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Maxwell's demon and the second law of thermodynamics


Moreno
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It is commonly assumed that perpetuum mobile of the second kind or maxwell's demon suppose to violate the second law of thermodynamics. And the second law of thermodynamics claims that:

 

"In every real process the sum of the entropies of all participant bodies is increased. In the idealized limiting case of a reversible process, the sum remains unchanged."

(Wikipedia)

 

However, does this "idealized limiting case", as they said, contradicts to nature's laws? If not, then doesn't it means that in theory perpetuum mobile of the second kind could be created? Let assume we have eternal non-dissipative oscillations in some system during which entropy increases in one part of a system and decreases for exactly the same value in the other part. It happens constantly during oscillations. Total amount of entropy in the universe is unaffected by this system. Thus we have a perfectly reversible process. Then during entropy reduction in one part of the system a useful work could be performed each time.

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I'm going to beat the Second Law

 

I'm going to beat the Second Law

 

I'm going to beat the Second Law

 

Let assume we have eternal non-dissipative oscillations in some system during which entropy increases in one part of a system and decreases for exactly the same value in the other part. It happens constantly during oscillations. Total amount of entropy in the universe is unaffected by this system. Thus we have a perfectly reversible process. Then during entropy reduction in one part of the system a useful work could be performed each time.

 

 

As far as the underlined part you are correct.

 

But in attacking the Second Law, you have forgotten the First Law (which is the underlined bit).

 

Did you have such a system in mind that we can analyse in full?

 

Simple versions of such systems are easy to describe. The (transfer of) work eventually damps out the oscillations.

 

Indeed study of the problem leads to the conclusion that an isolated system that can only undergo reversible processes cannot proceed by itself from one equilibrium state to another.

Edited by studiot
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Actually, you can't get the second law from classical (or quantum) mechanics. There's a really good book that details this result and proves the possibility of both a classical and a quantum Maxwellian Demon.

 

Here it is on Amazon.

 

I suppose that your route to the second law must depend upon which formulation you want to reach.

 

But I don't see this question is really about Maxwell's demon.

 

In any event, thank you for the reference I might look at it as it seems interesting.

But there are quite a few books about the philosophical underpinnings of the second law.

 

Professor's Atkins little book, Four Laws that Drive the Universe, is particularly good as it does this for all four of them.

 

:)

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Slightly off topic but if an antimatter particle and its counter part collide, although there will be some energy emitted, if that energy is used up independently I.E it doesnt effect anything else in the universe, such that the exertion force didnt reach an object with which to act upon or if the gamma particles wind up in a black hole so the photons dont reach anything either. Wouldnt that count as defying the 2nd law? matter is destroyed and the entropy decreases?

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I suppose that your route to the second law must depend upon which formulation you want to reach.

The problem is that there isn't a route. You can't get to either the absolute or the probabilistic formulation of the second law by way of either classical or quantum mechanics.

Professor's Atkins little book, Four Laws that Drive the Universe, is particularly good as it does this for all four of them.

I'll check it out.

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ydoaPs

The problem is that there isn't a route. You can't get to either the absolute or the probabilistic formulation of the second law by way of either classical or quantum mechanics.

 

 

I don't think this thread is meant to be about the proof or substantiation of the Second Law.

 

I don't know what moreno really wants since although he has been back to this thread he has chosen not to clarify his original post as I invited him to do.

 

The Wiki quote is inapplicable to classical themodynamics, which was clearly stated for a cyclic process only.

Further the Wiki statement, which is basically the inequality of Clausius, is useless in cases where q = 0.

Discussion about entropy in these cases is like discussing the direction the zero vector points in.

 

Moreno's proposed a case which can be theoretically realised in the mechanical world (q=0) and I am happy to delve more deeply into the thermodynamic mathematics of this so long as he is not asking us to do all the work.

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My impression was that he was asking if we can use the one swing of a REVERSIBLE cyclic process to extract useable work.

I thought it obvious that if work is extracted, the process can no longer be reversed.

 

Until we know exactly what process moreno has in mind, how can we decide if it is reversible or not?

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I'm going to beat the Second Law

 

I'm going to beat the Second Law

 

I'm going to beat the Second Law

 

 

As far as the underlined part you are correct.

 

But in attacking the Second Law, you have forgotten the First Law (which is the underlined bit).

 

Did you have such a system in mind that we can analyse in full?

 

Simple versions of such systems are easy to describe. The (transfer of) work eventually damps out the oscillations.

