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# Youtube says the 2nd Law is Broken.

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59 minutes ago, sethoflagos said:

There is now an exchange of work between particles and box.

The collisions are elastic and the box mass is infinite. Ideal system, remember? If there is no displacement, there is no work.

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It has become your very own 'undisclosed piston'. Do you now see and understand the issue

These collisions would increase, not decrease, the volume of the box, so no.

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6 minutes ago, swansont said:

The collisions are elastic and the box mass is infinite. If there is no displacement, there is no work.

If the centre of mass of the particles remains stationary at the centre of the box, then it's possible that no nett work has been performed on the gas.

Now explain to me how that centre of mass can move towards the sides of the box to any reasonable degree without the box performing work on the gas.

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2 minutes ago, sethoflagos said:

If the centre of mass of the particles remains stationary at the centre of the box, then it's possible that no nett work has been performed on the gas.

Now explain to me how that centre of mass can move towards the sides of the box to any reasonable degree without the box performing work on the gas.

When m goes to infinity, the mass of the gas is irrelevant. The CoM will be the same, no matter what happens to the gas. It’s a non-issue for this explanation

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5 minutes ago, swansont said:

When m goes to infinity, the mass of the gas is irrelevant. The CoM will be the same, no matter what happens to the gas. It’s a non-issue for this explanation

I specifically stated the centre of mass of the gas because this is precisely the quantity that you are trying to move around relative to the CoM of the box to establish your case.

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Explain the CoM problem for one ball in my example. Is the CoM fixed?

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1 minute ago, swansont said:

Explain the CoM problem for one ball in my example. Is the CoM fixed?

The overall CoM is essentially fixed by your constraint of infinite box mass. The total linear momentum of the system with respect to the total CoM is equal to the linear momentum of the ball wrt the total CoM. What happens at collision with the box wall is observer dependent. An observer travelling with the ball will see the ball suddenly accelerate from rest and deduce that work has been done on the ball providing it with a kinetic energy it did not previously have. An observer stationary wrt the system CoM may (per your assumption) see a specular collision where the kinetic energy of the ball is preserved. A third observer may well see the ball suddenly come to a standstill. All intermediate variations are possible. Single particle thermodynamics really has no meaning due to this observer bias.

It isn't until we get into multiple particle systems where particle-particle collisions are possible, that we can shake off this observer bias and establish the concept of thermodynamic equilibrium - one requirement of which will be to set constraints on the motion of the centre of mass of the particles in the system (excluding the box).

Maybe I should give you a heads up of where we are heading with this.

I'm using a combination of the 1st Law and conservation of linear momentum to establish the necessity for a stationary gas CoM as a priliminary condition of thermodynamic equibrium in, for now, a defined NVE state.

The next stage will be to recruit the conservation of angular momentum to establish that there can be no nett flow towards or away from the CoM at equilibrium ie that in the absence of input of external work or torque, the gas remains evenly distributed throughout the available volume.

After that, well I trust that any residual ideas of macroscopic systems hopping willy-nilly between unrelated, random microstates will have dissipated.

I thank you all in advance for forcing me to think through all this stuff carefully stage by stage. It's over 40 years since I sat through my last thermo lecture!

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I get what you're saying but isn't this completely negligible compared to the difference between the energy levels, of all particles randomly appearing on one side of the box... vs them being forced there by work being done on them?

(in the category of nit-picking an idealization? )

Edited by J.C.MacSwell

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1 minute ago, J.C.MacSwell said:

I get what you're saying but isn't this completely negligible compared to the difference between the energy levels, of all particles randomly appearing on one side of the box... vs them being forced to it with work being done on them?

I do not believe it is possible for all particles to be on one side of the box in the absence of shaftwork or equivalent energy input. We're not able to compute system evolution collision by collision, but what we do know for sure is that the outcomes of each collision are very far from random - the entire discipline of physics is based on the symmetries of conserved quantities through such events. That deterministic principle does not suddenly vanish simply because we have insufficient computing power to handle the long term dynamics of macroscopic systems: those single event symmetries remain intact, instant by instant, eon by eon.

