Entropy is a measure of energy availiable for work ?

[Rap]

Entropy is a measure of energy availiable for work ?

"Entropy is a measure of energy availiable for work". Can someone explain this to me? Give some examples that show in what sense it is true. It has to come with a lot of caveats, proviso's etc. because its simply not true on its face.

I mean, if I have a container of gas at some temperature above 0 K, then I can extract all of its internal energy as work, just let it quasistatically expand to infinity.

 

[Studiot]

I mean, if I have a container of gas at some temperature above 0 K, then I can extract all of its internal energy as work, just let it quasistatically expand to infinity.

Such a container has internal pressure P.

Expanding from P into a vacuum does no work.
Expanding from P against an external pressure P' < P does work, but as this happens P diminishes until P = P' when the system is in equilibrium.

How would you proceed from this equilibrium to infinity, where P = 0 ?

 

[Rap]

Such a container has internal pressure P.

Expanding from P into a vacuum does no work.
Expanding from P against an external pressure P' < P does work, but as this happens P diminishes until P = P' when the system is in equilibrium.

How would you proceed from this equilibrium to infinity, where P = 0 ?

Yes, you would need initial pressure (P) greater than zero, and ambient pressure (P') equal to zero, i.e. in outer space. But the point remains, the statement that "entropy is a measure of energy unavailable for work" is contradicted by this example.

 

[Studiot]

Expanding from P into a vacuum does no work.

This is fundamental, pushing against something that offers no resistance does no work.

 

[russ_watters]

"Entropy is a measure of energy availiable for work".

Where did you get that quote? It is wrong: it is missing the word "not"!

Entropy is ... the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.

http://en.wikipedia.org/wiki/Entropy

 

[Studiot]

Where did you get that quote? It is wrong: it is missing the word "not"!

Agreed - I think that was a typo -, but we also need to dispel the misconception in the proposed counterexample that follows.

Once that is done it is easy to explain the correct reasoning.

 

[Rap]

Yes, sorry, I misquoted it. It should be something like "Entropy is a measure of the energy NOT available for useful work". I also see the point that the example I gave is not good. In order for the expansion to be slow, there has to be an opposing force almost equal to the pressure force, and that force has to diminish as the volume increases (and pressure decreases). This opposing force would be the mechanism by which work was done on the environment. Something like a mass in a gravitational field, in which the mass is slowly being reduced by removal.

I found a website http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node48.html which seems to give an explanation, I will have to look at it closely.

Also the Wikipedia quote says it is "energy per unit temperature" not available for work, which I cannot immediately decipher.

 

[Rap]

Looking at the above site, it seems to me what it is saying is that the amount of energy unavailable for useful work in a Carnot cycle is equal to the entropy extracted from the hot reservoir (which is equal to the entropy deposited in the cold reservoir) times the temperature of the cold reservoir. How you get from this to the idea that "Entropy is a measure of the energy unavailable for work" still eludes me.

Entropy of what? I tried assuming that the hot reservoir was actually a hot system with a finite amount of energy and entropy. Again, you can prove that if you extract entropy ∆S from the hot body, the amount of unavailable energy is Tc ∆S where Tc is the temperature of the cold reservoir. But you can only extract so much entropy from the hot body using a working body at the cold reservoir temperature. If their temperatures are very close, you can extract very little entropy and energy. The great majority of the internal energy of the hot body is unavailable for work in this case.

If you set the cold reservoir to zero degrees, then the amount of energy unavailable for work is zero. You could extract all of the internal energy of the hot body as work. (right?).

I still don't get it.

 

[Andrew Mason]

Looking at the above site, it seems to me what it is saying is that the amount of energy unavailable for useful work in a Carnot cycle is equal to the entropy extracted from the hot reservoir (which is equal to the entropy deposited in the cold reservoir) times the temperature of the cold reservoir. How you get from this to the idea that "Entropy is a measure of the energy unavailable for work" still eludes me.

Entropy of what? I tried assuming that the hot reservoir was actually a hot system with a finite amount of energy and entropy. Again, you can prove that if you extract entropy ∆S from the hot body, the amount of unavailable energy is Tc ∆S where Tc is the temperature of the cold reservoir. But you can only extract so much entropy from the hot body using a working body at the cold reservoir temperature. If their temperatures are very close, you can extract very little entropy and energy. The great majority of the internal energy of the hot body is unavailable for work in this case.

If you set the cold reservoir to zero degrees, then the amount of energy unavailable for work is zero. You could extract all of the internal energy of the hot body as work. (right?).

I still don't get it.

We have hashed this over before. The problem with "entropy is a measure of the energy unavailable for work" is that standing alone it is not clear and open to a number of interpretations. It requires a special definition of "energy unavailable for work" as the explanation shows. Without that explanation it can easily lead to misinterpretation.

For example, between two temperatures, Th and Tc, the heat flow Q is capable of producing an amount of work, W = Q(1-Tc/Th) ie. with a Carnot engine. So there is energy E = QTc/Th energy that is, in that sense, "unavailable" for doing work. Yet, as we know, ΔS = 0 for a Carnot engine. So one could well ask, how is 0 a measure of QTc/Th? The answer is: "well, that is not what we mean by 'energy unavailable to do work'. We really mean 'lost work' which is the potential work that could be extracted minus the work that was actually extracted, or the amount of additional work required to restore the system and surroundings to their original state if you saved the output work and used it to drive the process in reverse." Hence the confusion.

So, as I have said before, this particular statement should not be used to introduce the concept of entropy. By itself it explains nothing and leads to great confusion.

AM

 

[Rap]

We have hashed this over before. The problem with "entropy is a measure of the energy unavailable for work" is that standing alone it is not clear and open to a number of interpretations. It requires a special definition of "energy unavailable for work" as the explanation shows. Without that explanation it can easily lead to misinterpretation.

For example, between two temperatures, Th and Tc, the heat flow Q is capable of producing an amount of work, W = Q(1-Tc/Th) ie. with a Carnot engine. So there is energy E = QTc/Th energy that is, in that sense, "unavailable" for doing work. Yet, as we know, ΔS = 0 for a Carnot engine. So one could well ask, how is 0 a measure of QTc/Th? The answer is: "well, that is not what we mean by 'energy unavailable to do work'. We really mean 'lost work' which is the potential work that could be extracted minus the work that was actually extracted, or the amount of additional work required to restore the system and surroundings to their original state if you saved the output work and used it to drive the process in reverse." Hence the confusion.

So, as I have said before, this particular statement should not be used to introduce the concept of entropy. By itself it explains nothing and leads to great confusion.

AM

Well, I kind of thought that was case, but there was also the possibility I was missing something. Thanks for the clarification.

 

[Studiot]

The time for hand waving is over, here is some mathematics.

Consider a universe consisting of a system contained in a heat bath or reservoir at uniform constant temperature T.

Consider changes in the function Z = entropy of bath plus entropy of the system = entropy of this universe.

dZ = dS_b + dS_s....1

Where b refers to the bath and s refers to the system.

If the system absorbs heat dq the same amount of heat is lost by the bath so the entropy change

dS_b =-dq/T......2

substituting this into equation 1

dZ = dS_s - dq/T.....3

Now consider a change of state of the system from state A to state B

By the first law

dU_s = dq - dw .....4

Combining equations 3 and 4 and rearranging

dS_s = (dU_s+dw)/T + dZ

re-arranging

dw = TdS_s - dU_s -TdZ

Since TdZ is always a positive quantity or zero

dw ≤ TdS_s - dU_s

Where dw is the work done by the system

This is the principle of maximum work and calculates the maximum work that can be obtained from the system. As you can see it has two components vis from the entropy created and from the change in internal energy and these act in opposite directions so in this sense the TdS_s term reduces the amount of work obtainable from the internal energy of the system and accounts for unavailable energy since TdS_s has the dimensions of energy.

Note the usual caveat
The inequality refers to irreversible processes, the equality to reversible ones.