 

Indeed study of the problem leads to the conclusion that an isolated system that can only undergo reversible processes cannot proceed by itself from one equilibrium state to another.

I didn't claim I'm going to beat the second law. I said if we believe in current second law definition it seems like we may construct (in theory!) a perpetuum mobile of the second kind which works WITHIN the second law.

 

What is concerning to some particular example, it's pretty difficult to do because we need to agree what is a truly closed system and find it in nature. The Earth may not be a truly closed system and its extremely difficult to isolate any system from surrounding environment.

 

But let try. For example there is a Belousov chemical reactions which seem to be oscillation of some chemical solution from one chemical state to another and back.

https://en.wikipedia.org/wiki/Belousov%E2%80%93Zhabotinsky_reaction

I have no idea if such type of oscillations could be truly eternal and non-dissipative. But if they do, let imagine a chemical system which exist in thermodynamic equilibrium with surrounding environment. Such system would be divided in two main parts which experience oscillations between these two parts. When one part of the system transfers from amorphous to crystalline state entropy reduces and it cools down. It shrinks in size and absorbs heat from surrounding environment. The other part of the system in the same time experience opposite transition - it transfers from crystalline to amorphous state, releases heat, increases in size and capable produce some useful work, for example, to push a piston. After work is done heat released to environment. However, entire system constantly draws the same amount of heat from environment as it releases. So, it exist in thermodynamic equilibrium with nature.

 

If according to second law there could exist some processes which remain total amount of entropy in universe unchanged (I assume that under word "process" they also mean conversion of some type of energy into some kind of a useful work), and we could build some engine which harnesses such type of process, then our ability to perform some useful work should never diminish. And we would have a perpetuum mobile. If our ability to perform work diminishes then it means entropy increases during the process. But the second law doesn't state it is the only outcome possible.

 

This one article mentions some similar ides (though not necessarily the same).

http://phys.org/news/2014-09-physicists-zero-friction-quantum.html

Edited by Moreno
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I don't have thermodynamic (enthalpy) data for the B-Z reaction and the only source I could find wants $40 for the information.

 

http://www.sciencedirect.com/science/article/pii/004060319180455R

 

Most articles concentrate on the chemical and ensuing rate non linear differential equations that cause the oscillations rahter than addressing the calorimetry.

 

So if any professional chemist can supply this data it would be vary helpful to progress calculations.

 

Meanwhile here is an outline of a mechanical system that acts as you describe up to the underlined part of my quote in post4.

 

 

 

Consider a sealed adiabatic cylinder, C, separated into two chambers A (equilibrium volume Va) and B (equilibrium volume Vb) by a frictionless system.

Both chambers are filled with inert ideal gas.

 

 

Cylinder C forms a system but to analysise its action we must split it into two sub systems A and B.

 

Let us start with the piston at the equilibrium position and imagine it drawn slightly aside and then released.

Note no heat has entered the system and no heat can transfer from one chamber to the other.

 

If Sa is the original entropy of A then

[math]d{S_a} = \frac{{d{U_a}}}{{{T_a}}} + \frac{{{P_a}}}{{{T_a}}}d{V_a}[/math]

 

In the absence of heat transfer

[math]d{U_a} = - Pd{V_a}[/math]

 

Substituting leads to

[math]d{S_a} = 0[/math]

 

Similarly for chamber B.

Thus the overall system entropy change is Sc = Sa + Sb = 0

 

Since we have specified no dissipative forces system C will oscillate forever as specified, using the input mechanical energy as the source of the oscillation energy.

 

If we now connect an adiabatic rod to the piston and use the motion to extract work via some external system D, the oscillation will die away as the work is extracted.

There will be no entropy change as a result of this for systems A, B and C or for the wrok extracted.

However entropy will increase in the surroundings and possibly D.

Edited by studiot
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I don't have thermodynamic (enthalpy) data for the B-Z reaction and the only source I could find wants $40 for the information.

 

http://www.sciencedirect.com/science/article/pii/004060319180455R

 

Most articles concentrate on the chemical and ensuing rate non linear differential equations that cause the oscillations rahter than addressing the calorimetry.

 

So if any professional chemist can supply this data it would be vary helpful to progress calculations.

 

Meanwhile here is an outline of a mechanical system that acts as you describe up to the underlined part of my quote in post4.

 

 

 

Consider a sealed adiabatic cylinder, C, separated into two chambers A (equilibrium volume Va) and B (equilibrium volume Vb) by a frictionless system.

Both chambers are filled with inert ideal gas.

