for the Insight articles, and the present version of my article is well over 1000 words. Here is the introductory text that I cut out:

In this author's judgement, the primary cause of the (students') confusion is the poor manner in which these concepts are taught in textbooks and courses.

The standard approach is to present the chronological development of the subject in a straight line from beginning to end. Although this is the way that the subject had developed historically, it is not necessarily the best way to teach the subject. It is much more important for the students to gain a solid understanding of the material by whatever means possible than to adhere to a totally accurate account of the chronological sequence. Therefore, in the present document, we have created a somewhat fictionalized account of the historical sequence of events in order to minimize the historical discussion, focus more intently on the scientific findings, and make the concepts clearer and less confusing to students.

Another shortcoming of existing developments is that the physical situations they discuss are not specified precisely enough, and the mathematical relationships likewise lack proper constraint on their applicability and limitations (particularly the so-called Clausius Inequality). There is also a lack a concise mathematical statement of the second law of thermodynamics in such a way that it can be confidently applied to practical situations and problem solving. In the present development, we have endeavored to overcome these shortcomings.
​

I agree that most people have a very hard time grasping entropy and the second law of thermodynamics. But I am not sure I understand why your article keeps referring to reversible processes and adiabatic idealizations. In natural systems, the entropy production rate of every process is always positive (ΔS > 0) or zero (ΔS = 0). But only idealized adiabatic (perfectly insulated) and isentropic (frictionless, non-viscous, pressure-volume work only) processes actually have an entropy production rate of zero. Heat is produced, but not entropy. In nature, this ideal can only be an approximation, because it requires an infinite amount of time and no dissipation.

This is an example of one of those instances I was referring to in which the constraints on the equations is not spelled out clearly enough, and, as a result, confusion can ensue. The situation you are referring to here with the inequality (ΔS > 0) and equality (ΔS = 0) applies to the combination of the system and the surroundings, and not just to a closed system. Without this qualification, the student might get the idea that for a closed system, ΔS≥0 always, which is, of course, not the case.

Even though reversible processes are an idealization, there is still a need for beginners to understand them. First of all they provide an important limiting case with which irreversible processes can be compared. In geometry, there is no such thing as a perfect circle, a perfect rectangle, a perfect square, etc., but yet we still study them and apply their concepts in our work and lives. Secondly, some of the processes that occur in nature and especially in industry can approach ideal reversible behavior. Finally, and most importantly, reversible processes are the only vehicle we have for determining the change in entropy between two thermodynamic equilibrium states of a system or material.

You hardly mention irreversible processes. An irreversible process degrades the performance of a thermodynamic system, and results in entropy production. Thus, irreversible processes have an entropy production rate greater than zero (ΔS > 0), and that is really what the second law is all about (beyond the second law analysis of machines or devices). Every naturally occurring process, whether adiabatic or not, is irreversible (ΔS > 0), since friction and viscosity are always present.

I'm sorry that impression came through to you because that was not my intention. I feel that it is very important for students to understand the distinction between real irreversible processes paths and ideal reversible process paths. Irreversible process paths are what really happens. But reversible process paths are what we need to use to get the change in entropy for a real irreversible process path.

Here is my favorite example of an irreversible thermodynamic process, the Entropy Rate Balance Equation for Control Volumes:

This equation applies to the more general case of an open system for which mass is entering and exiting, and I was trying to keep things simple by restricting the discussion to closed systems. Also, entropy generation can be learned by the struggling students at a later stage.

And here are are a couple of other important things you did not mention about entropy:

1) Entropy is a measure of molecular disorder in a system. According to Kelvin, a pure substance at absolute zero temperature is in perfect order, and its entropy is zero. This is the less commonly known Third Law of Thermodynamics.

2) "A system will select the path or assemblage of paths out of available paths that minimizes the potential or maximizes the entropy at the fastest rate given the constraints." This is known as the Law of Maximum Entropy Production. "The Law of Maximum Entropy Production thus has deep implications for evolutionary theory, culture theory, macroeconomics, human globalization, and more generally the time-dependent development of the Earth as a ecological planetary system as a whole." http://www.lawofmaximumentropyproduction.com/

As I said above, I was trying to limit the scope exclusively to what the beginning students needed to understand in order to do their homework.