 

[Darwin123]

"Entropy is a measure of energy availiable for work". Can someone explain this to me? Give some examples that show in what sense it is true. It has to come with a lot of caveats, proviso's etc. because its simply not true on its face.

I mean, if I have a container of gas at some temperature above 0 K, then I can extract all of its internal energy as work, just let it quasistatically expand to infinity.

"Entropy is a measure of energy availiable for work". Can someone explain this to me? Give some examples that show in what sense it is true. It has to come with a lot of caveats, proviso's etc. because its simply not true on its face.

I mean, if I have a container of gas at some temperature above 0 K, then I can extract all of its internal energy as work, just let it quasistatically expand to infinity.

What you said is only possible if the gas expands both adiabatically and reversibly. In an adiabatic and reversible expansion, the change in entropy of the gas is zero. Under that condition, one could turn all the internal energy into work.

Any deviation from the conditions of adiabatic and reversible would result in some internal energy not being turned to work.


First, I prove that one can extract all the internal energy from a monotonic ideal gas using an expansion that is BOTH adiabatic and reversible.

Suppose one were to take an ideal gas in a closed chamber and expand it both adiabatically and slowly, so that it is in a state near thermal equilibrium at all times. No entropy goes in or out of the chamber.

At the end of the expansion, even in the limit of infinite volume, you would end up with a gas of finite temperature.

The ideal gas law is:
1) PV=nRT
where P is the pressure of the gas, V is the volume of the chamber, n is the molarity of the gas, R is the gas constant and T is the temperature.

The internal energy of the ideal gas is:
2) U=(3/2)nRT
where U is the internal energy of the gas and everything else is the same.


Substituting equation 2 into equation 1:
3)U=(3/2)PV

Before the gas starts expanding, let P=P0, V=V0, T=T0, and U=U0. The chamber is closed, so "n" is constant the entire time. There are three degrees of freedom for each atom in a mono atomic gas. Therefore, for a mono atomic gas:
3) U0=1.5 P0 V0

The adiabatic expansion of an mono-atomic gas is:
4) P0 V0 ^(5/3)= P V^(1.5)

Therefore,
5) P = P0 (V0/V)^(5/3)

The work, W, done by the gas is
6) W = ∫[V0→∞] P dV

Substituting equation 5 into equation 6:
7) W = (P0 V0^(5/3)))∫[V0→∞] V^(-5/3) dV

Evaluating the integral in equation 6:
8) W = (3/2)(P0 V0^5/3)V0^(-2/3)

9) W=(1.5) P0 V0

The expression for W in equation 9 is the same as the expression for U0 in equation 3. Therefore, the internal energy has been taken out completely.

I’ll solve the problem later (a week or so) for an expansion with sliding friction. There, the increase in entropy characterizes the amount of internal energy not turned into work. However, I will set up the problem. I will show the two equations that makes the case with sliding friction different from the reversible condition.

One remove the reversibility condition by adding sliding friction,
10) Wf = ∫[V0→Vf] (P-Pf) dV
where Pf is the pressure due to sliding friction and Vf is the volume when the P=Pf. The chamber stops expanding when P=Pf.


However, another expression is necessary to describe how much entropy is created.
11) dQ = Pf dV

Equation merely says that the energy used up by the sliding friction causes entropy to be created. The heat energy, dQ, is the energy used up by friction.


I don’t have time now, so I leave it as an exercise. Honest, moderator, I promise to get back to it. However, he wants an example where the creation of entropy limits the work that can be extracted. This is a good one.


Spoiler

Wf<W. Not all the internal energy is turned into work with sliding friction included. Let Q be the work done by the sliding force alone. The increase in entropy is enough to explain why the internal energy is not being turned into work.

 

[Studiot]

Any deviation from the conditions of adiabatic and reversible would result in some internal energy not being turned to work.

Do you not agree that the maximum possible work is extracted in a reversible isothermal expansion?

The adiabatic expansion of an mono-atomic gas is:
4) P0 V0 ^(5/3)= P V^(1.5)

Are you sure you mean this: you have different values of gamma on each side?

 

[Darwin123]

Do you not agree that the maximum possible work is extracted in a reversible isothermal expansion?

In an isothermal expansion, energy is entering the gas from a hot reservoir. Therefore, one can't say that one is extracting the energy from the ideal gas. Most of the energy is being extracted from the hot reservoir, not from the ideal gas.

In the corresponding isothermal expansion, the ideal gas is acting like a conduit for energy and entropy. The ideal gas is not acting as a storage matrix for the energy.

The OP was saying that the work was being extracted from the internal energy that was initially embedded in the ideal gas. All work energy comes from the container of gas, not outside reservoirs. So the walls of the container have to be thermal insulators.

If you allow heat energy to conduct through the walls of the container, then the work may exceed the initial value of the internal energy. The hot reservoir can keep supplying energy long after the internal energy is used up.

So I stick to my guns with regards to the specificity of the OPs hypothesis. He was unconsciously assuming that the expansion is both adiabatic and reversible.

I diagree with those people who said that the gas would remain the same temperature during the expansion, and that not all the energy could be extracted from the internal energy. I showed that the internal energy could be entirely extracted for an ideal monotonic gas under adiabatic and reversible conditions.

The big problem that I have with the OP's question is with the word "quasistatic". I conjecture that the OP thought that "quasistatic" meant "both adiabatic and reversible".

A quasistatic process can be both nonadiabatic and irreversible. However, "quasistatic" is a useful hypothesis. The word "quasistatic" implies spatial uniformity. In this case, the word quasistatic implies that the temperature and the pressure of the ideal gas is uniform in the chamber.

Quasistatic implies that enough time has passed between steps that both temperature and pressure are effectively constant in space. Thus, temperature is not a function of position. Pressure is not a function of position.

Are you sure you mean this: you have different values of gamma on each side?

I think this is correct. I didn't spend much time checking my work. If you see an arithmetic blunder, feel free to correct me.

Also, I specified a specific case to simplify the problem. So even if I did it correctly, the gamma value that I used was atypical. Next time, I will let gamma be a parameter of arbitrary value.

I specified a monotonic gas. The gas is comprise of individual atoms. There are no internal degrees of freedom in these atoms. Any correlation between coefficients may be due to my choice.

There are many ways an expansion can be irreversible. I think the one most people think of is where the expansion is not quasistatic. Suppose the gas is allowed to expand freely, so that temperature and pressure are not uniform. This is irreversible. However, the mathematics is way beyond my level of expertise.

The problem is tractable if the process is quasistatic. So, I think the best thing would be to show how it works with an quasistatic but irreversible. For instance, what happens if one turns on the sliding friction and the static friction in this expansion. That would result in a process where entropy is created. In other words, friction would result in an irreversible expansion even under quasistatic conditions.

I don't have time now. I will post a solution to that later. For now, just remember that not all quasistatic processes are reversible.

 

[Studiot]

I wanted to be more subtle and polite but


{P_1}{V_1}^\gamma = {P_2}{V_2}^\gamma


Whereas you have


{P_0}{V_0}^{\frac{5}{3}} = {P_1}{V_1}^{\frac{3}{2}}

I diagree with those people who said that the gas would remain the same temperature during the expansion

Is this a rejection of Joule's experiment and the definition of an ideal gas?
If so you should make it clear that your view is not mainstream physics.

It should be noted that Joules experiment was both adiabatic and isothermal and has been repeated successfully many times.

You are correct in observing that during an isothermal expansion neither q nor w are zero.
However what makes you think the internal energy is the same at the beginning and end, in the light of your above statement?

One definition (or property derivable from an equivalent definition) of an ideal gas is that its internal energy is a function of temperature alone so if the temperature changes the internal energy changes. To remove all the internal energy you would have to remove all the kinetic energy of all its molecules.

 

[Andrew Mason]

I wanted to be more subtle and polite but {P_1}{V_1}^\gamma = {P_2}{V_2}^\gammaWhereas you have{P_0}{V_0}^{\frac{5}{3}} = {P_1}{V_1}^{\frac{3}{2}}

I think it was a typo. Darwin did write: P = P0 (V0/V)^(5/3) a little farther down.