In my experience, truly random behaviour is rarely observed and then only in a very few, very special scenarios. The rest is merely 'complicated'.

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10 hours ago, sethoflagos said:

The overall CoM is essentially fixed by your constraint of infinite box mass.

You had specifically focused on the CoM of the gas. You went out of your way to point this out.

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The total linear momentum of the system with respect to the total CoM is equal to the linear momentum of the ball wrt the total CoM. What happens at collision with the box wall is observer dependent. An observer travelling with the ball will see the ball suddenly accelerate from rest and deduce that work has been done on the ball providing it with a kinetic energy it did not previously have. An observer stationary wrt the system CoM may (per your assumption) see a specular collision where the kinetic energy of the ball is preserved. A third observer may well see the ball suddenly come to a standstill. All intermediate variations are possible. Single particle thermodynamics really has no meaning due to this observer bias.

An observer traveling with the ball isn't really a viable approach, knowing full well we are going to move on to multiple balls, because we’ve already done this. Plus the view prior to this was an observer in the lab frame. You seem reticent to answer the question, which is quite obviously “no, the CoM of a single ball is not fixed”

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It isn't until we get into multiple particle systems where particle-particle collisions are possible, that we can shake off this observer bias and establish the concept of thermodynamic equilibrium - one requirement of which will be to set constraints on the motion of the centre of mass of the particles in the system (excluding the box).

Indeed. One wonders what the point of introducing it was.

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Maybe I should give you a heads up of where we are heading with this.

Yes, of course, all if this posturing suggested an agenda.

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I'm using a combination of the 1st Law and conservation of linear momentum to establish the necessity for a stationary gas CoM as a priliminary condition of thermodynamic equibrium in, for now, a defined NVE state.

The next stage will be to recruit the conservation of angular momentum to establish that there can be no nett flow towards or away from the CoM at equilibrium ie that in the absence of input of external work or torque, the gas remains evenly distributed throughout the available volume.

After that, well I trust that any residual ideas of macroscopic systems hopping willy-nilly between unrelated, random microstates will have dissipated.

I thank you all in advance for forcing me to think through all this stuff carefully stage by stage. It's over 40 years since I sat through my last thermo lecture!

This “next stage” will be in a new topic, right?

Can we get back to the discussion at hand?

9 hours ago, sethoflagos said:

I do not believe it is possible for all particles to be on one side of the box in the absence of shaftwork or equivalent energy input.

Belief is not a physics argument. It is obviously possible when N is small. What mechanism makes it impossible when N is large?

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11 hours ago, sethoflagos said:

I do not believe it is possible for all particles to be on one side of the box in the absence of shaftwork or equivalent energy input

Of course it is possible.

There are two lines of inquiry I was intending to pursue.
Swansont raised one of them before I did so I will say something on that one first, although the better and natural order would be to consider the classical approach first.

You did not reply to my question

Which version of the second law are you considering ?

There reason for this will become apparent in due course.

So let us consider the activity of a single particle (molecule if you like).

But let us  simplify matters even further.

Let us place the particle in a cubical box so that it bounces normally back and fore between opposite faces ABCD and EFGH.

As a result of this it always follows the same track as in the third sketch.

The consequences of this are that no force is exerted on any of the other four faces of the box.

Now the thermodynamic pressure is defined as the average force of impact on all the faces of the box and being the same in all directions.
Yet only two of the six faces experience any pressure force at all.

Further when the particles is somewhere between faces no face at all experiences a pressure force.

Pressure is an intensive variable, which should be the same throught the box or it is not defined.

What about volume, an extensive variable ?

Well the particle, being confined to its track, cannot access most of the volume of the box.
So is volume a defined property either?

Or is this a question of these Caratheodory microstates in his definition of the second law?

Finally how can a particle striking a boundary wall be in equilibrium?

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2 hours ago, swansont said:

An observer traveling with the ball isn't really a viable approach, knowing full well we are going to move on to multiple balls, because we’ve already done this. Plus the view prior to this was an observer in the lab frame. You seem reticent to answer the question, which is quite obviously “no, the CoM of a single ball is not fixed”

No, the CoM of the ball (wrt the system CoM) is not fixed.

