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# Variations and consequences of the Laws of Thermodynamics

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This thread was inspired by the following comment in this query about air conditioning.

Quote

2nd law: In an isolated system entropy will increase or remain the same.

But  is that so ?

The change in entropy going from state A to state B is always the same, irrespective of the path between A and B since entropy is a state variable and thus a function of the state of the system alone.

It make no difference whether that path is reversible or irreversible.

Only in the case of a reversible path is the entropy given by the expression

$\Delta S = {S_B} - {S_A} = \int_A^B {\frac{{dq}}{T}}$

In order to calculate the entropy change for an irreversible path we must find an alternative reversible method of going from state A to state B and then use this to calculate the entropy change.

For any completely isolated system we are restricted to adiabatic processes since no heat can either enter or leave the sytem.

For a reversible process in any such system dq=0, hence ΔS is also zero, which means that S is a constant.

Thus if one part of the system increases in entropy another part must decrease by the same amount.

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1 hour ago, studiot said:

This thread was inspired by the following comment in this query about air conditioning.

Quote

Carrock  :2nd law: In an isolated system entropy will increase or remain the same.

But  is that so ?

The change in entropy going from state A to state B is always the same, irrespective of the path between A and B since entropy is a state variable and thus a function of the state of the system alone.

It make no difference whether that path is reversible or irreversible.

...............................

For any completely isolated system we are restricted to adiabatic processes since no heat can either enter or leave the sytem.

For a reversible process in any such system dq=0, hence ΔS is also zero, which means that S is a constant.

Thus if one part of the system increases in entropy another part must decrease by the same amount.

You cover "in an isolated system entropy need not increase."

I'll assume ideal, classical conditions, including an ideal gas.

An (old) example of "in some isolated systems entropy can increase:"

You have an x cubic units sealed container of ideal gas, inside an empty sealed container of y cubic units.

A side of the smaller container is 'instantaneously' removed without otherwise affecting the system.

The gas irreversibly expands and comes to equilibrium in the y cubic units container. No work has been done so the temperature has not changed.

The entropy has increased by the factor ln(y/x).

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2 hours ago, Carrock said:

You cover "in an isolated system entropy need not increase."

I'll assume ideal, classical conditions, including an ideal gas.

An (old) example of "in some isolated systems entropy can increase:"

You have an x cubic units sealed container of ideal gas, inside an empty sealed container of y cubic units.

A side of the smaller container is 'instantaneously' removed without otherwise affecting the system.

The gas irreversibly expands and comes to equilibrium in the y cubic units container. No work has been done so the temperature has not changed.

The entropy has increased by the factor ln(y/x).

Yes, it's true that if you physically change the system, the new system can have greater entropy than the original.

The original system can never gain entropy if isolated.

Do you understand why this is and that this is why the old formulations were made in terms of cyclic processes?

Edited by studiot
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5 hours ago, studiot said:

Yes, it's true that if you physically change the system, the new system can have greater entropy than the original.

The original system can never gain entropy if isolated.

Maybe a misunderstanding? If the original system is in equilibrium, then it can never gain entropy if isolated.

In my example the side of the small box was removed 'instantaneously,' taken out of the larger box and the larger box sealed 'instantaneously,' before any ideal gas atoms could escape. That was the original isolated system, which then spontaneously changed itself and increased its entropy.

5 hours ago, studiot said:

Do you understand why this is and that this is why the old formulations were made in terms of cyclic processes?

No and yes.

Rather than pretend I didn't refresh my understanding, I'll just quote a bit of Wikipedia e.g.:

Quote

The repeating nature of the process path allows for continuous operation, making the cycle an important concept in thermodynamics.

The value of this concept can often be seen during problematic, non cyclic engine startup, before the startup transients die down.

Enough for today!

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22 minutes ago, Carrock said:

Rather than pretend I didn't refresh my understanding,

It's more important and more subtle than that.

Your proposed experiment uses an irreversible process.

This means that since the original system is isolated and remains so it cannot interact with its surroundings.