Is this a rejection of Joule's experiment and the definition of an ideal gas?
If so you should make it clear that your view is not mainstream physics.

I am not sure that you are both talking about the same process. Darwin was referring to a quasi-static adiabatic expansion of an ideal gas. Temperature is given by the adiabatic condition:

TV^{(\gamma - 1)} = \text{constant}

So, if volume changes this cannot be isothermal. Temperature has to change.

I think you (Studiot) may be talking about free expansion, not quasi-static expansion, in which case T is constant for an ideal gas.

AM

 

[Rap]

Studiot - I agree with your derivation but with regard to the OP, the correct statement would then be: "an infinitesimal change in entropy is a measure of the minimum infinitesimal amount of energy unavailable for work given a particular ambient temperature."

Darwin123 - Thank you for clarifying the muddled OP. Depending on conditions, all, some, or none of a system's internal energy can be converted to work, and so the statement "Entropy is a measure of the energy unavailable for work" is ambiguous at best, wrong at worst. You are right, I was assuming adiabatic and reversible. Adiabatic or else you are potentially using energy from somewhere else to do the work. Reversible, because it will give the miniumum amount of work unavailable. I should have said that instead of quasistatic. I take quasistatic to mean, by definition, a process can be described as a continuum of equilibrium states.

 

[Darwin123]

I wanted to be more subtle and polite but


{P_1}{V_1}^\gamma = {P_2}{V_2}^\gamma

Oops. My typo. I meant,

I {P_0}{V_0}^{\frac{5}{3}} = {P_1}{V_1}^{\frac{5}{3}}

Is this a rejection of Joule's experiment and the definition of an ideal gas?
If so you should make it clear that your view is not mainstream physics.

No, it was a typo. Certainly not mainstream physics. However, I did not insert that mistake into my later equations.

It should be noted that Joules experiment was both adiabatic and isothermal and has been repeated successfully many times.

Good for Joule! More power to him!

You are correct in observing that during an isothermal expansion neither q nor w are zero.

However what makes you think the internal energy is the same at the beginning and end, in the light of your above statement?

The internal energy of an ideal gas varies only with its temperature. The internal energy is not explicitly determined by either its pressure or its temperature. If you know how many atoms are in a molecule of the ideal, the number of molecules and the temperature, then you can uniquely determine the internal energy.

The quantity of the ideal gas in the closed container is constant. The number of atoms per molecule is constant. For an isothermal expansion, the temperature of the ideal gas is constant.

Therefore, the internal energy of the ideal gas is constant for an isothermal expansion. The internal energy never changes during the entire expansion, even in the limit of infinite volume. However, work is being done for the entire time. Therefore, the work can't come from the internal energy.

An isothermal expansion is actually the most inefficient way to use the internal energy of the ideal gas. None of the internal energy of the ideal gas becomes work in an isothermal expansion. All heat energy absorbed by the ideal gas instantly turns into work on the surroundings.

In an isothermal expansion, not a single picojoule of work comes from the internal energy of the gas. It all comes from the heat reservoir connected to the ideal gas.

One should note that the total amount of work done by the gas in the isothermal expansion is infinite. The work done by the gas increases with the logarithm of volume. So an infinite volume means that an infinite amount of work is performed. Do the calculations for yourself.

Obviously, the infinite energy that is going to become work is not in the internal energy of the gas. In fact, the internal energy of the gas remains the same even in the extreme limit of infinite work.

The energy for work is being supplied by the heat reservoir, not the ideal gas. In order to maintain a constant temperature, the container has to be in thermal contact with

One definition (or property derivable from an equivalent definition) of an ideal gas is that its internal energy is a function of temperature alone so if the temperature changes the internal energy changes. To remove all the internal energy you would have to remove all the kinetic energy of all its molecules.

Therefore, the kinetic energy of the molecules in the gas can not change during an isothermal expansion.

If you want to extract all the internal energy of the ideal gas to work on the surroundings, then you have to decrease the temperature to absolute zero. This can be done in an adiabatic expansion. It can't be done in an isothermal expansion.

Entropy can change only two ways. It can move or it can be created. In an adiabatic process, entropy can't move. The temperature changes instead. In an isothermal process, the entropy moves in such a way as to keep the temperature constant.

 

[Rap]

Entropy can change only two ways. It can move or it can be created. In an adiabatic process, entropy can't move. The temperature changes instead. In an isothermal process, the entropy moves in such a way as to keep the temperature constant.

Note, that is only for a simple system (homogeneous).

For complex systems contained inside a thermally insulating boundary, entropy may move around inside the system, driven by temperature differences inside the system, equalizing them when possible, but can never be transferred across the boundary. During these internal sub-processes, entropy may also be created. In an isothermal process, the boundary is thermally open, and entropy may move across the boundary, again driven by temperature differences between the system and the environment, in such a way as to equalize internal temperatures at the constant temperature of the environment, when possible.

Entropy transfer goes hand in hand with energy transfer via dU=T dS. If a process is converting energy to work, and you want to know how much of that energy is converted to work, in order to keep the bookwork straight, you cannot bring in energy or entropy from somewhere else to accomplish that work. To ensure this, the process has to be adiabatic, i.e. inside a thermally insulating boundary which prevents entropy and energy coming in from somewhere else.

For finite temperature differences, transfer of entropy across a thermally open boundary causes creation of entropy at the boundary which is transferred to the lower temperature system. The transfer of entropy is of order ∆T, the creation of entropy at the boundary is of order ∆T^2, so, in the limit of small ∆T, entropy may be transferred without creation at the boundary. If ∆T is identically zero, there will be no transfer of entropy, since only temperature differences will drive entropy transfer.

 

[Andrew Mason]

Entropy can change only two ways. It can move or it can be created. In an adiabatic process, entropy can't move. The temperature changes instead. In an isothermal process, the entropy moves in such a way as to keep the temperature constant.

I think you have to qualify this statement. You have to speaking about a reversible process. Entropy can and certainly does change in an adiabatic irreversible expansion.

And I wouldn't say it moves because entropy is not a conserved quantity such as energy or momentum. We can speak of energy or momentum transfer because the loss of energy/momentum must result in the gain of energy/momentum of some other body so it behaves as if it moves. Entropy does not behave like. So I would suggest that the concept of entropy moving is not a particularly helpful one.

In a reversible isothermal process total entropy change is 0. In a real isothermal process, the entropy of the system + surroundings inevitably increases. It is not entropy that moves. It is energy. And the faster the energy moves, the greater the increase in entropy. So I might suggest that entropy increase is related more to the speed of energy transfer (heat flow) than to the fact that a body remains at the same temperature.

AM

 

[Rap]

I think you have to qualify this statement. You have to speaking about a reversible process. Entropy can and certainly does change in an adiabatic irreversible expansion.

And I wouldn't say it moves because entropy is not a conserved quantity such as energy or momentum. We can speak of energy or momentum transfer because the loss of energy/momentum must result in the gain of energy/momentum of some other body so it behaves as if it moves. Entropy does not behave like. So I would suggest that the concept of entropy moving is not a particularly helpful one.

In a reversible isothermal process total entropy change is 0. In a real isothermal process, the entropy of the system + surroundings inevitably increases. It is not entropy that moves. It is energy. And the faster the energy moves, the greater the increase in entropy. So I might suggest that entropy increase is related more to the speed of energy transfer (heat flow) than to the fact that a body remains at the same temperature.

AM

When entropy is conserved, i.e. when a process is reversible, the concept of entropy moving is just as valid as the concept of conserved energy moving, and is therefore very helpful. If you can identify the location at which entropy is created (e.g. at the thermally open boundary between two systems with a finite temperature difference), and identify into which system the created entropy is deposited, then I see no conceptual problem with Dirac123's statement resulting from the non-conservation of entropy. Not yet being totally certain of this concept, I would be interested in any specific process you can think of where this concept is inappropriate.

