# Understanding Entropy and the 2nd Law of Thermodynamics

## Introduction

The second law of thermodynamics and the associated concept of entropy have been sources of confusion to thermodynamics students for centuries.  The objective of the present development is to clear up much of this confusion.  We begin by first briefly reviewing the first law of thermodynamics, in order to introduce in a precise way the concepts of thermodynamic equilibrium states, heat flow, mechanical energy flow (work), and reversible and irreversible process paths.

## First Law of Thermodynamics

A thermodynamic equilibrium state of a system is defined as one in which the temperature and pressure are constant, and do not vary with either location within the system (i.e., spatially uniform temperature and pressure) or with time (i.e., temporally constant temperature and pressure).

Consider a closed system (no mass enters or exits) that, at initial time $t_i$, is in an initial equilibrium state, with internal energy $U_i$, and, at a later time $t_f$, is in a new equilibrium state with internal energy $U_f$.  The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let $\dot{q}(t)$ represent the rate of heat addition across the interface at time t, and let $\dot{w}(t)$ represent the rate at which the system does work at the interface at time t. According to the first law (basically conservation of energy),
$$\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W$$
where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

The time variation of $\dot{q}(t)$ and $\dot{w}(t)$ between the initial and final states uniquely defines the so-called process path. There is an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

A reversible process path is defined as one for which, at each instant of time along the path, the system is only slightly removed from being in thermodynamic equilibrium with its surroundings.  So the path can be considered as a continuous sequence of thermodynamic equilibrium states.  As such, the temperature and pressure throughout the system along the entire reversible process path are completely uniform spatially.  In order to maintain these conditions, a reversible path must be carried out very slowly so that $\dot{q}(t)$ and $\dot{w}(t)$ are both very close to zero over then entire path.

An irreversible process path is typically characterized by rapid rates of heat transfer $\dot{q}(t)$ and work  being done at the interface with the surroundings $\dot{w}(t)$.    This produces significant temperature and pressure gradients within the system (i.e., the pressure and temperature are not spatially uniform throughout), and thus, it is not possible to identify specific representative values for either the temperature or the pressure of the system (except at the initial and the final equilibrium states). However, the pressure ##P_{Int}(t)## and temperature ##T_{Int}(t)## at the interface can always be measured and controlled using the surroundings to impose whatever process path we desire.  (This is equivalent to specifying the rate of heat flow and the rate of doing work at the interface $\dot{q}(t)$ and $\dot{w}(t)$).

Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
$$\dot{w}(t)=P_{Int}(t)\dot{V}(t)$$
where, again, ##P_{Int}(t)## is the pressure at the interface with the surroundings, and where $\dot{V}(t)$ is the rate of change of system volume at time t.

If the process path is reversible, the pressure P throughout the system is uniform, and thus matches the pressure at the interface, such that

$$P_{Int}(t)=P(t)\mbox{ (reversible process path only)}$$

Therefore, in the case of a reversible process path, $$\dot{w}(t)=P(t)\dot{V}(t)\mbox{ (reversible process path only)}$$

This completes our discussion of the First Law of Thermodynamics.

## Second Law of Thermodynamics

In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate $\dot{q}(t)$ and the rate of doing work $\dot{w}(t)$ as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:
$$Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}$$
$$W=\int_{t_i}^{t_f}{\dot{w}(t)dt}$$
In the present section, we will be introducing a third integral of this type (involving the heat transfer rate $\dot{q}(t)$) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in a complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800s.

(The following is a somewhat fictionalized account, designed to minimize the historical discussion, and focus more intently on the scientific findings.) Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_{Int}(t)}dt}$$
where ##T_{Int}(t)## is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing:  For any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) is not arbitrary; instead, there is a unique upper bound to the value of the integral. Clausius also found that this observation is consistent with all the “word definitions” of the Second Law.

Clearly, if there is an upper bound for this integral, this upper bound has to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there exists a well-defined subset, each member of which gives exactly the same maximum value for the integral. This subset consists of all the reversible process paths. Thus, to determine the change in entropy between two equilibrium states, one must first “dream up” a reversible path between the two states and then evaluate the integral over that path. Any other process path will give a value for the integral lower than the entropy change.  (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.  Thus, for example, if the actual process path were adiabatic, the reversible path would not need to be adiabatic.)

So, mathematically, we can now state the Second Law as follows:

$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_{Int}(t)}dt}\leq\Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}$$
where $\dot{q}_{rev}(t)$ is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the system temperature at time t (which, for a reversible path, matches the temperature at the interface with the surroundings ##T_{Int}(t)##). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.  The relationship is referred to as the Clausius Inequality.