But the only way to restore the system to its original state is by way of an external intervention or interaction.

In truth the original state is not accessible to the new system after the irreversible process.

This is because there are really two systems the original system and the new or transformed system.

The state of either of these systems is inaccessible to the other.

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5 minutes ago, studiot said:

It's more important and more subtle than that.

Really?

7 minutes ago, studiot said:

Your proposed experiment uses an irreversible process.

This means that since the original system is isolated and remains so it cannot interact with its surroundings.

The second sentence is true whether or not the process is irreversible.

11 minutes ago, studiot said:

But the only way to restore the system to its original state is by way of an external intervention or interaction.

Entropy has been increased so of course.

44 minutes ago, studiot said:

In truth the original state is not accessible to the new system after the irreversible process.

Did you mean: the original state of the isolated system cannot be restored after the irreversible process has changed the state of the system?

Entropy has been increased so of course.

16 minutes ago, studiot said:

In truth the original state is not accessible to the new system after the irreversible process.

This is because there are really two systems the original system and the new or transformed system.

The state of either of these systems is inaccessible to the other.

Clarify please. I don't see any sense in the second or third sentence.

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11 hours ago, Carrock said:

Really?

Yes really.

11 hours ago, Carrock said:

The second sentence is true whether or not the process is irreversible.

Agreed, but you are focusing on the wrong part of that sentence.

11 hours ago, Carrock said:

Entropy has been increased so of course.

Again agreed, but a state is not defined by one variable alone but by the inclusion of every state variable. Other state variables are also different between the new and old systems.

11 hours ago, Carrock said:

Did you mean: the original state of the isolated system cannot be restored after the irreversible process has changed the state of the system?

Entropy has been increased so of course.

I meant (and said) that the irreversible process has changed the system. So the state of the new system (being a list of values of all state variables) differs from that of the original system.

11 hours ago, Carrock said:

Clarify please. I don't see any sense in the second or third sentence.

Consider two conditions on the original system.

In one case the original sytem is truly isolated.

In the second case the original system is not isolated an so can interact with its surroundings.

In the second case and following the removal of the internal wall it is possible to restore the resulting system to the same conditions (list of state variables) as pertained in the original system, by outside intervention or if you prefer interaction with the surroundings.

So it can truly be said that a system in state A can be taken to state B and returned to state A as the same system, but with changes in the surroundings.

It is the emboldened phrase that is the usual way of putting this, but that is all too often ignored.

However with the isolated system interaction is barred so the return is not possible, and would still not be possible regardless of the entropy issue.

Entropy is a measure of an observational truth, not the other way round.

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2 hours ago, studiot said:

I meant (and said) that the irreversible process﻿ has change﻿d the system. So the state of the new system (being a list of values of all state variables) differs from that of the original system.

When you change a system how does that become a new system rather than a modified system? You seem not to be distinguishing between a system and its state. This is unconventional and confusing.

1. Would you disagree with "So the modified state of the system (being a list of values of all state variables) differs from the original state of the system?"

2 hours ago, studiot said:

So it can truly be said that a system in state A can be taken to state B and returned to state A as the same system, but with changes in the surroundings.

It is the emboldened phrase that is the usual way of p﻿utting this, but that is all too often ignored.

I specified an isolated system with entropy increasing so that these and some other issues would be irrelevant.

Edited by Carrock
changed 'new' to 'modified'
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Thermodynamics, in its general meaning, is always equilibrium Thermodynamics. Calculations of processes assume that the systems go through a series of equilibrium states during the process, which is called a quasi-static process. Reversibility is not required for studiot's entropy change equation in the first post. The equation is fully applicable to bringing two otherwise isolated systems with different temperatures into thermal contact, which I will use as an example:

In the theory of thermodynamic processes, both systems' states change to the final state through a series of individual equilibrium states (*). Because of their different temperatures and conservation of energy (and because/if the higher-temperature system is the one losing heat to the colder) the sum of entropies increases. In the final state, both systems can be considered as two sub-volumes of a common system that is in thermal equilibrium. Since the common system is in equilibrium (and isolated), it has a defined entropy. Since entropy is extensive, it can be calculated as the sum of the two entropies of the original systems' end states. As far as I understand it, removing barriers between two parts of a container is essentially the same as bringing two systems in thermal contact. Except that the two systems can exchange particles instead of heat.