 

[Andrew Mason]

Note, that is only for a simple system (homogeneous).
Entropy transfer goes hand in hand with energy transfer via dU=T dS. If a process is converting energy to work, and you want to know how much of that energy is converted to work, in order to keep the bookwork straight, you cannot bring in energy or entropy from somewhere else to accomplish that work. To ensure this, the process has to be adiabatic, i.e. inside a thermally insulating boundary which prevents entropy and energy coming in from somewhere else.

Is that really correct? It seems to me that a thermally insulating boundary does not prevent entropy or energy "coming in" from somewhere else: it only prevents heat flow across the boundary.

When entropy is conserved, i.e. when a process is reversible, the concept of entropy moving is just as valid as the concept of conserved energy moving, and is therefore very helpful. If you can identify the location at which entropy is created (e.g. at the thermally open boundary between two systems with a finite temperature difference), and identify into which system the created entropy is deposited, then I see no conceptual problem with [STRIKE]Dirac123[/STRIKE]'s [Darwin123] statement resulting from the non-conservation of entropy. Not yet being totally certain of this concept, I would be interested in any specific process you can think of where this concept is inappropriate.

It is not necessarily inappropriate and it may be a helpful concept when analysing reversible processes in which entropy does not change.

But I think that, overall, the concept of entropy "moving" just adds to confusion about an already very difficult concept. It gives the impression that entropy is a physical quantity. Entropy is a statistical. It is a bit like temperature in that respect. We would not say that temperature moves between bodies or is lost or created.

We do say that Q (= mass x temperature x heat capacity) moves or is lost or created (heat flow) but we relate it to U and W, which together are always conserved. You can't do that with entropy.

AM

 

[Rap]

Is that really correct? It seems to me that a thermally insulating boundary does not prevent entropy or energy "coming in" from somewhere else: it only prevents heat flow across the boundary.

It is not necessarily inappropriate and it may be a helpful concept when analysing reversible processes in which entropy does not change. AM

I think it is correct: dQ=TdS means the flow of heat energy into a boundary is the temperature of the "in" side of the boundary times the entropy flow into the boundary. The flow of heat energy out of a boundary is temperature of the "out" side of the boundary times the entropy flow out of the boundary. No heat flow (dQ=0) means no entropy flow (dS=0) and vice versa. Energy into the boundary equals energy out, so if the temperatures are different, entropy in does not equal entropy out - entropy is being created at the boundary and deposited in the low temperature system.

But I think that, overall, the concept of entropy "moving" just adds to confusion about an already very difficult concept. It gives the impression that entropy is a physical quantity. Entropy is a statistical. It is a bit like temperature in that respect. We would not say that temperature moves between bodies or is lost or created.

We do say that Q (= mass x temperature x heat capacity) moves or is lost or created (heat flow) but we relate it to U and W, which together are always conserved. You can't do that with entropy.

AM

Temperature is also statistical, but it is an intensive quantity, not extensive. The concepts of "conserved" and "created" and "destroyed" do not apply to intensive quantities like temperature, pressure, chemical potential. They do apply to extensive quantities. For a simple system undergoing a reversible process,

1. Volume is conserved in a mechanically isolated system and volume changes are driven by pressure differences alone.

2. Particle number is conserved in a materially isolated system and particle number changes are driven by chemical potential differences alone.

3. Similarly, entropy is conserved in a thermally isolated system, entropy flows are driven by temperature differences alone.

If X is intensive, then differences in X drive changes in an extensive Y, where X and Y are conjugate variables. T and S, P and V, µ and N, are conjugate variables. The product of X and dY has units of energy and the fundamental law states that the sum of all those products is a conserved, extensive internal energy dU=TdS-PdV+µdN. (No worry about the sign of PdV)

If you relax the above constraints on the systems, conservation becomes more problematical. Volume is still always conserved in a mechanically isolated system. For a homogeneous system in which there may be chemical reactions, particle number is not conserved in a materially isolated system (although conservation of "component particles" is). Similarly, in an irreversible process, entropy is not conserved, it is created, in a thermally isolated system.

I think the simple fact that entropy is extensive implies that the concept of entropy flow, entropy creation at particular locations, entropy density, etc. is viable. I don't think it adds confusion, I think it brings insight and clarity to the concept of "classical entropy". If its true, I fully expect statistical mechanics to verify this, rather than muddying the waters.

 

[Darwin123]

I think it is correct: dQ=TdS means the flow of heat energy into a boundary is the temperature of the "in" side of the boundary times the entropy flow into the boundary. The flow of heat energy out of a boundary is temperature of the "out" side of the boundary times the entropy flow out of the boundary. No heat flow (dQ=0) means no entropy flow (dS=0) and vice versa. Energy into the boundary equals energy out, so if the temperatures are different, entropy in does not equal entropy out - entropy is being created at the boundary and deposited in the low temperature system.



Temperature is also statistical, but it is an intensive quantity, not extensive. The concepts of "conserved" and "created" and "destroyed" do not apply to intensive quantities like temperature, pressure, chemical potential. They do apply to extensive quantities. For a simple system undergoing a reversible process,

1. Volume is conserved in a mechanically isolated system and volume changes are driven by pressure differences alone.

2. Particle number is conserved in a materially isolated system and particle number changes are driven by chemical potential differences alone.

3. Similarly, entropy is conserved in a thermally isolated system, entropy flows are driven by temperature differences alone.

This I disagree with, at least in the OP's scenario.

The OP said that the expansion is quasistatic. This means that the expansion is so slow, the ideal gas is in a state infinitesimally close to thermal equilibrium. This means that variation of temperature and pressure within the ideal gas are negligible. If there is a significant amount of entropy created, it is not from the temperature differences in the gas.

Similarly, there can't be any coherent sound energy. The gas can't "squeak". A coherent sound wave would imply that the macroscopic pressure and the macroscopic temperature were inhomogeneous.

The only thing that we can be sure of from his description is that the expansion is adiabatic. Otherwise, work won't entirely come from the internal energy of the gas. Therefore, I presume that the container of ideal gas is comprised of a piston and cylinder made of some thermal insulator with zero heat capacity. The pressure outside the container doesn't really matter. It could be zero.

In the OP's scenario, there could be friction between the piston surface and the cylinder surface that contains the gas. Because of heat conduction from surface to gas, the "heat energy" created by the friction goes right back into the ideal gas. Because the expansion is adiabatic, neither entropy nor heat energy can be conducted out of the container. Therefore, work done by the friction can't affect the internal energy of the ideal gas.

In the simplest case that I can imagine, the friction is a combination of sliding friction and static friction. Simple formulas for sliding friction and static friction are taught in introductory physics classes. Although real friction is more complicated, I think this approximation is good enough as an illustration.

I intend to show a calculation for this situation, where all entropy created comes from sliding fiction. In this example, there will be no aerodynamic friction or temperature differences in the ideal gas. The expansion will be adiabatic and quasistatic, but not reversible.

 

[Rap]

This I disagree with, at least in the OP's scenario.

The OP said that the expansion is quasistatic. This means that the expansion is so slow, the ideal gas is in a state infinitesimally close to thermal equilibrium. This means that variation of temperature and pressure within the ideal gas are negligible. If there is a significant amount of entropy created, it is not from the temperature differences in the gas.

Similarly, there can't be any coherent sound energy. The gas can't "squeak". A coherent sound wave would imply that the macroscopic pressure and the macroscopic temperature were inhomogeneous.

The only thing that we can be sure of from his description is that the expansion is adiabatic. Otherwise, work won't entirely come from the internal energy of the gas. Therefore, I presume that the container of ideal gas is comprised of a piston and cylinder made of some thermal insulator with zero heat capacity. The pressure outside the container doesn't really matter. It could be zero.