Chet

 

[Jimster41]

That was great Chet. It helps to know the purpose and scope. Hey, can you explain to a confused student why the change in entropy in a closed system is not always greater than or equal to 0? I think I know (Poincare' recurrence?) but I also think I'm probably wrong.

 

[Chestermiller]

That was great Chet. It helps to know the purpose and scope. Hey, can you explain to a confused student why the change in entropy in a closed system is not always greater than or equal to 0? I think I know (Poincare' recurrence?) but I also think I'm probably wrong.

Suppose you compress a gas isothermally and reversibly in a closed system. To hold the temperature constant, do you have to add heat or remove heat? After you compress the gas to a smaller volume at the same temperature, are the number of quantum states available to it greater or fewer?

You are aware that, in thermodynamics, there is a difference between a closed system and an isolated system, correct?

Chet

 

[Jimster41]

Suppose you compress a gas isothermally and reversibly in a closed system. To hold the temperature constant, do you have to add heat or remove heat? After you compress the gas to a smaller volume at the same temperature, are the number of quantum states available to it greater or fewer?

You are aware that, in thermodynamics, there is a difference between a closed system and an isolated system, correct?

Chet

No sir, I was not clear on that precise difference of terms! Now I am. I believe you need to remove heat. Hmm, the quantum states. That one really makes me think, with great confusion, which is not good, since the answer should probably be obvious. In the closed system that has been isothermically compressed (heat removed), I would say the number of states is fewer? But it 's basically a guess. I don't know how to decompose that question, with any confidence. I think I know something about the parts, but probably have way too many questions and misconceptions tangled up in it. Please do illuminate!

I say fewer because the volume is less, and so the available "locations" are reduced. But this does not seem very satisfactory, right, or clear.

 

[Chestermiller]

No sir, I was not clear on that precise difference of terms! Now I am. I believe you need to remove heat. Hmm, the quantum states. That one really makes me think, with great confusion, which is not good, since the answer should probably be obvious. In the closed system that has been isothermically compressed (heat removed), I would say the number of states is fewer? But it 's basically a guess. I don't know how to decompose that question, with any confidence. I think I know something about the parts, but probably have way too many questions and misconceptions tangled up in it. Please do illuminate!

I say fewer because the volume is less, and so the available "locations" are reduced. But this does not seem very satisfactory, right, or clear.

Both your answers are correct. You remove heat from the system in an isothermal reversible compression, so ΔS < 0 (q is negative). The number of states available to the system is fewer, so, by that criterion also, ΔS < 0.

A closed system is one that cannot exchange mass with its surroundings, but it can exchange heat and mechanical energy (work W). An isolated system is one that can exchange neither mass, heat, nor work.

Chet

 

[insightful]

I find your "temperature at the interface with the surroundings" confusing in that to me it implies an average temperature between the system and the surroundings at that point. Would it be more clear to say "temperature of the surroundings at the interface" or am I missing something?

 

[Chestermiller]

I find your "temperature at the interface with the surroundings" confusing in that to me it implies an average temperature between the system and the surroundings at that point. Would it be more clear to say "temperature of the surroundings at the interface" or am I missing something?

What you're missing is that, at the interface, the local temperature of the system matches the temperature of the surroundings. There is no discontinuity in temperature (or in force per unit area) at the interface. However, in an irreversible process, the temperature within the system varies with distance from the interface.

Chet

 

[insightful]

So you're saying there is a temperature gradient between the "bulk" system and the interface, but no temperature gradient between the "bulk" surroundings and the interface?

 

[Chestermiller]

So you're saying there is a temperature gradient between the "bulk" system and the interface

Yes. With an irreversible process, there is a temperature difference between the average temperature in the system and the temperature at the interface. However, at the very interface, the local system temperature matches the local surroundings temperature.

, but no temperature gradient between the "bulk" surroundings and the interface?

Not necessarily. I've tried to get us focused primarily on the system. I'm assuming that we are not concerning ourselves with the details of what is happening within the surroundings, except at the interface, where we are assuming that either the heat flux or the temperature is specified. (Of course, more complicated boundary conditions can also be imposed, and are included within the framework of our methodology). Thus, the "boundary conditions" for work and heat flow on the system are applied at the interface.