The two systems that are brought into contact are not isolated - they are brought into contact. If one insists on calling the two systems a single, unified system right after contact, then this unified system is not in an equilibrium state (**). And I believe this is exactly where your disagreement lies: Does this unified, non-equilibrated state have an entropy? I do not know. I am tempted to go with studiot and say that entropy in the strict sense is a state variable of thermal equilibrium states - just from a gut feeling. On the other hand, in these "bring two sub-volumes together"-examples the sum of the two original entropies under a thermodynamic process seem like a good generalization of the state variable and converges to the correct state value at the end of the process.

(*): I really want to point out that this is merely a process in the theory framework of equilibrium Thermodynamics. It is most certainly not what happens in reality, where a temperature gradient along the contact zone is expected.

(**): In the absence of a theory for non-equilibrium states this kind of means that it is not a defined thermodynamic state at all. But since there obviously is a physical state, I will ignore this for this post.

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Thank you for your contribution Timo, you poneyed up with many of many underlying thoughts, in particular that Thermo can get quite tricky when you start to think deeply about it.

What you have said deserves some serious thought and consideration and may well lead to a better version of what I am saying.

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I thank you also Timo.

I reread this thread in light of your post and saw a comment I missed and should have addressed.

On 5/19/2019 at 12:11 PM, studiot said:

In order to calculate the entropy change for an irreversible path we must find an alternative reversible method of going from state A to state B and then use this to calculate the entropy change.﻿

I disagree with 'must.'

It is IMO a very good calculation method. I don't think its impossibility re my isolated system would prevent its use with the working assumption that my system is temporarily not isolated. But it's not the only method.

Save mouse wheel....

On 5/19/2019 at 3:50 PM, Carrock said:

You cover "in an isolated system entropy need not increase."

I'll assume ideal, classical conditions, including an ideal gas.

An (old) example of "in some isolated systems entropy can increase:"

You have an x cubic units sealed container of ideal gas, inside an empty sealed container of y cubic units.

A side of the smaller container is 'instantaneously' removed without otherwise affecting the system.

The gas irreversibly expands and comes to equilibrium in the y cubic units container. No work has been done so the temperature has not changed.

The entropy has increased by the factor ln(y/x).

I used this example for simple maths.

The crucial issues.

6 hours ago, timo said:

If one insists on calling the two systems a single, unified system right after contact, then this unified system is not in an equilibrium state (**). And I believe this is exactly where your disagreement lies: Does this unified, non-equilibrated state have an entropy?﻿﻿ I do not know. I am tempted to go with studiot and say that entropy in the strict sense is a state variable of thermal equilibrium states - just from a gut feeling.

The gas in x and the empty rest of y are each in equilibrium states until the first atom leaves the x volume. I believe the entropy of the entire system can be calculated when each part is in self equilibrium.

The following seems to confirm that view.

7 hours ago, timo said:

In the theory of thermodynamic processes, both systems' states change to the final state through a series of individual equilibrium states (*). Because of their different temperatures and conservation of energy (and because/if the higher-temperature system is the one losing heat to the colder) the sum of entropies increases.

After the side is removed but before any atom has passed the position of the missing side calculate the entropy of the gas still in equilibrium in the volume x. See e.g. http://hyperphysics.phy-astr.gsu.edu/hbase/Therm/entropgas.html (latexphobia.). The entropy of the empty part of y is 0.

The subsequent expansion is not isothermal but the final equilibrium temperature of the gas is the same as before it expanded.

The new entropy of the gas in volume y can be calculated.(latexphobia.)

Entropy change:

delta S = nK ln(y/x)

The important difference between my and timo's examples is that in mine there is equilibrium only in the initial and final states and there are always equilibrium states in his.