In the OP's scenario, there could be friction between the piston surface and the cylinder surface that contains the gas. Because of heat conduction from surface to gas, the "heat energy" created by the friction goes right back into the ideal gas. Because the expansion is adiabatic, neither entropy nor heat energy can be conducted out of the container. Therefore, work done by the friction can't affect the internal energy of the ideal gas.

In the simplest case that I can imagine, the friction is a combination of sliding friction and static friction. Simple formulas for sliding friction and static friction are taught in introductory physics classes. Although real friction is more complicated, I think this approximation is good enough as an illustration.

I intend to show a calculation for this situation, where all entropy created comes from sliding fiction. In this example, there will be no aerodynamic friction or temperature differences in the ideal gas. The expansion will be adiabatic and quasistatic, but not reversible.

I guess I don't know which point you disagree with in the quote, but I will try to make the same calculation. A quasistatic expansion of a gas in a thermally and materially isolated system, but mechanically open, with sliding friction. Specifically a cylinder with one circular end (the piston) a movable boundary. I'm guessing the force of friction can be considered constant and opposing the motion of the piston. So the heat energy created by friction is K F x where K is a constant, F is the friction force and x is the distance moved. If the force of the gas on the piston is ever less than F, the piston stops. I will try to do that calculation.

 

[Andrew Mason]

I think it is correct: dQ=TdS means the flow of heat energy into a boundary is the temperature of the "in" side of the boundary times the entropy flow into the boundary.

The dQ is the reversible heat flow, not the actual heat flow.

The flow of heat energy out of a boundary is temperature of the "out" side of the boundary times the entropy flow out of the boundary. No heat flow (dQ=0) means no entropy flow (dS=0) and vice versa.

If the reversible heat flow, dQrev = 0, dS = 0. But for an irreversible adiabatic process ΔS>0.

3. Similarly, entropy is conserved in a thermally isolated system, entropy flows are driven by temperature differences alone.

Entropy is conserved only if all processes within the thermally isolated system are reversible.

AM

 

[Rap]

The dQ is the reversible heat flow, not the actual heat flow.

If the reversible heat flow, dQrev = 0, dS = 0. But for an irreversible adiabatic process ΔS>0.

Yes, and looking at what I said, I failed to specify that the process needed to be quasistatic. Then there is no entropy created in the hot body as a result of energy flowing from the hot body into the boundary. Likewise, there is no entropy created in the cold body as a result of energy flowing out of the boundary into the cold body. In other words, both bodies' equilibria are not disturbed by the energy flow across the boundary. Their state parameters may change, but the state parameters stay homogeneous in both bodies.

There is no entropy creation inside either body, yet the drop in entropy of the hot body is less than the increase of entropy of the cold body as a result of the entropy transfer, so the process is irreversible. Entropy is created in or at the boundary.

If the process were not quasistatic, there would be temperature gradients inside the bodies as well (rather than just at the boundary) and temperature gradients create entropy, so entropy would be created in the bodies themselves, not just at the boundary.

Entropy is conserved only if all processes within the thermally isolated system are reversible.

Right - I stated before I listed those 3 points "For a simple system undergoing a reversible process,"

 

[Andrew Mason]

Yes, and looking at what I said, I failed to specify that the process needed to be quasistatic. Then there is no entropy created in the hot body as a result of energy flowing from the hot body into the boundary. Likewise, there is no entropy created in the cold body as a result of energy flowing out of the boundary into the cold body. In other words, both bodies' equilibria are not disturbed by the energy flow across the boundary. Their state parameters may change, but the state parameters stay homogeneous in both bodies.

There is no entropy creation inside either body, yet the drop in entropy of the hot body is less than the increase of entropy of the cold body as a result of the entropy transfer, so the process is irreversible. Entropy is created in or at the boundary.

I am just trying to understand what you are saying here. It appears that you saying that entropy is created as the result of a quasi-static heat flow. I don't follow this. The only way to make heat flow quasi-static is to have an infinitessimal temperature difference. If that is the case, ΔS = 0 so I don't see entropy being created anywhere. If there is entropy "created" at a boundary, it is not a quasi-static process.

If the process were not quasistatic, there would be temperature gradients inside the bodies as well (rather than just at the boundary) and temperature gradients create entropy, so entropy would be created in the bodies themselves, not just at the boundary.

Temperature gradients are not the only sources of increased entropy. Heat does not have to flow to create entropy eg. the mixing of two different gases.

AM

 

[Darwin123]

I am just trying to understand what you are saying here. It appears that you saying that entropy is created as the result of a quasi-static heat flow. I don't follow this. The only way to make heat flow quasi-static is to have an infinitessimal temperature difference. If that is the case, ΔS = 0 so I don't see entropy being created anywhere. If there is entropy "created" at a boundary, it is not a quasi-static process.

Help us out, here. Maybe you can find us a formal definition of the word "quasistatic".

I assumed that the word “quasistatic” used by the OP was the same as near-equilibrium. A later post by the OP confirmed that he was talking about a process which was the sum of a series of small processes where the state of the ideal gas was near equilibrium.

Entropy can be created even if a system is “near equilibrium” at each infinitesimal step. Maybe you are right in that there is a temperature gradient where the entropy is created. At the point of contact between two surfaces, the temperature gradient magnitude can be very large. However, the inverse of the temperature gradient may not be completely macroscopic. On the length scale determined by the dimensions of the container of gas, the inverse gradient is microscopic. On the length scale determined by the dimensions of a molecule, the inverse gradient is rather large.

Frictional forces do create entropy. Most of the container of gas in an adiabatic expansion can be near thermal equilibrium. The important thing about a system that is “near equilibrium” is that the intensive quantities will be uniform over most of the system. There may be a large temperature in a 1 cubic micron volume at the point of contact. However, the temperature of the ideal gas over most of the volume will be at a single value, T. Similarly, the pressure at the point of contact may be huge, resulting in a large stress at the point of contact. However, one can approximate the pressure over most of the system by a single value, P.

The following article addresses the issue of how friction is treated in a thermodynamic analysis. Note that there was at least one investigator, Rymuza, who examined sliding friction in a near equilibrium process.


http://www.mdpi.com/1099-4300/12/5/1021
“On the Thermodynamics of Friction and Wear―A Review
Abstract: An extensive survey of the papers pertaining to the thermodynamic approach to tribosystems, particularly using the concept of entropy as a natural time base, is presented with a summary of the important contributions of leading researchers.
…
Friction is an energy transformation process. Using a near-equilibrium analysis, one can
demonstrate (see Section 4) how sliding energy is dissipated. Rymuza [31] considers friction as a process that transforms the external mechanical energy to the energy of internal processes. Rymuza proposes that the traditional ‘laws’ of friction is incapable of reflecting its energetic nature and suggests a new parameter called ‘coefficient of friction losses’ (CFL), so as to reflect both the dissipative nature of friction process and simultaneously provide a useful formulation for applicationin engineering practice.
…
As discussed in Sections 4 and 5, the temperature, and particularly temperature gradient within the mating bodies plays an important role in assessment of entropy generation in a tribosystem. Both theoretical and experimental methods have been developed for determination of temperature rise at the contact surface. Blok [69] is credited to be the first researcher who proposed a model for determination of the temperature rise at the surfaces of contacting bodies under boundary lubricated condition.”

The authors point out that the generation of entropy is caused by the large temperature gradient at the point of contact. However, the temperature gradient is caused by friction. The inhomogeneity in temperature is confined to “the point of contact”. I find it reasonable to assume that this is a “quasistatic” case.

Maybe if you find a formal definition of quasistatic, then I would be forced to agree with you. If quasistatic were formally defined by dS=0, then the OP and I are making a mistake. However, I didn’t know that quasistatic was defined so precisely.