Chet

 

[Jimster41]

Both your answers are correct. You remove heat from the system in an isothermal reversible compression, so ΔS < 0 (q is negative). The number of states available to the system is fewer, so, by that criterion also, ΔS < 0.

A closed system is one that cannot exchange mass with its surroundings, but it can exchange heat and mechanical energy (work W). An isolated system is one that can exchange neither mass, heat, nor work.

Chet

I realize why the quantum states question confuses me. Probably it is an issue of specific terms.

If I picture the piston and cylinder made of graph paper cells, containing 1's and 0's, with the volume of the uncompressed cylinder as an area of zeros 0's mixed with 1's (representing the uncompressed gas in the cylinder) this area then surrounded by some more 1's representing the boundaries of the cylinder, including the piston. If I then compress the gas, by changing some of the cylinder volume cells to 1's, I haven't changed the number of states in the system (the graph paper hasn't shrunk or lost cells) I have just added information assigning some of the cells of the cylinder volume with specific values. So I guess by "available QM states" you mean those that are uncertain, or "free" to be randomly set to 1or 0.

Maybe it's a bad metaphor, because I get even more confused when I think that to expand the "cylinder" I still have to add information, changing a set of "fixed cells" to be "free".

 

[Chestermiller]

I realize why the quantum states question confuses me. Probably it is an issue of specific terms.

If I picture the piston and cylinder made of graph paper cells, containing 1's and 0's, with the volume of the uncompressed cylinder as an area of zeros 0's mixed with 1's (representing the uncompressed gas in the cylinder) this area then surrounded by some more 1's representing the boundaries of the cylinder, including the piston. If I then compress the gas, by changing some of the cylinder volume cells to 1's, I haven't changed the number of states in the system (the graph paper hasn't shrunk or lost cells) I have just added information assigning some of the cells of the cylinder volume with specific values. So I guess by "available QM states" you mean those that are uncertain, or "free" to be randomly set to 1or 0.

Maybe it's a bad metaphor, because I get even more confused when I think that to expand the "cylinder" I still have to add information, changing a set of "fixed cells" to be "free".

My goal was to emphasize the classical approach to entropy in my development, and to generally skip the statistical thermodynamic perspective.

Chet

 

[Jimster41]

Thanks Chet. Didn't mean to take you off track. Just trying to leverage my confusion opportunity, to figure out what I'm getting wrong when I think of it. Maybe yournext one. Or I could talk it to a new thread?

 

[nitsuj]

What does the ∫ symbol in the thermodynamics equations mean? (i know what + - * / mean :) Oh and brackets)

 

[Chestermiller]

What does the ∫ symbol in the thermodynamics equations mean?

It is an integral sign. Apparently, you haven't had calculus yet. You are not going to be able to understand and apply much of thermodynamics without the basic tool of calculus.

(i know what + - * / mean :) Oh and brackets)

Brackets are a kind of parenthesis.

Chet

 

[nitsuj]

It is an integral sign. Apparently, you haven't had calculus yet. You are not going to be able to understand and apply much of thermodynamics without the basic tool of calculus.

Brackets are a kind of parenthesis.

Chet

thanks Chet, so ∫ is more complicated than it looks

 

[Chestermiller]

thanks Chet, so ∫ is more complicated than it looks

Hopefully not to those who have had calculus.

 

[wvphysicist]

I thought it was a lower bound, not an upper bound and the lower bound is zero. If you go from one state to another with a perfectly reversible process then the entropy generated is zero.

 

[Chestermiller]

I thought it was a lower bound, not an upper bound and the lower bound is zero. If you go from one state to another with a perfectly reversible process then the entropy generated is zero.

Nope, for a closed system undergoing a reversible change, the entropy change of the system is an upper bound. And for a perfectly reversible process, the entropy change for the system (which is clearly stated as the focus of my article) is not necessarily zero; in fact, it can even be less than zero.

For the combination of system and surroundings, the entropy generated in a reversible process is zero, provided that the surroundings are also handled reversibly.

Chet

 

[wvphysicist]

OK, I understand a little more and accept the last sentence. I think primarily about heat engines.