IMO this isn't a problem...

3 hours ago, studiot said:

....  Thermo can get quite tricky when you start to think deeply about it.﻿

Definitely agree.

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7 hours ago, Carrock said:

The important difference between my and timo's examples is that in mine there is equilibrium only in the initial and final states and there are always equilibrium states in his.

All states are "equilibrium states".

A condition of non equilibrium is not defined/definable.

This arises because the full description of a state comprises a list of the values of all state variables, each of which which must by definition represent the entire system.

Edited by studiot
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On 5/19/2019 at 3:50 PM, Carrock said:

You have an x cubic units sealed container of ideal gas, inside an empty sealed container of y cubic units.

A side of the smaller container is 'instantaneously' removed without otherwise affecting the system.

The gas irreversibly expands and comes to equilibrium in the y cubic units container. No work has been done so the temperature has not changed.

The entropy has increased by the factor ln(y/x).

What about the work done in the expansion?

"No work has been done"

Well look at this

Call the chamber with the gas A and the empty chamber B so the pressures are

Initially PA and PB=0.

Now however quickly some gas transfers itself, it is not possible for the final pressure to be achieved instantaneously.

It takes time for the pressure to equalise throughout the combined chamber.

So the pressure in A is constantly declining and the pressure in B is continually rising, both towards the equilibrium condition.

It is true that as gas expands into B initially it faces zero pressure

But

As soon as a small amout of gas has transferred, at time δt, the pressure in B is no longer zero but some small positive value.

Now any further expansion will be against this (albeit small) positive pressure so work must be done.
Of course PA is now diminished and continues to do so.

As this process continues ie in the next δt, this work will increase as the pressure in B rises.

Edited by studiot
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3 hours ago, studiot said:

All states are "equilibrium states".

A condition of non equilibrium is not defined/definable.

This arises because the full description of a state comprises a list of the values of all state variables, each of which which must by definition represent the entire system.

I don't agree with part or all of this but I suppose I can see its relevance for the following post.

1 hour ago, studiot said:

What about the work done in the expansion?

"No work has been done"

Well look at this....

I tried to indicate I was only concerned with initial and final conditions in my example of how an isolated system can increase its entropy from one well defined value to another well defined value. The total work done in getting from the low entropy state to the final high entropy state, which is an "equilibrium state" where the entire volume y (or A plus B) is in equilibrium, is zero.

While transitioning between states work is done.

2 hours ago, studiot said:

Now any further expansion will be against this (albeit small) positive pressure so work must be done﻿.

Also the temperature of the gas is lowered and it acquires kinetic energy.

As the volume y comes to equilibrium all this work is transformed to heat.

At equilibrium the temperature of the gas in the volume y is the same as it was in x.

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11 hours ago, Carrock said:

I tried to indicate I was only concerned with initial and final conditions in my example of how an isolated system can increase its entropy from one well defined value to another well defined value.

Here you have a problem with the initial conditions, if you conside the system to include the second chamber (B).

Let us look at the description of the 'system'  more closely.

Scenario (1)  -  Your description (please correct this if I misunderstood)

The system comprises the contents of chambers A and B.

The system boundary is formed with isolating adiabatic impenetrable walls.
But thus boundary must sorround both chambers A and B at all times.

However this system is not in Thermodynamic equilibrium.

Classical neither a pressure (state variable) nor a temperature (state variable) can be defined for this system.
This is because pressure and temperature are  intensive properties.

If you think this can be done please indicate how since the values must include those of chamber B.

Yes a volume (state variable) can be defined since it is an extensive property.

Scenario (2)

The system comprises the contents of chamber A alone.

Chamber B forms part of the surroundings of the system.

Now both temperature and pressure variables may be defined and the system is in Thermodynamic equilibrium, since its contribution to extensive properties need not be considered.

Your internal partition wall is not now part of the system but part of the system boundary.
When it is removed, the system looses its isolated status and mass is exchanged with the surroundings in the form of chaamber B.