Scientists established early that friction can generate caloric (another word for entropy). That famous cannon investigated by Count Rumford was probably an excellent example of a quasistatic system. The cannon bore was placed in water, so that the system was isothermal. The temperature of the cannon and the water it was immersed in was 100 degrees centigrade, and did not fluctuate enough for scientists to measure. Except at the point of contact, that cannon remained at the boiling point of water. Obviously, the temperature of the iron must have been much higher in the region near the point of contact. This contact temperature could not and probably cannot be measured by ordinary thermometry. However, it is enough to know that friction creates entropy.

http://en.wikipedia.org/wiki/Entropy
“Carnot based his views of heat partially on the early 18th century "Newtonian hypothesis" that both heat and light were types of indestructible forms of matter, which are attracted and repelled by other matter, and partially on the contemporary views of Count Rumford who showed (1789) that heat could be created by friction as when cannon bores are machined.
…
In the 1850s and 1860s, German physicist Rudolf Clausius objected to this supposition, i.e. that no change occurs in the working body, and gave this "change" a mathematical interpretation by questioning the nature of the inherent loss of usable heat when work is done, e.g. heat produced by friction.”

The following link provides a free copy of the article. However, you have to request it.

This looks like a great article. I have never seen anyone handle the problem of friction in thermodynamics so thoroughly. I may have to read it carefully before I get back into the thread.

 

[Rap]

I am just trying to understand what you are saying here. It appears that you saying that entropy is created as the result of a quasi-static heat flow. I don't follow this. The only way to make heat flow quasi-static is to have an infinitessimal temperature difference. If that is the case, ΔS = 0 so I don't see entropy being created anywhere. If there is entropy "created" at a boundary, it is not a quasi-static process.

Temperature gradients are not the only sources of increased entropy. Heat does not have to flow to create entropy eg. the mixing of two different gases.

AM

All reversible processes are quasistatic but not vice versa. Reversible processes do not create entropy, quasistatic processes may or may not. Quasistatic processes are described by a continuum of equilibrium states. (A curve on a PV or TS diagram). The only way to make heat flow REVERSIBLE is to have an infinitesimal temperature difference, because entropy is not created. I'm calling the process quasistatic because it is so slow that the two systems are always in practical equilibrium, but irreversible because entropy is being created.

I sound like I know exactly what I am talking about, but I'm still trying to piece this together, so I am looking for cases that challenge this viewpoint that Darwin123 put forth, that the entropy of classical thermodynamics can be treated sort of like a fluid that is both transported and created at particular places, but never destroyed. Like entropy of mixing - there are no temperature or pressure gradients, only a chemical potential gradient, yet entropy is created. I guess entropy transport is driven by temperature differences, but entropy creation is not limited to a temperature gradient.

 

[Studiot]

My history books have 'caloric' in use before Clausius was a gleam in his father's eye let alone entropy being a gleam in Clausius' eyes.

Originally there term calor or caloric or calorie was used to represent both temperature and heat since these were not properly distinguished.

Black enunciated the difference in "Lectures on the Elements of Chemistry" (published in 1803 four years posthumously).

I can agree with Rap's definition of a quasistatic process as a process that is plottable as a continuous curve on an indicator diagram.
By itself this does not make it reversible or irreversible since reversible changes can always be plotted and irreversible changes sometimes plotted.

It is also worth noting the difference in the definitions of entropy between Caratheodory and previous workers such as Clausius and Kelvin.

 

[Studiot]

From a post deleted ? or lost? by Darwin123
There most definitely are a pressure gradient when two gases mix. In fact, there are two pressure gradients

Rap mentioned fluids not gases mixing. It is possible to devise fluid mixing that occurs under a concentration gradient, with neither pressure nor temperature involved.

 

[Darwin123]

My history books have 'caloric' in use before Clausius was a gleam in his father's eye let alone entropy being a gleam in Clausius' eyes.

Originally there term calor or caloric or calorie was used to represent both temperature and heat since these were not properly distinguished.

I read English translations of Carnot and Clausius essays. Carnot does not use the word caloric to designate energy. Caloric is a fluid that carries energy, which is also heat. He never confuses temperature with caloric. Temperature appears analogous to pressure. His equations distinguish between temperature and caloric.

Clausius starts to use the word "heat" to refer to some type of energy. However, he also makes it clear that entropy can carry energy. Entropy is an intensive property. This means that it is localized. Every bit of entropy is located at a spatial point. Entropy that is created also exists at a spatial point. This is why entropy can move.

Clausius argues that heat is a form of motion rather than a fluid. This is based explicitly on the fact that friction creates entropy. However, the equations that he writes are consistent with entropy flowing.

I think it is useful to think of entropy as a fluid analog with temperature a pressure analog. Temperature is the pressure that the entropy is under. Or if you like electrodynamics, entropy is analogous to electric charge. Temperature is analogous to electrical potential. Entropy flows from a high to low temperature the way positive electric charge flows from high to low electric potential. The temperature is a monotonically increasing function of entropy density. If the density of entropy is high, then the temperature is high.

The motion of entropy is entirely consistent with the creation of entropy. Motion is a consequence of the fact that entropy is an intensive property. The motion of entropy has nothing to do with whether or not it is conserved. Fluids don't have to be conserved in order to flow. Chemical reactions can change the concentration of fluids even while they are flowing.

In the case of friction, entropy is created in a region where there is a nonzero gradient of some thermodynamic quantity. However, the temperature at the point of contact is very high. Therefore, entropy flows to a region of lower temperature.

In the case of a mixture of dissimilar gases, it is incorrect to say that "the" pressure is constant throughout the process. The sum of the partial pressures may be constant. However, the partial pressures are each changing while the mixture is going on. In fact, all the partial pressures are decreasing.The gradients of each partial pressure is a nonzero vector. Therefore, there are gradients that are creating entropy.

Each partial pressure is a thermodynamic quantity. The equation of state explicitly includes with each partial pressure. One can express the equation of state as a function of partial pressures. The important thing to notice in the case of "isobaric mixing" is that only the sum of pressures is constant.

The two biggest thing that makes entropy different from electric charge is that entropy can be created, and that entropy has only one sign. Electric charge is conserved, but entropy can be created. Electric charge can be positive or negative. The third law of thermodynamics shows that there is an minimum to the absolute entropy of a system. However, both electric charge and entropy are intensive properties. The fact that they are intensive implies that they can move.

 

[Andrew Mason]

Help us out, here. Maybe you can find us a formal definition of the word "quasistatic".

I assumed that the word “quasistatic” used by the OP was the same as near-equilibrium. A later post by the OP confirmed that he was talking about a process which was the sum of a series of small processes where the state of the ideal gas was near equilibrium.

Here is how I would define a quasistatic process: it is one that moves at an arbitrarily slow rate so that all components of the system and the surroundings are
a) in thermal equilibrium internally;
b) are arbitrarily close to thermal equilibrium with all components with which they are in thermal contact; and
c) are in, or are arbitrarily close to, dynamic equilibrium at all times with each other.

So, for example a Carnot engine operates using quasistatic processes. Whether the Carnot engine processes are reversible depends on whether the work produced during the process is turned into heat flow. If it is, the work cannot be used to reverse the Carnot engine to return the system and surroundings to its initial state. This is a subtle distinction between quasistatic and reversible that is not always made clear.

AM

 

[Rap]

My history books have 'caloric' in use before Clausius was a gleam in his father's eye let alone entropy being a gleam in Clausius' eyes.

Originally there term calor or caloric or calorie was used to represent both temperature and heat since these were not properly distinguished.

Black enunciated the difference in "Lectures on the Elements of Chemistry" (published in 1803 four years posthumously).

I can agree with Rap's definition of a quasistatic process as a process that is plottable as a continuous curve on an indicator diagram.
By itself this does not make it reversible or irreversible since reversible changes can always be plotted and irreversible changes sometimes plotted.

It is also worth noting the difference in the definitions of entropy between Caratheodory and previous workers such as Clausius and Kelvin.

Yes, I didn't clarify that. I always draw quasistatic lines on an indicator diagram as a solid line, non-quasistatic, but with some state function remaining constant, as a dotted line. (e.g. Joule expansion). Reversible is a subset of quasistatic, so yes, reversible can always be plotted with a solid line. Non-quasistatic, with no state function constant, its just two unconnectable points. (Can't think of an example right now).