I have two questions about closed systems. Consider two closed systems, both have a chemical reaction area which releases a small amount of heat and are initially at the freezing point. One has water and no ice and the other has ice. I expect after the chemical reaction the water system will absorb the heat with a tiny change in temperature and the other will convert a small amount of ice to water. Is there any difference in the increase of energy? Suppose I choose masses to enable the delta T of the water system to go toward zero. Is there any difference?

I have another problem with entropy. Some folks say it involves information. I have maintained that only energy is involved. Consider a system containing two gasses. The atoms are identical except half are red and the other are blue. Initially the red and blue are separated by a card in the center of the container. The card is removed and the atoms mix. How can there be a change in entropy?

Oh, one more please. Can you show an example where the entropy change is negative like you were saying?

 

[Chestermiller]

I have two questions about closed systems. Consider two closed systems, both have a chemical reaction area which releases a small amount of heat and are initially at the freezing point. One has water and no ice and the other has ice. I expect after the chemical reaction the water system will absorb the heat with a tiny change in temperature and the other will convert a small amount of ice to water. Is there any difference in the increase of energy? Suppose I choose masses to enable the delta T of the water system to go toward zero. Is there any difference?

I don't have a clear idea of what this equation is about. Let me try to articulate my understanding, and you can then correct it. You have an isolated system containing an exothermic chemical reaction vessel in contact with a cold bath. In one case, the bath contains ice floating in water at 0 C. In the other case, the bath contains only water at 0 C. Is there a difference in the energy transferred from the reaction vessel to the bath in the two cases. How is this affected if the amount of water in the second case is increased? (Are you also asking about the entropy changes in these cases?)

I have another problem with entropy. Some folks say it involves information. I have maintained that only energy is involved. Consider a system containing two gasses. The atoms are identical except half are red and the other are blue. Initially the red and blue are separated by a card in the center of the container. The card is removed and the atoms mix. How can there be a change in entropy?

This question goes beyond the scope of what I was trying to cover in my article. It involves the thermodynamics of mixtures. I'm trying to decide whether to answer this in the present Comments or write an introductory Insight article on the subject. I need time to think about what I want to do. Meanwhile, I can tell you that there is an entropy increase for the change that you described and that the entropy change can be worked out using energy with the integral of dq_rev/T.

Oh, one more please. Can you show an example where the entropy change is negative like you were saying?

This one is easy. Just consider a closed system in which you bring about the isothermal reversible compression of an ideal gas, so that the final temperature is equal to the initial temperature, the final volume is less than the initial volume, and the final pressure is higher than the initial pressure.

Chet

 

[DocZaius]

To make sure I understand the end of your article clearly:

(1) A system can go from an initial equilibirum state to a final equilibrium state through a reversible or irreversible process.
(2) Whichever process it undergoes, its change in entropy will be the same.
(3) That change in entropy can be determined by evaluating the following integral over any reversible process path the system could have gone through: \int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}.
(4) If the system goes through an irreversible process path, this integral: \int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_{Int}(t)}dt} will yield a lesser value than the reversible path integral, but the change in entropy would still be equal to the (greater) evaluation of the reversible path integral.

Is that right?

 

[Chestermiller]

To make sure I understand the end of your article clearly:

(1) A system can go from an initial equilibirum state to a final equilibrium state through a reversible or irreversible process.
(2) Whichever process it undergoes, its change in entropy will be the same.
(3) That change in entropy can be determined by evaluating the following integral over any reversible process path the system could have gone through: \int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}.
(4) If the system goes through an irreversible process path, this integral: \int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_{Int}(t)}dt} will yield a lesser value than the reversible path integral, but the change in entropy would still be equal to the (greater) evaluation of the reversible path integral.

Is that right?

Perfect.

 

[wvphysicist]

I have two questions about closed systems. Consider two closed systems, both have a chemical reaction area which releases a small amount of heat and are initially at the freezing point. One has water and no ice and the other has ice. I expect after the chemical reaction the water system will absorb the heat with a tiny change in temperature and the other will convert a small amount of ice to water. Is there any difference in the increase of energy? Suppose I choose masses to enable the delta T of the water system to go toward zero. Is there any difference?

I don't have a clear idea of what this equation is about. Let me try to articulate my understanding, and you can then correct it. You have an isolated system containing an exothermic chemical reaction vessel in contact with a cold bath. In one case, the bath contains ice floating in water at 0 C. In the other case, the bath contains only water at 0 C. Is there a difference in the energy transferred from the reaction vessel to the bath in the two cases. How is this affected if the amount of water in the second case is increased? (Are you also asking about the entropy changes in these cases?)