Note this is no different from considering a chemical reaction in which gas is evolved and collected in rubber (ie flexible) balloon.
In this case the evolved gas 'pushes back' some of the atmosphere and some volume is incorporated in the system.
Work is done and needs to be considered in the energy balance of the system.

Another way to put this is to say that the system has (been) changed.

As far as I can see scenario (2) concurs with what Timo said, but scenario (1) does not.

11 hours ago, Carrock said:

The total work done in getting from the low entropy state to the final high entropy state,

What exactly is your understanding of the relationship between work and entropy change?

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What's tricky is knowing which equation to use when and how this relates to the system definition.

Each way brings its own challenges to overcome.

Engineers have one way to do this, based on open systems and a control volume.

You can also obtain your expression of entropy change using statistical mechanics.

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10 minutes ago, studiot said:

What's tricky is knowing which equation to use when and how this relates to the system definition.

Each way brings its own challenges to overcome.

Engineers have one way to do this, based on open systems and a control volume.

You can also obtain your expression of entropy change using statistical mechanics.

I'll respond to this and the previous post eventually (tomorrow at earliest). I have had limited time, combined with getting back into latex (inspired by comments from you a few months ago); there's some good looking perfectly safe free software but sorting out dependencies etc is a pain and I've been putting it off.

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We are having a good conversation so take your time as needed.

I look forward to your response.

Just be aware that you have made difficulties for yourself by demanding an isolated system.

There can be entropy changes, but not in an isolated one.

As to Tex or MathML

You can use the greek characters available on your computer within the fonts.

In windows run Charmap.exe and copy the required character directly into the typing.
I don't know hoqw to access the fonts in linux, but they are still there.

Alternatively free wisywig maths editors that allow you to build up a maths expression for direct copy and pasting into the forum are available at

or

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On 5/22/2019 at 1:01 PM, studiot said:

However this system is not in Thermodynamic equilibrium.

Classical neither a pressure (state variable) nor a temperature (state variable) can be defined for this system.
This is because pressure and temperature are  intensive properties.

If you think this can be done please indicate how since the values must include those of chamber B.

Yes a volume (state variable) can be defined since it is an extensive property.

I regard this as a definition issue. I had assumed that, as in many other fields, thermodynamics could have two or more subsystems comprising one system, but all I could find was systems.

Quote

When the state of its content varies in space, the system can be considered as many systems located next to each other, each being a different thermodynamical system.

I'll refer to your diagram and labelling to avoid confusion.

The system A (i.e. the contents of chamber A) has a volume of x cubic units.

The system B (i.e. the contents of chamber B) has a volume of y-x cubic units.

The system C, which is the contents of chamber A plus the contents of chamber B, has a volume of y cubic units.

On 5/23/2019 at 2:03 PM, studiot said:

You can also obtain your expression of entropy change using statistical mechanics.

System A contains ideal gas in equilibrium at a temperature   $T= \frac {2U}{ 3Nk}$.

System B is a vacuum.

Just before the start of my scenario the barrier between system A and system B is almost instantaneously removed and taken outside both systems with no significant effect (at that instant) on any of the three systems. More plausible scenarios for this action can be devised. Classically, such things can be done with an arbitrarily small effect on the system.

Without the barrier, at the start of my scenario, system A, still instantaneously in equilibrium, has entropy $\displaystyle S = Nk \Bigg(\ln\bigg(\frac{V_x}{N}\Big(\frac{4 \pi mU}{3Nh^2}\Big)^\frac{3}{2}\bigg)+\frac{5}{2}\Bigg)$

System B has entropy $0\frac{J}{K}$.

On 5/22/2019 at 1:01 PM, studiot said:

Yes a volume (state variable) can be defined since it is an extensive property﻿.

As entropy is an extensive property﻿, the entropy of System C is the sum of the entropies of system A and system B. Do you agree?

Ignoring intermediate steps for now, system C eventually reaches thermal equilibrium.