Paper of interest for further learning:
http://arxiv.org/abs/cond-mat/9901352

Tags:
74 replies
1. Khashishi says:

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.

2. Jano L. says:

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
[SIZE=4][I]
The time variation of q˙(t) and w˙(t) between the initial and final states uniquely defines the so-called process path
[/I]

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.[/SIZE]

3. Chestermiller says:

[QUOTE=”DrDu, post: 5120849, member: 210532″]Should read “mathematically precise”.[/QUOTE]
[I]In my judgement[/I], 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, [I]more precisely[/I], 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

4. Chestermiller says:

[QUOTE=”DrDu, post: 5119931, member: 210532″]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.[/quote]
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 [I]beginning thermodynamics students[/I] 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.
[quote]
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.[/QUOTE]
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

5. Chestermiller says:

[QUOTE=”nothingkwt, post: 5106070, member: 479324″]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?[/QUOTE]
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 [I]isolated[/I] 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

6. Chestermiller says:

[QUOTE=”wvphysicist, post: 5105767, member: 451472″]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[SUB]rev[/SUB]/T.
I cannot understand the symbols in the integral.[/QUOTE]
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

7. Chestermiller says:

[QUOTE=”wvphysicist, post: 5105765, member: 451472″]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?[/QUOTE]
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

8. Chestermiller says:

[QUOTE=”DocZaius, post: 5102237, member: 60142″]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, [I]but the change in entropy would still be equal to the (greater) evaluation of the reversible path integral[/I].

Is that right?[/QUOTE]
Perfect.

9. DocZaius says:

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, [I]but the change in entropy would still be equal to the (greater) evaluation of the reversible path integral[/I].

Is that right?

10. Chestermiller says:

[QUOTE=”wvphysicist, post: 5101674, member: 451472″]
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?[/quote]
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?)

[quote]
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?[/quote]
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[SUB]rev[/SUB]/T.
[quote]
Oh, one more please. Can you show an example where the entropy change is negative like you were saying?[/QUOTE]
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

11. Chestermiller says:

[QUOTE=”wvphysicist, post: 5100183, member: 451472″]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.[/QUOTE]
Nope, for a closed system undergoing a reversible change, the entropy change [B][I]of the system[/I][/B] is an upper bound. And for a perfectly reversible process, the entropy change [I][B]for the system[/B][/I] (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

12. wvphysicist says:

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.

13. Chestermiller says:

[QUOTE=”nitsuj, post: 5099732, member: 315327″]thanks Chet, so ∫ is more complicated than it looks :smile:[/QUOTE]
Hopefully not to those who have had calculus.:smile:

14. nitsuj says:

[QUOTE=”Chestermiller, post: 5099362, member: 345636″]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[/QUOTE]

thanks Chet, so ∫ is more complicated than it looks :smile:

15. Chestermiller says:

[QUOTE=”nitsuj, post: 5099280, member: 315327″]What does the ∫ symbol in the thermodynamics equations mean? [/quote]
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.
[quote]
(i know what + – * / mean :) Oh and brackets)[/QUOTE]
Brackets are a kind of parenthesis.

Chet

16. nitsuj says:

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

17. Jimster41 says:

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?

18. Chestermiller says:

[QUOTE=”Jimster41, post: 5097215, member: 517770″]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”.[/QUOTE]
My goal was to emphasize the classical approach to entropy in my development, and to generally skip the statistical thermodynamic perspective.

Chet

19. Jimster41 says:

[QUOTE=”Chestermiller, post: 5095658, member: 345636″]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[/QUOTE] 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".

20. Chestermiller says:

[QUOTE=”insightful, post: 5095750, member: 552180″]So you’re saying there is a temperature gradient between the “bulk” system and the interface [/quote]
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.
[quote]
, but no temperature gradient between the “bulk” surroundings and the interface?[/QUOTE]
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

21. Chestermiller says:

[QUOTE=”insightful, post: 5095723, member: 552180″]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?[/QUOTE]
What you’re missing is that, at the interface, the local temperature of the system [I]matches[/I] 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

22. Chestermiller says:

[QUOTE=”Jimster41, post: 5095614, member: 517770″]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 sytem 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.[/QUOTE]
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

23. Jimster41 says:

[QUOTE=”Chestermiller, post: 5095596, member: 345636″]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 [I]isolated system[/I], correct?

Chet[/QUOTE]
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 sytem 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.

24. Chestermiller says:

[QUOTE=”Jimster41, post: 5095578, member: 517770″]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 sytem is not always greater than or equal to 0? I think I know (Poincare’ recurrence?) but I also think I’m probably wrong.[/QUOTE]
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 [I]isolated system[/I], correct?