2 hours ago, swansont said:

Indeed. One wonders what the point of introducing it was.

It was you who introduced the single ball case

2 hours ago, swansont said:

Yes, of course, all if this posturing suggested an agenda.

Needlessly offensive phrasing. My 'agenda' is simply to respond to points you raise in support of arbitrary random changes of state.

2 hours ago, swansont said:

Belief is not a physics argument. It is obviously possible when N is small. What mechanism makes it impossible when N is large?

There's a principle we use in Chem Eng that the transition between any two thermodynamic states is path independent i.e. when a body of gas transforms from one PVT state to a new PVT state, it is perfectly admissible to break it down into a sequence of smaller idealised steps, knowing that the overall changes remain fixed for purposes of calculating the enthalpy changes for instance. Hence, in the case in question, I analyse the change as an ideal isentropic compression (external work performed with no change in entropy) followed by a possible reversible cooling stage (shedding heat and entropy into the environment).

The Youtube presenters haven't specified the final temperature of the system, but they have stated that the entropy has fallen, therefore the cooling stage is a valid means of estimating the minimum transfer of entropy. My 40 years work experience screams out 'THE ENTROPY OF THE ENVIRONMENT HAS RISEN BY AT LEAST AN EQUAL AMOUNT'. Must have. Otherwise all my design calculations over the last 40 years have been wrong and numerous production facilities around the world are fundamentally unsafe. A little over-dramatic maybe, but my hackles have been raised.

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30 minutes ago, sethoflagos said:

It was you who introduced the single ball case

Yes. I didn’t introduce “single observer bias”

30 minutes ago, sethoflagos said:

Needlessly offensive phrasing. My 'agenda' is simply to respond to points you raise in support of arbitrary random changes of state.

“agenda” is offensive? You’re the one who stated a plan, not included in the OP.

30 minutes ago, sethoflagos said:

There's a principle we use in Chem Eng that the transition between any two thermodynamic states is path independent i.e. when a body of gas transforms from one PVT state to a new PVT state, it is perfectly admissible to break it down into a sequence of smaller idealised steps, knowing that the overall changes remain fixed for purposes of calculating the enthalpy changes for instance. Hence, in the case in question, I analyse the change as an ideal isentropic compression (external work performed with no change in entropy) followed by a possible reversible cooling stage (shedding heat and entropy into the environment).

But there is no compression, and no heat flow.

30 minutes ago, sethoflagos said:

The Youtube presenters haven't specified the final temperature of the system, but they have stated that the entropy has fallen, therefore the cooling stage is a valid means of estimating the minimum transfer of entropy. My 40 years work experience screams out 'THE ENTROPY OF THE ENVIRONMENT HAS RISEN BY AT LEAST AN EQUAL AMOUNT'. Must have. Otherwise all my design calculations over the last 40 years have been wrong and numerous production facilities around the world are fundamentally unsafe. A little over-dramatic maybe, but my hackles have been raised.

The entropy hasn’t fallen. AFAIK they’re wrong about that.

Will you now, finally, address the outstanding questions:

What is the mechanism that prevents more balls being on one side of the box*? At what value of N does the mechanism manifest itself?

*This is a big issue, because if we have symmetry, I can draw this line along any axis. If it can’t happen, then doesn’t the number in each octant have to be constant? If it’s not, then some half has more than the other

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2 hours ago, studiot said:

Which version of the second law are you considering ?

The 2nd Law is the 2nd Law. Let us assume dS/dt >= 0 for an isolated system.

2 hours ago, studiot said:

Let us place the particle in a cubical box so that it bounces normally back and fore between opposite faces ABCD and EFGH.

You're in the same trap as swansont. At each bounce, there is an interchange of momentum with the box. The box is actively participating in the thermodynamic process, so you now have to consider it's inertia, thermal capacity, temperature and entropy.

2 hours ago, studiot said:

Now the thermodynamic pressure is defined as the average force of impact on all the faces of the box and being the same in all directions.
Yet only two of the six faces experience any pressure force at all.