Regarding definitions of entropy, I think Lieb and Yngvason have taken the next step beyond Caratheodory. Caratheodory's definition of classical entropy is restricted to quasistatic transformations from equilibrium state 1 to equilibrium state 2, while Lieb and Yngvason's definition of classical entropy removes the quasistatic constraint. Their papers are rather hairy, but Thess gives a more user friendly description - see https://www.amazon.com/dp/3642133487/?tag=pfamazon01-20 - but I only get the general idea of Lieb-Yngvason and Caratheodory, I still haven't mastered either. My hunch is that Lieb and Yngvason are on to something.

Rap mentioned fluids not gases mixing. It is possible to devise fluid mixing that occurs under a concentration gradient, with neither pressure nor temperature involved.

Yes, the "entropy of mixing" problem, where you have A particles on one side, B particles on the other, separated by a partition, both sides at the same temperature and pressure, but obviously not at the same chemical potentials. Then you quickly remove the partition. When equilibrium finally occurs, the two types are fully mixed and the total entropy is larger than the sum of the two original entropies and the chemical potentials are uniform. Usually done for two non-reacting ideal gases, so that the final entropy can actually be calculated. Unlike Joule expansion, you could assume LTE, where any small volume element is a thermodynamic system in equilibrium with a universally constant T and P, and actually solve the problem for two ideal gases as it develops in time. It would be interesting to calculate the entropy density as a function of position as it develops in time. I've been thinking of doing this, just to get a better grasp of entropy creation.

 

[Studiot]

@Darwin

My point was that caloric was a word used in the english speaking world and the concept already dispelled by the time of Clausius,
Clausius was a mid 19th century worker, Black a later 18th century one.

@Rap

I am not suggesting Caratheodory as the most recent authority. I raised his definition as it seems to me the most pertinent to this discussion on lines on indicator diagrams and mathematical continuity.
Workers prior to Caratheodory all specified what we now call the second law in terms of cyclic processes ie closed loops on the indicator diagram composed of several lines. That is from State A to State B and back again. Caratheodory was the first to offer a definition that could be applied to a single line ie from State A to State B.

My English translation has

"In the neighbourhood of any equilibrium state of a system there are states that are inaccessible by an adiathermal process"

 

[Darwin123]

Yes, I didn't clarify that. I always draw quasistatic lines on an indicator diagram as a solid line, non-quasistatic, but with some state function remaining constant, as a dotted line. (e.g. Joule expansion). Reversible is a subset of quasistatic, so yes, reversible can always be plotted with a solid line. Non-quasistatic, with no state function constant, its just two unconnectable points. (Can't think of an example right now).

Regarding definitions of entropy, I think Lieb and Yngvason have taken the next step beyond Caratheodory. Caratheodory's definition of classical entropy is restricted to quasistatic transformations from equilibrium state 1 to equilibrium state 2, while Lieb and Yngvason's definition of classical entropy removes the quasistatic constraint. Their papers are rather hairy, but Thess gives a more user friendly description - see https://www.amazon.com/dp/3642133487/?tag=pfamazon01-20 - but I only get the general idea of Lieb-Yngvason and Caratheodory, I still haven't mastered either. My hunch is that Lieb and Yngvason are on to something.



Yes, the "entropy of mixing" problem, where you have A particles on one side, B particles on the other, separated by a partition, both sides at the same temperature and pressure, but obviously not at the same chemical potentials. Then you quickly remove the partition. When equilibrium finally occurs, the two types are fully mixed and the total entropy is larger than the sum of the two original entropies and the chemical potentials are uniform. Usually done for two non-reacting ideal gases, so that the final entropy can actually be calculated. Unlike Joule expansion, you could assume LTE, where any small volume element is a thermodynamic system in equilibrium with a universally constant T and P, and actually solve the problem for two ideal gases as it develops in time. It would be interesting to calculate the entropy density as a function of position as it develops in time. I've been thinking of doing this, just to get a better grasp of entropy creation.

It should be note that the process of mixing is driven by the gradient of the partial pressure. The gradient of partial pressure is sometimes considered the first contribution of entropy production.

Making an unsupported claim is against the rules of the forum. I neglected to provide references and links before in my previous links. Therefore, here are a few references and links that show how gradients of partial pressure created entropy.


http://www.landfood.ubc.ca/soil200/components/air.htm
“Diffusion - moving force is gradient of partial pressure of any constituent member of air to migrate from a zone of higher to lower pressure, even while air as a whole may remain stationary. In other words, through diffusion each gas moves in a direction determined by its own partial pressure.”



http://www.wiley-vch.de/books/sample/352731024X_c01.pdf
“Momentum transfer arguments lead to the conclusion that for diffusion in a gas mixture the gradient of partial pressure should be regarded as the fundamental .driving force,. since that formulation remains valid even under non-isothermal conditions.”

http://www.chem.ntnu.no/nonequilibrium-thermodynamics/pub/getfilee173.pdf
“The small magnitude of the gradient of partial pressure along the airways suggests that the first contribution of entropy production can be neglected in the simplified model.”


http://books.google.com/books?id=rY...page&q="gradient of partial pressure"&f=false
“The rate of diffusion of a gas is proportional to the gradient of partial pressure…”

http://books.google.com/books?id=1w...page&q="gradient of partial pressure"&f=false
“As the permeate pressure is decreased, there occurred an increase in the gradient of partial pressure or chemical potential across a dried, nonswollen layer. As a result, when the permeate components become easily vaporized, the flux or total amount of product is increased.”


http://www.decompression.org/maiken/Bubble_Decompression_Strategies.htm
“The gradient of partial pressure across the bubble surface. G = T - Pb, where T is tissue tension and Pb is the pressure inside the bubble.”


No link here. However, Sears and Zemansky is a a classic textbook on thermodynamics. The following pages describe the process of how when different gases mix.
“Heat and Thermodynamics, an Intermediate Testbook – 6th edition” by Mark Zemansky and Richard Dittman, (Mcgraw Hill, 1979) pp 361-363.

 

[Studiot]

It should be note that the process of mixing is driven by the gradient of the partial pressure. The gradient of partial pressure is sometimes considered the first contribution of entropy production.

Not under the conditions I outlined or Rap agreed with in the extract you quoted from his post.

Here is another situation for you to consider:-

Consider a salt or sugar solution a few degrees above freezing point in an open vessel under constant pressure P.

Now add a cube of ice at freezing point, of just sufficient mass that the result in the vessel will be a more dilute solution at exactly freezing point.

What is the entropy of mixing?

 

[Rap]

It should be note that the process of mixing is driven by the gradient of the partial pressure. The gradient of partial pressure is sometimes considered the first contribution of entropy production.

I don't think its partial pressure differences that drive particle transport, its chemical potential differences. If you had a boundary that passed particles slowly but did not pass heat or work, you could have the partial pressures the same on each side, but different temperatures, and therefore different densities (higher density on the cold side), and there would be a net flow of particles to equalize density. You can see this because on the hot side, there would be fewer particles passing, but with greater momentum per particle, while on the cold side, more would be passing but with less momentum per particle. There would be particle transfer from the cold side to the hot side. The chemical potential on either side, would not be the same, the chemical potential on the cold side would be higher than on the hot side.

Chemical potential gradients will also cause entropy production.

 

[Andrew Mason]

I don't think its partial pressure differences that drive particle transport, its chemical potential differences. If you had a boundary that passed particles slowly but did not pass heat or work, you could have the partial pressures the same on each side, but different temperatures, and therefore different densities (higher density on the cold side), and there would be a net flow of particles to equalize density. You can see this because on the hot side, there would be fewer particles passing, but with greater momentum per particle, while on the cold side, more would be passing but with less momentum per particle. There would be particle transfer from the cold side to the hot side. The chemical potential on either side, would not be the same, the chemical potential on the cold side would be higher than on the hot side.

Chemical potential gradients will also cause entropy production.