 

 

Cylinder C forms a system but to analysise its action we must split it into two sub systems A and B.

 

Let us start with the piston at the equilibrium position and imagine it drawn slightly aside and then released.

Note no heat has entered the system and no heat can transfer from one chamber to the other.

 

If Sa is the original entropy of A then

[math]d{S_a} = \frac{{d{U_a}}}{{{T_a}}} + \frac{{{P_a}}}{{{T_a}}}d{V_a}[/math]

 

In the absence of heat transfer

[math]d{U_a} = - Pd{V_a}[/math]

 

Substituting leads to

[math]d{S_a} = 0[/math]

 

 

Similarly for chamber B.

 

Thus the overall system entropy change is Sc = Sa + Sb = 0

 

Since we have specified no dissipative forces system C will oscillate forever as specified, using the input mechanical energy as the source of the oscillation energy.

 

If we now connect an adiabatic rod to the piston and use the motion to extract work via some external system D, the oscillation will die away as the work is extracted.

There will be no entropy change as a result of this for systems A, B and C or for the wrok extracted.

However entropy will increase in the surroundings and possibly D.

 

As I said, if during some energy conversion process entropy in the universe in total will increase, our ability to perform useful work will diminish. Therefore in order for perpetuum mobile of the second kind to function increase of entropy in some place should be compensated by its decrease somewhere else (and simultaneously!) for exactly the same value. I'm not sure particularly about the cylinder you described, but think that if cylinder looses thermal energy to the environment it would need to draw the same amount of thermal energy from environment to compensate looses. So, you describe situation when entropy increases in total, which doesn't suit a hypothetical conditions I mentioned.

Edited by Moreno
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As I said, if during some energy conversion process entropy anywhere in the universe will increase, our ability to perform useful work will diminish. Therefore in order for perpetuum mobile of the second kind to function increase of entropy in some place should be compensated by its decrease somewhere else (and simultaneously!) for exactly the same value. I'm not sure particularly about the cylinder you described, but think that if cylinder looses thermal energy to the environment it would need to draw the same amount of thermal energy from environment to compensate looses. So, you describe situation when entropy increases in total, which doesn't suit a hypothetical conditions I mentioned.

 

 

 

And as I said, adiabatic.

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And as I said, adiabatic.

Then it is an insufficient condition by itself to work as PM. I also have a question about Maxwell's demon. In a classical example it is described as a mechanism which without energy looses gathers hot molecules in one reservoir and cold molecules in the other. But what happens to entropy of these two reservoirs in comparison to initial state? Since entropy is an additive value we could say entropy of initial reservoir was=X, and after it was divided in two and molecules were sorted by their speed, what would be entropy of Y+Z, the same as X in total or larger?

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Please explain this sentence.

To be perfectly adiabatic doesn't seem to be sufficient to be a perpetuum mobile. But in any case you assumed it could only loose energy to environment, but never gain it.

Edited by Moreno
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To be perfectly adiabatic doesn't seem to be sufficient to be a perpetuum mobile. But in any case you assumed it could only loose energy to environment, but never gain it.

 

That was pretty meaningless, and certainly little to do with thermodynamics.

Would you like to say that again in English?

 

Note I put some substantial effort into supporting your proposal and I don't like being taken for a fool.

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I do not know, but common sense tells that if the universe exist in the form as we know it, then either first or the second law of thermodinamics supposed to be violated somewhere in the past. Or not?

 

Interesting. Why do you think that?

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Interesting. Why do you think that?

If universe existed eternally, and the second law of TD worked eternally as well, then we suppose to be in the state of thermal death already.

 

Is presence of large thermal fluctuations sufficient to extract useful energy and produce useful work? For example, modern science doesn't put theoretical limit on size of atoms (or even elementary particles). If we would have a giant atom or elementary particle with 1 kg mass, will it experience Brownian motion similar to usual atoms? Atoms of air move with an average speed 500 m/sec at room temperature. If huge atom will behave in the same way, we could take a hollow tube, create an ideal vacuum inside and put only this one huge atom inside. Imagine a hollow tube inside which 1 kg of mass moves with speed 500 m/s. If huge atom would have some large magnetic moment we would be able to make AC generator out of it and generate AC current. Or not? (I do not really count on it) , but still...

Edited by Moreno
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Checking through your posts, I think you are suffering under a misunderstanding of some of the basics of Thermodynamics.

 

In popular writing entropy is often portrayed as energy that is somehow unavailable. Nothing could be further from the truth.