Further when the particles is somewhere between faces no face at all experiences a pressure force.

Pressure is an intensive variable, which should be the same throught the box or it is not defined.

Your system is an almost perfect vacuum. You want to discuss vacuums? They're pretty well defined, I think.

2 hours ago, studiot said:

Well the particle, being confined to its track, cannot access most of the volume of the box.
So is volume a defined property either?

It's rapidly becoming irrelevant in your idealised case. I'm more interested at this point in what the temperature of your box is, since this is now setting the conditions necessary for thermal equilibrium.

2 hours ago, studiot said:

Or is this a question of these Caratheodory microstates in his definition of the second law?

Is it? Never heard of it.

2 hours ago, studiot said:

Finally how can a particle striking a boundary wall be in equilibrium?

Your box is an active part of the thermodynamic system, which is why I ask what its temperature is, because that will define the equilibrium state. I suspect with only one single particle the equilibrium temperature is to all intents and purposes absolute zero, so your particle will shed its last little shred of momentum to the box and become stationary wrt the system CoM. Equilibrium has been reached.

54 minutes ago, swansont said:

But there is no compression, and no heat flow.

Real world processes tend to follow non-analytic PVT paths, which is why we characterise them by a sequence of analytic ideal steps, often an infinite number. There may well be no physical piston or heat exchanger involved but is entirely appropriate to fabricate a few for calculation purposes.

"A thermodynamic process path is the path or series of states through which a system passes from an initial equilibrium state to a final equilibrium state and can be viewed graphically on a pressure-volume (P-V), pressure-temperature (P-T), and temperature-entropy (T-s) diagrams.

There are an infinite number of possible paths from an initial point to an end point in a process. In many cases the path matters, however, changes in the thermodynamic properties depend only on the initial and final states and not upon the path."

54 minutes ago, swansont said:

The entropy hasn’t fallen. AFAIK they’re wrong about that.

At last!

54 minutes ago, swansont said:

What is the mechanism that prevents more balls being on one side of the box*?

Conservation of linear momentum. It's not permissible in this simple NVE case to conveniently transfer some to the box because that is the equivalent of introducing external Q and W terms.

54 minutes ago, swansont said:

*This is a big issue, because if we have symmetry, I can draw this line along any axis. If it can’t happen, then doesn’t the number in each octant have to be constant? If it’s not, then some half has more than the other

Precisely!

54 minutes ago, swansont said:

At what value of N does the mechanism manifest itself?

A low one. You objected to my earlier suggestion of 42, but it wasn't entirely flippant. Enough to constitute a system that isn't too swamped by quantum effects. Having said that, if you take the st. dev. of momentum from the Maxwell-Boltzmann distribution and plug it into the Heisenberg Inequality, some very interesting expressions arise that look very 2nd Law-ish to me. I suspect a very deep link in there somewhere, which actually strongly enforces my trust in the 2nd Law.

Edited by sethoflagos
missing words

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26 minutes ago, sethoflagos said:

The 2nd Law is the 2nd Law. Let us assume dS/dt >= 0 for an isolated system.

There declares the man who also writes

27 minutes ago, sethoflagos said:
2 hours ago, studiot said:

Or is this a question of these Caratheodory microstates in his definition of the second law?

Is it? Never heard of it.

in the same post.

This was a far better statement as it leads to yet another version of the second law (The Chemist's version)

1 hour ago, sethoflagos said:

My 40 years work experience screams out 'THE ENTROPY OF THE ENVIRONMENT HAS RISEN BY AT LEAST AN EQUAL AMOUNT'.

Of course there were at least three different versions of the second law by those who originally wrote it and they are the subject of my first line of enquiry.

However since my kinetic line was so ill received I don't know whether to bother.

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Caratheodory

In the neighbourhood of any state of a system in equilibrium there exist states which are inaccessible by adiabatic processes alone

The original three statements, which are the classical one's most used by Chemical Engineers contain an extra condition that is all to often forgotten, and you seem to have forgotten it this time. It is that condition that I wanted to discuss first, before going off down a kinetic/statistical track.