Its entropy is $\displaystyle S = Nk \Bigg(\ln\bigg(\frac{V_y}{N}\Big(\frac{4 \pi mU}{3Nh^2}\Big)^\frac{3}{2}\bigg)+\frac{5}{2}\Bigg)$

The change in entropy is $\Delta S = nK \ln\big(\frac {y}{x}\big)$

Its temperature is the same as system A's original temperature i.e. $T= \frac {2U}{ 3Nk}$, since neither U nor N has changed. No net work has been done.

Intermediate steps:

On 5/19/2019 at 12:11 PM, studiot said:

In order to calculate the entropy change for an irreversible path we must find an alternative reversible method of going from state A to state B and then use this to calculate the entropy change.

For any completely isolated system we are restricted to adiabatic processes since no heat can either enter or leave the sytem.

For a reversible process in any such system dq=0, hence ΔS is also zero, which means that S is a constant.

Thus if one part of the system increases in entropy another part must decrease by the same amount.

dU = T dS - P dV from http://www.splung.com/content/sid/6/page/secondlaw

and

On 5/23/2019 at 2:37 PM, studiot said:

There can be entropy changes, but not in an isolated [system].

You seem to be saying that if e.g. the left wall of the chamber was rigid, adiabatic and movable, entropy increase would happen since compressing the gas from volume y back down to volume x would be possible i.e. reversing the process; if the left wall is not movable the expansion process is IMO unchanged but entropy cannot increase since the process is irreversible....

From your source http://www.splung.com/content/sid/6/page/secondlaw

Quote

This defines entropy as a mathematical construct which only remains constant in a perfectly efficient (but hypothetical) closed thermodynamic cycle.

i.e. since my example is neither perfectly efficient at producing work nor a closed thermodynamic cycle entropy increases.

IMO some of the above quotes are inconsistent with this:

On 5/20/2019 at 12:00 PM, studiot said:

Entropy is a measure of an observational truth, not the other way round.

I'm not sure if all these are your views or partly representation of Timo's views.

I'm avoiding quoting Timo since we seem to agree with him but draw different conclusions.

On 5/21/2019 at 8:36 PM, studiot said:

All states are "equilibrium states".

A condition of non equilibriu﻿m is not defined/definable.

This arises because the full descriptio﻿n of a state comprise﻿s a list of the values of all state variables, each of which which must by definition represent the entire system.

The above quote is arguably inconsistent with

On 5/22/2019 at 1:01 PM, studiot said:

In this case the evolved gas 'pushes back' some of the atmosphere and some volume is incorporated in the system.

Work is done and needs to be considered in the energy balance of the system.

I took that a bit casually; of course the gas cooling was not work. All work done ends up becoming heat as the system approaches equilibrium.

In short, the system evolution is not in practice describable with any accuracy but the entropy increases monotonically from the initial state to the final equilibrium state.

More I could say but not now...

Edited by Carrock
added entropy change
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Thank you for this detailed reply.

Here are my initial thoughts, more detail to follow as it is late here.

The point I am making is that entropy increases when you consider the entropy of the system plus the entropy of the surroundings.

System B is (by definition) part of the surroundings to system A at the beginning of your process, since it is not part of system A.
System B is also isolated from system A.

During the process systems A and B amalgamate to form system C, which is again isolated from the rest of the universe.

This is therefore not the system you started with, unless you choose to so adopt it, instead of system A.

This approach has the advantage that isolation (of system C = the system) is maintained at all times.

But the volume of system C does not change so ln(V2/V1) = 0

and classical thermodynamics is upheld.

Alternatively you can start with system A alone as the system.

In this case you loose the isolation the moment you remove the barrier between system A and some of its surroundings.

If you calculate the number of microstates in the enlarged system from Boltzman's Law then you will find that the increase in W more than compensates for the loss of entropy in system A alone as a result of the mass efflux.

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• 4 weeks later...

I think it is disappointing that laws of our universe do not permit to create a practical engines which would work without entropy change (and therefore wouldn't require constant fuel supply). Even though such engines may not contradict to the formal 3 laws of thermodynamics. Or they do?

Edited by Moreno

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