 

Entropy is not a form of energy.

 

Entropy is heat energy divided by temperature.

 

To suggest that they are the same you might as well suggest that mass is the same as force. After all, mass is force divided by acceleration! Or that specific heat is a form of energy, since it is also heat energy divided by temperature.

 

Note carefully that heat is only one form of energy; there are many forms that may be gained, lost or possessed by a system, one being converted to another by thermodynamic or other processes.

 

Now in order to model such processes we have to be able to specify the system and most importantly define the boundary around the system. I cannot stress how important the boundary is.

 

The boundary is important because we have two laws that describe what happens when energy crosses that boundary and enters or leaves the system.

The First and Second Laws of Thermodynamics.

Neither apply directly within the system alone, which bring me to your next misunderstanding.

Adiabatic does not mean no energy crosses the boundary.

It means no heat energy crosses the boundary.

This heat energy appears as a variable in both the First and Second Laws, but does not include any non heat energy.

 

You wish to discuss energy, entropy and perpetual motion.

 

Firstly let us be clear that perpetual motion is not forbidden by the Laws of Physics.

On the contrary it is expressly required in appropriate circumstances.

For example.

 

Newton’s First Law

 

An undisturbed body continues in its state of motion forever.

 

Earnshaw’s Theorem (abbreviated as appropriate)

 

A system containing more than one charge cannot remain at rest.

 

So perpetual motion would be the natural state of every material thing in the universe if left undisturbed.

 

What about you idea of powered perpetual motion?

 

Engines (and you are describing an engine) must work in cycles to continue operation.

Something that can only work for one cycle or part thereof cannot sustain perpetual motion against opposing forces.

 

The first law requires that

 

Net Work Transfer = Net Heat Transfer

 

W = Q

 

In other words if there is no heat input the system will eventually run out of energy to drive the perpetual motion so it will stop.

 

But the second law requires that in order to transfer heat there must be a temperature difference.

 

Finally where does entropy come into this?

 

Well the original heat engines were steam engines and they had a mechanism that could draw a chart of the pressure v volume of the system. These primitive chart recorders were called ‘indicators’ and the diagram they traced out were called ‘indicator diagrams’.

The area traced out on these diagrams directly represented the mechanical work done (i.e. the mechanical non heat energy transferred).

It was realised that a similar energy diagram for heat energy was desirable and so a suitable variable to pair with temperature was devised.

This variable became known as entropy (symbol S) and T- S indicator diagram have attained great importance in mechanical engineering.

 

Indicator diagrams have come to mean the chart of any pair of variables whose product is energy.

 

A look at some engine cycles shows a comparison of P - V and T – S diagrams where it can be immediately seen that entropy for the whole system decreases during parts of the cycles.

 

post-74263-0-62363700-1450530259.jpg

 

 

 

This is common to many types of working machinery and the cycle for a compressor is displayed below.

So what you are asking for is already done, but it has to be done in cycles.

 

post-74263-0-60193300-1450530260_thumb.jpg

 

 

 

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  • 3 weeks later...

 

...[in other words if there is no heat input the system will eventually run out of energy to drive the perpetual motion so it will stop.

 

But the second law requires that in order to transfer heat there must be a temperature difference.]...

 

 

 

 

 

 

Maxwell's demon suppose to consume heat energy from the surrounding environment. So, there suppose to be a heat input.

​Yes, it is postulated that heat cannot transfer spontaneously from colder body to a hotter one. But why exactly? Sometimes it is said that it would reduce total entropy and therefore contradict to the tendency of entropy to increase or stay the same. But is it always a case? For example if we separate a container filled with gas by membrane and gather slow molecules in one part and fast molecules in the other, then entropy in the hot part of container suppose to increase, but entropy in the cold part of container suppose to decrease and total entropy of container suppose to remain the same? In this case heat transfer from colder body to hotter body doesn't necessarily suppose to reduce total entropy?

Edited by Moreno
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  • 1 year later...

One more attack on the second law was attempted:

 

"The erasure mechanism can be used to design generalized heat engines operating under the reservoirs of multiple conserved quantities such as a thermal reservoir and a spin reservoir," Bedkihal said. "For example, one may design heat engines using semiconductor quantum dot systems where lattice vibrations constitute a thermal reservoir and nuclear spins constitute a polarized spin reservoir. Such heat engines go beyond the traditional Carnot heat engine that operates under two thermal reservoirs."

 

Read more at: https://phys.org/news/2017-02-physicists-erasing-energy.html#jCp

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