Mixing is driven simply by the randomness of molecular motion. The only thing that can prevent it is a barrier.

Consider two identical volumes and quantities of ideal chemically inert gas molecules, say He and Ne, and both are at the same temperature and pressure but are separated by a wall. If you remove the wall the gas molecules will mix. And you can wait forever and the molecules will not un-mix so it is not reversible. Entropy will increase.

AM

 

[Darwin123]

I don't think its partial pressure differences that drive particle transport, its chemical potential differences.

Here is a link that says that they are basically the same thing in the special case of an ideal gas, or ideal solute.

http://books.google.com/books?id=1wA...ure"&f=false
“As the permeate pressure is decreased, there occurred an increase in the gradient of partial pressure or chemical potential across a dried, nonswollen layer. As a result, when the permeate components become easily vaporized, the flux or total amount of product is increased.”

I understand that chemical potential and partial pressure aren't precisely the same thing for other than ideal gases and solutes. I am not trying to promote my own definition of partial pressure.

However...

The specific scenarios discussed in this thread involve ideal gases. So until we get away from ideal gases, chemical potential of a component and the partial pressure of a component can be discussed interchangeably. Furthermore, the chemical potential of a component generally increases with the partial pressure of a component even for components that are not ideal gases.

 

[GravitatisVis]

Would it be correct to say that TS is the energy not available for doing useful work?

When you write down any thermodynamic potential, you have the TS term, but the other terms (like PV, \muN, HM, etc) are the "work terms", which tell you how much useful energy you can get out of your system (depending on the constraints).

Can anyone comment on this?

 

[Andrew Mason]

Would it be correct to say that TS is the energy not available for doing useful work?

When you write down any thermodynamic potential, you have the TS term, but the other terms (like PV, \muN, HM, etc) are the "work terms", which tell you how much useful energy you can get out of your system (depending on the constraints).

Can anyone comment on this?

If you mean: T = Tc and S = ΔS, this would be correct.

When a thermodynamic process occurs that results in a heat flow Qh from temperature Th, heat flow Qc to temperature Tc and work output W = Qh-Qc, the energy that is "unavailable to do work" (which is also known as the "lost work") is the difference between the maximum work that can be produced (ie. in a reversible process) for that heat flow Qh from Th to Tc and the work actually produced. The "lost work" is TcΔS.

For a reversible process, ΔS = -Qh/Th + Qc/Tc = 0 so Qc/Qh = Tc/Th. For an irreversible process, -Qh/Th + Qc/Tc = ΔS > 0. So the reversible work is Qh-Qc = Qh - Qh(Tc/Th) = Qh(1-Tc/Th).

For the irreversible process involving the same Qh, the actual work is Qh-Qc'. The difference between the reversible work and the irreversible work (ie. the lost work) is:

W_L = (Q_h - Q_c) - (Q_h - Q_c&#039;) = Q_h(1-T_c/T_h) - (Q_h - Q_c&#039;)

= -Q_h(T_c/T_h) + Q_c&#039; = T_c(-Q_h/T_h + Q_c&#039;/T_c )= T_c(ΔS)


AM

 

[Rap]

Here is a link that says that they are basically the same thing in the special case of an ideal gas, or ideal solute.

http://books.google.com/books?id=1wA...ure"&f=false
“As the permeate pressure is decreased, there occurred an increase in the gradient of partial pressure or chemical potential across a dried, nonswollen layer. As a result, when the permeate components become easily vaporized, the flux or total amount of product is increased.”

I understand that chemical potential and partial pressure aren't precisely the same thing for other than ideal gases and solutes. I am not trying to promote my own definition of partial pressure.

However...

The specific scenarios discussed in this thread involve ideal gases. So until we get away from ideal gases, chemical potential of a component and the partial pressure of a component can be discussed interchangeably. Furthermore, the chemical potential of a component generally increases with the partial pressure of a component even for components that are not ideal gases.

The partial pressure and chemical potential are "equivalent" for ideal gases only if the temperatures are the same. The chemical potential for i-type particles in an ideal gas is \mu_i=\mu_{i0}+kT\ln(P_iT^{-5/2}) so you can see that the chemical potential is proportional to the logarithm of the partial pressure when temperatures are constant.

 

[Darwin123]

The partial pressure and chemical potential are "equivalent" for ideal gases only if the temperatures are the same. The chemical potential for i-type particles in an ideal gas is \mu_i=\mu_{i0}+kT\ln(P_iT^{-5/2}) so you can see that the chemical potential is proportional to the logarithm of the partial pressure when temperatures are constant.

According to your formula, the chemical potential increases with partial pressure under isothermal conditions. Therefore, if the gradient of partial pressure is a nonzero vector then the gradient of chemical potential is a nonzero vector pointing in the same direction. If the gradient of partial pressure is a zero vector, then the gradient of chemical potential is a zero vector.

Of course, this leaves the problem of systems that are not isothermal.

Entropy is created in the regions of space where the gradient of chemical potential is not zero. That is what you said. Someone else said that entropy is only created in places where the temperature has a nonzero potential.

According to your formula, the two statements are not equivalent. In a region where both temperature and partial pressure is nonzero, it may be possible to find points where the chemical potential is zero. For instance, the temperature and the partial pressure can be pointing in opposite directions with equal magnitudes.

My current question is:
1) What type of gradients cause the generation of entropy? However, entropy is still an intensive quantity. Therefore, entropy is created at specific points regardless of what gradient generates it. It has to be created in the region where that gradient is nonzero.

Entropy acts like an indestructible fluid, where temperature is like the pressure of the entropy. It can be created in a particular spot by a gradient, although I am not sure precisely what type of gradient that is. I am trying to find out. I haven't found any references where the author spells it out.

 

[Studiot]

So I am open to suggestions.

You receive (proper) suggestions in a discussion by addressing others' (proper) remarks to yourself.

 

[Rap]

Entropy is created in the regions of space where the gradient of chemical potential is not zero. That is what you said. Someone else said that entropy is only created in places where the temperature has a nonzero potential..

Well, I don't remember saying that, but if I did, then I've changed my mind. I'm sure I said a temperature gradient creates entropy, but as to what else, I too am trying to figure that out.

According to your formula, the two statements are not equivalent. In a region where both temperature and partial pressure is nonzero, it may be possible to find points where the chemical potential is zero. For instance, the temperature and the partial pressure can be pointing in opposite directions with equal magnitudes...

That sounds possible...

Entropy acts like an indestructible fluid, where temperature is like the pressure of the entropy. It can be created in a particular spot by a gradient, although I am not sure precisely what type of gradient that is. I am trying to find out. I haven't found any references where the author spells it out.

Agree with everything, except that because entropy can be created, but volume cannot, I only think of temperature being to entropy like pressure is to volume in the case where entropy is not created. When entropy is created, the analogy starts to puzzle me.

Regarding gradients producing entropy, I think there are other possibilities, like the friction of a piston doing work. The entropy is created at the points of contact between piston and cylinder and is added to the system. Energy added is T dS if you can assume the whole thing happens at a constant temperature.

You receive (proper) suggestions in a discussion by addressing others' (proper) remarks to yourself.

Did I miss something? If I did, I apologize. What did I miss?

 

[Darwin123]

Well, I don't remember saying that, but if I did, then I've changed my mind. I'm sure I said a temperature gradient creates entropy, but as to what else, I too am trying to figure that out.



That sounds possible...



Agree with everything, except that because entropy can be created, but volume cannot, I only think of temperature being to entropy like pressure is to volume in the case where entropy is not created. When entropy is created, the analogy starts to puzzle me.

Regarding gradients producing entropy, I think there are other possibilities, like the friction of a piston doing work. The entropy is created at the points of contact between piston and cylinder and is added to the system. Energy added is T dS if you can assume the whole thing happens at a constant temperature.



Did I miss something? If I did, I apologize. What did I miss?

I think that he was replying to my remark, not yours.
I apologize. I tried to correct the situation. In any case, you don't have to worry about it.