Chet

25. Chestermiller says:

[QUOTE=”INFO-MAN, post: 5095433, member: 553300″]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.[/quote]
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:

[INDENT]In this author’s judgement, the primary cause of the ([I]students'[/I]) 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.
[/INDENT]

[quote]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 (Δ[I]S[/I] > 0) or zero (Δ[I]S [/I]= 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[URL=’http://en.wikipedia.org/wiki/Isentropic_process’]. [/URL]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.[/quote]
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 (Δ[I]S[/I] > 0) and equality (Δ[I]S [/I]= 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.
[quote]
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 (Δ[I]S[/I] > 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 (Δ[I]S[/I] > 0), since friction and viscosity are always present.[/quote]
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 [I]irreversible[/I] processes paths and ideal [I]reversible[/I] 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.
[quote]
Here is my favorite example of an irreversible thermodynamic process, the Entropy Rate Balance Equation for Control Volumes:

[IMG]https://www.ecourses.ou.edu/ebook/thermodynamics/ch06/sec067/media/eq060701.gif[/IMG][/quote]
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.

[quote]
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) “[U]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[/U].” 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.” [URL]http://www.lawofmaximumentropyproduction.com/[/URL][/quote]

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

26. Jimster41 says:

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

[URL]http://arxiv.org/abs/cond-mat/9901352[/URL]

[SIZE=6][B]The Entropy Production Fluctuation Theorem and the Nonequilibrium Work Relation for Free Energy Differences[/B][/SIZE]
[URL=’http://arxiv.org/find/cond-mat/1/au:+Crooks_G/0/1/0/all/0/1′]Gavin E. Crooks[/URL]
(Submitted on 29 Jan 1999 ([URL=’http://arxiv.org/abs/cond-mat/9901352v1′]v1[/URL]), 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.

27. INFO-MAN says:

Hello Chestermiller.
[I]
“There have been nearly as many formulations of the second law as there have been discussions of it.”[/I]
~P. W. Bridgman

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.

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 (Δ[I]S[/I] > 0) or zero (Δ[I]S [/I]= 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[URL=’http://en.wikipedia.org/wiki/Isentropic_process’]. [/URL]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 (Δ[I]S[/I] > 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 (Δ[I]S[/I] > 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:

[IMG]https://www.ecourses.ou.edu/ebook/thermodynamics/ch06/sec067/media/eq060701.gif[/IMG]

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) “[U]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[/U].” 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.” [URL]http://www.lawofmaximumentropyproduction.com/[/URL]

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

28. Chestermiller says:

[QUOTE=”Jimster41, post: 5094494, member: 517770″]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 [USER=174065]@techmologist[/USER] 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.[/QUOTE]
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

29. Jimster41 says:

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 [USER=174065]@techmologist[/USER] 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.

30. Chestermiller says:

[QUOTE=”Jimster41, post: 5094209, member: 517770″]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?[/QUOTE]
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

31. Jimster41 says:

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?

32. Chestermiller says:

Thanks Jimster.

[QUOTE=”Jimster41, post: 5093960, member: 517770″]Thanks Chester.
Yes. I really did find that clear.

Which is not to say I understood it…

What is special about the reversible paths? [/quote]
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.
[quote]
Are all the other paths, the non-reversible ones, the same, or do some integrate to different values <= DeltaS than others?[/QUOTE] They integrate to different values < ΔS. The equal sign does not apply to irreversible paths. They are all less.Chet

33. anorlunda says:

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.”

34. Jimster41 says:

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?

35. DrDu says:

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

36. DrDu says:

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.

37. GM Jackson says:

Thanks for your definitive take on thermodynamics one and two.

38. nothingkwt says:

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?

39. wvphysicist says:

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[SUB]rev[/SUB]/T.I cannot understand the symbols in the integral.

40. wvphysicist says:

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?

41. wvphysicist says:

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?

42. insightful says:

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?

43. insightful says:

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?

44. Jimster41 says:

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

45. MexChemE says:

Excellent article! One of the best definitions of entropy and the second law I’ve ever read.

46. davidbenari says:

In the analysis of mixtures, we have that for ideal mixtures ##\Delta_mix H=0##. So I think it could be argued that the entropy change for ideal mixtures is zero, according to ##dS=\frac{dq}{dT}##. However, this is not the case and in fact the entropy is given by ##-nR\Sum_i x_i ln x_i ##

How can I resolve this?

I’m not sure if this is the kind of reply that is expected here, so I would look to know that too hehe.

Thanks.

47. Evanish says:

This is probably a stupid question but what is dt? It appears in most of the equations, but I can’t find it’s definition in the article.