Using identical reaction vessels the energy transferred is set the same. The question is about the entropy change. Will heating water a tiny delta T or melting ice result in the same entropy change?

 

[wvphysicist]

This question goes beyond the scope of what I was trying to cover in my article. It involves the thermodynamics of mixtures. I'm trying to decide whether to answer this in the present Comments or write an introductory Insight article on the subject. I need time to think about what I want to do. Meanwhile, I can tell you that there is an entropy increase for the change that you described and that the entropy change can be worked out using energy with the integral of dq_rev/T.
I cannot understand the symbols in the integral.

 

[Chestermiller]

I have two questions about closed systems. Consider two closed systems, both have a chemical reaction area which releases a small amount of heat and are initially at the freezing point. One has water and no ice and the other has ice. I expect after the chemical reaction the water system will absorb the heat with a tiny change in temperature and the other will convert a small amount of ice to water. Is there any difference in the increase of energy? Suppose I choose masses to enable the delta T of the water system to go toward zero. Is there any difference?

I don't have a clear idea of what this equation is about. Let me try to articulate my understanding, and you can then correct it. You have an isolated system containing an exothermic chemical reaction vessel in contact with a cold bath. In one case, the bath contains ice floating in water at 0 C. In the other case, the bath contains only water at 0 C. Is there a difference in the energy transferred from the reaction vessel to the bath in the two cases. How is this affected if the amount of water in the second case is increased? (Are you also asking about the entropy changes in these cases?)

Using identical reaction vessels the energy transferred is set the same. The question is about the entropy change. Will heating water a tiny delta T or melting ice result in the same entropy change?

Yes, provided the delta T of the water is virtually zero. Otherwise, the final reactor temperature will not be the same in the two cases. So the reactor would have a different entropy change.

Chet

 

[Chestermiller]

This question goes beyond the scope of what I was trying to cover in my article. It involves the thermodynamics of mixtures. I'm trying to decide whether to answer this in the present Comments or write an introductory Insight article on the subject. I need time to think about what I want to do. Meanwhile, I can tell you that there is an entropy increase for the change that you described and that the entropy change can be worked out using energy with the integral of dq_rev/T.
I cannot understand the symbols in the integral.

I don't understand what you are saying here. There are some basic concepts that need to be developed to describe mixtures. The starting point for mixtures goes back again to ideal gases, and discusses the thermodynamic properties of entropy, enthalpy, free energy, and volume of ideal gas mixtures, based on "Gibbs Theorem." This development enables you to determine the change in entropy in going from two pure gases, each at a certain pressure, to a mixture of the two gases at the same pressure. The partial pressures of the gases in the mixture are lower than their original pressures, so, to get to their final partial pressures, you would have to increase their volumes reversibly at constant temperature. This would give rise to an entropy increase for each of the gases. I'm uncomfortable going into it in more detail than this, because, to do it right, you need more extensive discussion. If you want to find out more about this, see Chapter 10 of Introduction to Chemical Engineering Thermodynamics by Smith and van Ness.

Chet

 

[nothingkwt]

I have always thought that a reversible process would give the minimum change in entropy, i.e. a lower bound for the integral. Is it not that the higher the entropy change the more energy it's dissipating from irreversibilities? In other words, why exactly is the integrand always less than or equal to and not greater than or equal to?

 

[Chestermiller]

I have always thought that a reversible process would give the minimum change in entropy, i.e. a lower bound for the integral. Is it not that the higher the entropy change the more energy it's dissipating from irreversibilities? In other words, why exactly is the integrand always less than or equal to and not greater than or equal to?

This is discussed in posts #28 and 29. As I clearly said in my article, I am referring to the entropy of a (closed) system, not to the entropy of the combination of system and surroundings (which constitutes an isolated system). Have you never seen the Clausius Inequality in the form that I presented it before?

All you need to do to convince yourself that what I said is correct is do a few sample problems for irreversible processes, where you compare with the integral of the heat flow at the boundary divided by the temperature at the boundary with the entropy change of the system between the same two initial and final equilibrium states. In an isolated system, the integral is zero (since no heat is passing across the boundary of an isolated system), but the entropy change is greater than zero (for an irreversible change).