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16 minutes ago, studiot said:

Caratheodory

In the neighbourhood of any state of a system in equilibrium there exist states which are inaccessible by adiabatic processes alone

Every fibre of my being feels this principle to be true. Though the name is unfamiliar to me. Thank you. I'll try to remember the spelling!

20 minutes ago, studiot said:

The original three statements, which are the classical one's most used by Chemical Engineers contain an extra condition that is all to often forgotten, and you seem to have forgotten it this time. It is that condition that I wanted to discuss first, before going off down a kinetic/statistical track.

Fire away, I'm all ears.

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1 hour ago, sethoflagos said:

Real world processes tend to follow non-analytic PVT paths, which is why we characterise them by a sequence of analytic ideal steps, often an infinite number. There may well be no physical piston or heat exchanger involved but is entirely appropriate to fabricate a few for calculation purposes.

"A thermodynamic process path is the path or series of states through which a system passes from an initial equilibrium state to a final equilibrium state and can be viewed graphically on a pressure-volume (P-V), pressure-temperature (P-T), and temperature-entropy (T-s) diagrams.

There are an infinite number of possible paths from an initial point to an end point in a process. In many cases the path matters, however, changes in the thermodynamic properties depend only on the initial and final states and not upon the path."

There is no change in P, V or T

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At last!

?

This was never part of what I had been discussing. I wasn’t aware you were waiting for me to address it.

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Conservation of linear momentum. It's not permissible in this simple NVE case to conveniently transfer some to the box because that is the equivalent of introducing external Q and W terms.

With one ball it’s not a problem, though.

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Precisely!

So the number of particles is equal in each octant?

What if there are an odd number of particles?

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A low one. You objected to my earlier suggestion of 42, but it wasn't entirely flippant.

You said 42 before I brought this argument up. I hope causality isn’t under attack here, too.

42 isn’t divisible by 8. Let’s make it 48 - 6 particles per octant. Why can’t that number change? Why can’t there be 7 in one octant, and 5 in another? What prevents that?

It’s not momentum. They travel at different speeds, and there’s a finite transit time.

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1 hour ago, swansont said:

There is no change in P, V or T

However all the particles are in one side of the box. So lets drop in a partition isolating a system of N particles on one side from an absolute vacuum on the other. Please don't reply 'there is no partition'.

1 hour ago, swansont said:

With one ball it’s not a problem, though

Yes it is. Ball and box is a two-body system with regular interchange of (at the very least) momentum between them. If you think that this is awkward, just wait until we start considering the regular interchange of torque.

1 hour ago, swansont said:

So the number of particles is equal in each octant?

Every action has an equal and opposite reaction. Every particle arrival or departure within a space brings with it changes in mass, linear momentum, orbital angular momentum, axial spin etc and each must be balanced precisely by an equal and opposite external action.

1 hour ago, swansont said:

What if there are an odd number of particles?d

Sit it stationary at the CoM and forget about it. It isn't a problem until something bumps into it.

1 hour ago, swansont said:

You said 42 before I brought this argument up. I hope causality isn’t under attack here, too.

I'll ask the mice.

1 hour ago, swansont said:

42 isn’t divisible by 8. Let’s make it 48 - 6 particles per octant. Why can’t that number change? Why can’t there be 7 in one octant, and 5 in another? What prevents that?

It’s not momentum. They travel at different speeds, and there’s a finite transit time.

The box is a mathematical artefact introduced solely to define the system geometrical constraints. A condition of it being considered non-interacting is that we must be able to delete it from the snapshot and see no nett change to the system. Therefore the only states it is admissible to consider are those where every single particle linear momentum vector is exactly balanced by an equal and opposite vector. Just as the box could transfer momentum to and from the system, so can the octants transfer momentum between each other. While it must be possible for one octant to momentarily grab an extra particle or two (I'll agree with you that far), there are one or two particles on the imminent verge of leaving it probability one ensuring that the system never departs by a measurable amount from equilibrium. Serious departures demand serious external work.