Chet

 

[GM Jackson]

Thanks for your definitive take on thermodynamics one and two.

 

[DrDu]

I fear that one may get from this article the impression that the concept of entropy can only be introduced under very restricing assumptions. Here some rethoric questions: Are the systems for which we can introduce entropy really restricted to those describable in terms of only T and P? How about chemical processes or magnetization? Can and work only enter through the boundaries? How about warming a glas of milk in the microwave, then? In this case, the pressure is constant, but we can't assign a unique temperature to the system, not even at the boundary, as the distribution of energy over the internal states of the molecules is out of equilibrium.
This already shows that the Clausius inequality is of restricted value, as the integrals aren't defined for most of the irreversible processes.

In fact, we don't need to break our heads about the complicated structure of non-equilibrium states. The point is that we can calculate entropy integrating over a sequence of equilibrium states. It plays no role whether we can approximate this integral by an actual quasistatic process.

 

[Chestermiller]

I fear that one may get from this article the impression that the concept of entropy can only be introduced under very restricing assumptions. Here some rethoric questions: Are the systems for which we can introduce entropy really restricted to those describable in terms of only T and P? How about chemical processes or magnetization? Can and work only enter through the boundaries? How about warming a glas of milk in the microwave, then? In this case, the pressure is constant, but we can't assign a unique temperature to the system, not even at the boundary, as the distribution of energy over the internal states of the molecules is out of equilibrium.
This already shows that the Clausius inequality is of restricted value, as the integrals aren't defined for most of the irreversible processes.

Thanks for your comment DrDu.

I fear that, perhaps, you have not read all my responses to the comments that have been made so far. I tried to make it clear what my motivation was for writing this article, particularly with regard to its limited scope. See, in particular, posts #12 and #31. To reiterate: My target audience was beginning thermodynamics students who are being exposed to the 1st and 2nd laws for the first time, but who, due to the poor manner in which the material is presented in most texts and courses, are so confused that they are unable to do their homework. I tried to keep the article "short and sweet" so that the students would not lose interest and stop reading before they reached the end. I did not want the article to be a complete treatise on thermodynamics. If you would like to expand on what I have written, you are welcome to write an Insights article of your own. I'm sure it would be very well received.

Along these same lines, I might mention that I am currently preparing another Insights article that focuses on the work done in reversible versus irreversible gas expansion/compression processes, and quantitatively identifies the fundamental mechanism by which the work in the two situations differ.

In fact, we don't need to break our heads about the complicated structure of non-equilibrium states. The point is that we can calculate entropy integrating over a sequence of equilibrium states. It plays no role whether we can approximate this integral by an actual quasistatic process.

I thought that I had covered this in my article when I wrote: "Note that the reversible process path used to determine the entropy change does not necessarily need to bear any resemblance to the actual process path."

Chet

 

[DrDu]

But you claimed your formulation of the Clausius inequality to be $\emph{mathematically precise}$, didn't you?

 

[DrDu]

But you claimed your formulation of the Clausius inequality to be $emph{mathematically precise}$, didn't you?

Should read "mathematically precise".

 

[Chestermiller]

Should read "mathematically precise".

In my judgement, this is sufficiently precise mathematically to address the students' needs at their introductory level (i.e., giving them the ability to understand and do their homework). As with any subject, additional refinement can be introduced at a later stage. For example, when we first learn about heat capacity, we are told that it is defined in terms of the path-dependent heat flow Q = C ΔT, but later learn that, more precisely, heat capacity is a function of state (not path), defined in terms of the partial derivatives of internal energy U or enthalpy H with respect to temperature. In my opinion, my judgement call is a considerably less blatant use of literary license than this.

Chet

 

[Jano L.]

I think your article may be helpful to students, but it would be good to put some more disclaimers to places where it simplifies a lot.

For example, you wrote

The time variation of q˙(t) and w˙(t) between the initial and final states uniquely defines the so-called process path


I think this is true for simple system whose thermodynamic state is determined by two numbers, say entropy and internal energy. But there may be more complicated situations, when one has magnetic and electric work in addition to volume work and then two numbers q˙ and w˙ are not sufficient to determine the path through the state space.

 

[Khashishi]

This is a good explanation, but personally I feel like the classical description of thermodynamics which defines entropy as some maximum value of an integral should be deprecated in light of our increasing knowledge of physics. The statistical mechanics definition of entropy is far superior. The classical definition is inherently confusing because entropy is a state variable, yet it is defined in terms of paths. Each path gives you a different integral. Experimentally, how do you determine what the maximum is out of an infinite number of possible paths? If the system is opened up in a way such that more paths become available, can the entropy increase?

The statistical mechanics definition (and the related information theory definition) makes it clear why it is a state variable, because it only depends on the states. The paths are irrelevant.

 

[Jano L.]

This is a good explanation, but personally I feel like the classical description of thermodynamics which defines entropy as some maximum value of an integral should be deprecated in light of our increasing knowledge of physics. The statistical mechanics definition of entropy is far superior. The classical definition is inherently confusing because entropy is a state variable, yet it is defined in terms of paths. Each path gives you a different integral. Experimentally, how do you determine what the maximum is out of an infinite number of possible paths? If the system is opened up in a way such that more paths become available, can the entropy increase?

The statistical mechanics definition (and the related information theory definition) makes it clear why it is a state variable, because it only depends on the states. The paths are irrelevant.

There is no one entropy to be defined by some optimal definition. In thermodynamics, (Clausius) entropy is defined through integrals. There is nothing confusing about it; to understand entropy in thermodynamics, the paths and integrals are necessary. It is hardly a disadvantage of the definition, since processes and integrals are very important things to understand while learning thermodynamics.

In statistical physics, there are several concepts that are also called entropy, but neither of these is the same concept as Clausius entropy. *Sometimes* the statistical concept has similar functional dependence on the macroscopic variables to thermodynamic entropy. But it is not the same concept as thermodynamic entropy. Any use of statistical physics for explanation of thermodynamics is based on the *assumption* that statistical physics applies to thermodynamic systems. It does not replace thermodynamics in any way.

 

[Chestermiller]

I think your article may be helpful to students, but it would be good to put some more disclaimers to places where it simplifies a lot.

For example, you wrote

The time variation of q˙(t) and w˙(t) between the initial and final states uniquely defines the so-called process path


I think this is true for simple system whose thermodynamic state is determined by two numbers, say entropy and internal energy. But there may be more complicated situations, when one has magnetic and electric work in addition to volume work and then two numbers q˙ and w˙ are not sufficient to determine the path through the state space.

Thanks Jano L.

I toyed with the idea of mentioning that there are other forms of work that might need to be considered also, but in the end made the judgement call not to. You read my motivation for the article in some of my responses to the comments and in the article itself. I just wanted to include the bare minimum to give the students what they needed to do most of their homework problems. I felt that, if I made the article too long and comprehensive, they would stop reading before they had a chance to benefit from the article. There are many other things that I might have included as well, such as the more general form of the first law, which also includes changes in kinetic and potential energy of the system.

I invite you to consider writing a supplement to my article in which you flesh things out more completely. Thanks for your comment.

Chet

 

[Chestermiller]

This is a good explanation, but personally I feel like the classical description of thermodynamics which defines entropy as some maximum value of an integral should be deprecated in light of our increasing knowledge of physics. The statistical mechanics definition of entropy is far superior.

That's your opinion.

The classical definition is inherently confusing because entropy is a state variable, yet it is defined in terms of paths. Each path gives you a different integral. Experimentally, how do you determine what the maximum is out of an infinite number of possible paths? If the system is opened up in a way such that more paths become available, can the entropy increase?

Did you not read the article? I made it perfectly clear that:

1. Entropy is a function of state
2. There are an infinite number of process paths between the initial and final equilibrium states of the system
3. The integral of dq/T over all these possible paths has a maximum value, and is thus a function of state
4. All reversible paths between the initial and final equilibrium states of the system give exactly the same (maximum) value of the integral, so you don't need to evaluate all possible paths
5. To get the change in entropy between the initial and final equilibrium states of the system, one needs only to conceive of a single convenient reversible path between the two states and integtrate dq/T for that path.

Try using the statistical mechanical definition of entropy to calculate the change in entropy of a real gas between two thermodynamic equilibrium states.

Chet