Edited by sethoflagos
sp

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2 hours ago, sethoflagos said:

Every fibre of my being feels this principle to be true. Though the name is unfamiliar to me. Thank you. I'll try to remember the spelling!

Fire away, I'm all ears.

So consider the following process as outlined on the PV diagram where working fluid moves from state 1 to state 4 in three reversible stages or legs.

Stage 1 -2 heat is accepted via an isothermal expansion at the lower temperature T1 and expansion work is therefore done.

Stage 2 - 3 work is again done but no heat exchanged during an adiabatic compression, buit the work is of the opposite sign to that of the first leg.

Stage 3 - 4 Heat is now rejected at a higher temperature T2 and  work done in an isothermal compression to reach a point where the combined work of stages 2 and 3 equal that of stage 1.

Thus exactly zero external work is performed, but a quantity of heat is transferred from a lower temperature reservoir to a higher temperature heat reservoir.

So can you explain why this does not contravene the Second Law  ?

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1 hour ago, sethoflagos said:

However all the particles are in one side of the box. So lets drop in a partition isolating a system of N particles on one side from an absolute vacuum on the other. Please don't reply 'there is no partition'.

No, you don’t get to change the conditions of the example.

Quote

Yes it is. Ball and box is a two-body system with regular interchange of (at the very least) momentum between them. If you think that this is awkward, just wait until we start considering the regular interchange of torque.

It’s an ideal system. You keep ignoring that.

Even if it isn’t, fine. If we’re at STP and have one mole of gas, so the volume of the container is 22.4L. The ideal gas has a mass of, say, 10 grams. The container is just under 30 cm a side and has a mass of, say, a kg. You have your momentum transfer, so the container can shift a small amount if needed.

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Every action has an equal and opposite reaction. Every particle arrival or departure within a space brings with it changes in mass, linear momentum, orbital angular momentum, axial spin etc and each must be balanced precisely by an equal and opposite external action.

How are those violated? Action/reaction arises from force. Arrival, departure, etc. are not force.

Nothing external. It’s all internal.

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The box is a mathematical artefact introduced solely to define the system geometrical constraints. A condition of it being considered non-interacting is that we must be able to delete it from the snapshot and see no nett change to the system. Therefore the only states it is admissible to consider are those where every single particle linear momentum vector is exactly balanced by an equal and opposite vector.

That’s not how gases behave. It’s not a 1-for-1 matchup. You could have 10 moving right with 1 unit of momentum and 1 moving left with 10 units, and momentum is zero.

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Just as the box could transfer momentum to and from the system

So you agree you don’t need the atoms to have their momentum cancel

Quote

, so can the octants transfer momentum between each other. While it must be possible for one octant to momentarily grab an extra particle or two (I'll agree with you that far), there are one or two particles on the imminent verge of leaving it probability one ensuring that the system never departs by a measurable amount from equilibrium. Serious departures demand serious external work.

Why are small fluctuations OK, but larger ones require work?

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27 minutes ago, studiot said:

So consider the following process as outlined on the PV diagram where working fluid moves from state 1 to state 4 in three reversible stages or legs.

Stage 1 -2 heat is accepted via an isothermal expansion at the lower temperature T1 and expansion work is therefore done.

Stage 2 - 3 work is again done but no heat exchanged during an adiabatic compression, buit the work is of the opposite sign to that of the first leg.

Stage 3 - 4 Heat is now rejected at a higher temperature T2 and  work done in an isothermal compression to reach a point where the combined work of stages 2 and 3 equal that of stage 1.

Thus exactly zero external work is performed, but a quantity of heat is transferred from a lower temperature reservoir to a higher temperature heat reservoir.

So can you explain why this does not contravene the Second Law  ?

Looks to me like you've just heated up some high pressure gas from T1 to T2 at roughly constant pressure. Close the cycle, put some preliminary numbers to it and we can discuss.

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1 hour ago, swansont said:

No, you don’t get to change the conditions of the example.

I've not changed the conditions one iota.

1 hour ago, swansont said:

It’s an ideal system. You keep ignoring that.

Even ideal systems have to observe basic symmetries.

1 hour ago, swansont said:

Even if it isn’t, fine. If we’re at STP and have one mole of gas, so the volume of the container is 22.4L. The ideal gas has a mass of, say, 10 grams. The container is just under 30 cm a side and has a mass of, say, a kg. You have your momentum transfer, so the container can shift a small amount if needed.

Better. Now sum all those microstates where the container remains unchanged from its original state, and we have a workable thermodynamic ensemble.

1 hour ago, swansont said:

How are those violated? Action/reaction arises from force. Arrival, departure, etc. are not force.

The mean free path in air is what? 100 nm? There may be a lag between arrival and collision, but it's trivial in a macroscopic system.

2 hours ago, swansont said:

That’s not how gases behave. It’s not a 1-for-1 matchup. You could have 10 moving right with 1 unit of momentum and 1 moving left with 10 units, and momentum is zero.

We don't need a 1-1 match up. The entire reaction vector to the momentum of a single particle can be split anyway you like between the remaining N-1 particles in the system. In a sense it is. So long as motion of one particle is exactly matched by some ensemble in contrary motion to fill the void, then sanity is preserved and you have the restoring force you've been asking for.

2 hours ago, swansont said:

So you agree you don’t need the atoms to have their momentum cancel.

Not when you've transferred nett momentum into your box. But now you have a non-canonical microstate. Do you really think you can both have your cake and eat it?

2 hours ago, swansont said:

Why are small fluctuations OK, but larger ones require work?

They all require work in a sense. However small fluctuations are reversed almost immediately by the inertial reaction mechanisms discussed above, and are inherently reversible in nature. Larger fluctuations imply a long term resistance to those reaction forces that increase in direct proportion to the scale of the disturbance and I see no internal mechanism that could account for such a resistance.

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1 hour ago, sethoflagos said:

Looks to me like you've just heated up some high pressure gas from T1 to T2 at roughly constant pressure. Close the cycle, put some preliminary numbers to it and we can discuss.

If you don't know how to answer my question, which did not mention a cycle or cycles, then please just say so and don't mess around wasting everyone's time.

If there was anything unclear in my description please just ask and I will amplify the point.

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10 minutes ago, studiot said:

If there was anything unclear in my description please just ask and I will amplify the point.

Do you think this is the first time I've ever seen a sketch for a heat pump that claims to break the 2nd Law? Really?

I've just worked through the expression for the shaft work of your blasted adiabatic compression stage wondering why the hell am I spending my time on this hair-brained stuff when I'm not being paid for it, and you have the gall to say I'm messing around wasting everyone's time.

HOW DARE YOU!

Let's be clear on this - you are the one who has come up with an invention that any patent office would chuck in the waste-paper basket unread!

I had about 12 hours work left to properly research and analyse your scheme so that I could help you through your misunderstanding. I've been helping my junior engineers overcome such hurdles for many years. But now shall I bother? What the hell do you think. Grow up!

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40 minutes ago, sethoflagos said:

I've not changed the conditions one iota.

You added a partition, where there was none. You added a piston earlier, and talked about changing variables that were fixed, and doing work.

You’ve done little but change the conditions.

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9 minutes ago, sethoflagos said:

Do you think this is the first time I've ever seen a sketch for a heat pump that claims to break the 2nd Law? Really?

I've just worked through the expression for the shaft work of your blasted adiabatic compression stage wondering why the hell am I spending my time on this hair-brained stuff when I'm not being paid for it, and you have the gall to say I'm messing around wasting everyone's time.

HOW DARE YOU!

Let's be clear on this - you are the one who has come up with an invention that any patent office would chuck in the waste-paper basket unread!

I had about 12 hours work left to properly research and analyse your scheme so that I could help you through your misunderstanding. I've been helping my junior engineers overcome such hurdles for many years. But now shall I bother? What the hell do you think. Grow up!

Wow, what a lot of invective just to dodge answering a question, similar to the one you asked in the OP.

The joke is that it is based on one from a textbook entitled

Thermodynamics for Chemical Engineers

written by three professors from the Dept of Chem Eng at Imperial College.

Edited by studiot

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