I This gas work violates 2nd law of thermodynamics?

[aliinuur]

TL;DR Summary

Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

 

[weirdoguy]

Show us your calculation that supports your claim.

 

[russ_watters]

TL;DR Summary: Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

No, that does not violate the 2nd Law of thermo. It's not even clear why you think it does because you haven't said anything about the 2nd Law of thermo here.

 

[aliinuur]

No, that does not violate the 2nd Law of thermo. It's not even clear why you think it does because you haven't said anything about the 2nd Law of thermo here.

The 2nd law says that heat always flows from hotter without external work. But in the example a piston is enough to transfer heat from gases which were initially at equal temperature.

 

[weirdoguy]

In case you didn't notice:

Show us your calculation that supports your claim.

 

[aliinuur]

In case you didn't notice:

You want me to calculate entropy change? It is not about entropy, it is about heat transfer from hotter to colder

 

[russ_watters]

But in the example a piston is enough to transfer heat from gases which were initially at equal temperature.

No it doesn't. It transfers energy between the two pistons via mechanical work.

 

[jack action]

There is no heat transfer.

But there is internal energy (gas A) converted into kinetic energy which is converted back into internal energy (gas B).

In thermodynamics, heat is the thermal energy transferred between systems due to a temperature difference. [...]

Heat is energy in transfer to or from a thermodynamic system, by a mechanism that involves the microscopic atomic modes of motion or the corresponding macroscopic properties. This descriptive characterization excludes the transfers of energy by thermodynamic work or mass transfer.

 

[Chestermiller]

Actually, the entropy of each of the gases is higher in the final state than in the initial state (due to viscous dissipation). This is certainly consistent with the 2nd law of thermodynamics.

 

[Philip Koeck]

TL;DR Summary: Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

View attachment 328132

Not sure if this has been pointed out: In this experiment work has been done on the system!
Somehow you have to achieve the situation shown on the left. For example you can pull the two pistons apart to create a vacuum between them.
Your device uses external work to transport heat from cold to hot.

 

[Philip Koeck]

TL;DR Summary: Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

View attachment 328132

More comments (this is actually quite interesting):

If gas A expands adiabatically it doesn't actually transfer any heat to gas B or to the surroundings and if gas B is compressed adiabatically it doesn't receive heat.
In that case there isn't any heat transfer from cold to hot anyway.

Another question is whether the collision between the pistons can be elastic the way the process is drawn. If they move together to their final position that looks more like an inelastic collision.

Maybe if the pistons are mass- and frictionless the collision could be elastic.
In that case I would say that the expansion of A is free (Joule expansion) and the temperature of gas A stays the same.

 

[Squizzie]

Maybe if the pistons are mass- and frictionless ...

The OP implied a "massy" piston as it acquired KE during the expansion.

 

[Philip Koeck]

The OP implied a "massy" piston as it acquired KE during the expansion.

Yes, he does.
My point was that the process shown in the first post can't really be as simple as discussed by the OP.
If the pistons have mass you have to decide what happens when they collide.
Is the collision elastic or inelastic? The result would be very different.
Clearly the pistons would not just stop by themselves as shown.
Some inelastic process would be required to stop them.

You could simplify the discussion by assuming mass-less pistons, but then you probably would get a very different result.
A correct analysis should show that all laws of thermodynamics are obeyed.

If you want to try the analysis, at least I would be interested in what you get.

 

[Chestermiller]

The OP implied a "massy" piston as it acquired KE during the expansion.

So. What is your solution to this problem for the final pressures, volumes, temperatures, and entropy change?

 

[Squizzie]

If you want to try the analysis, at least I would be interested in what you get.

I think I'll pass on that.
Coding the physics behind the latch preventing the RH piston from expanding would be tricky!

 

[Chestermiller]

I think I'll pass on that.
Coding the physics behind the latch preventing the RH piston from expanding would be tricky!

Then just tell us in works. how you would solve it.

 

[Squizzie]

Then just tell us in works. how you would solve it.

I would need to know what is the nature of the mechanism ensuring "piston B can only move to the right"

 

[Chestermiller]

I would need to know what is the nature of the mechanism ensuring "piston B can only move to the right"

How would you envision this could be done?

 

[Squizzie]

How would you envision this could be done?

I think that question should be addressed to the OP, @aliinuur

 

[Chestermiller]

I think that question should be addressed to the OP, @aliinuur

Are you really saying that you are unable to conceive of how this could be done?

 

[Squizzie]

Are you really saying that you are unable to conceive of how this could be done?

There are many ways: Each one would generate its unique solution.
I'm not inclined to work through them all - or indeed any of them, until the requirement is specified.

 

[Chestermiller]

There are many ways: Each one would generate its unique solution.
I'm not inclined to work through them all - or indeed any of them, until the requirement is specified.

Let's see any one of your options.

 

[Squizzie]

Let's see any one of your options.

The reason I'm not inclined to do so is that I wish to avoid a "number of angels on a pin" argument about the mechanism that you are asking me to invent.
As I said, that is the OP's responsibility.
[EDIT] In any consulting assignment, it is always the client's responsibility to set the requirements.

 

[Philip Koeck]

TL;DR Summary: Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

View attachment 328132

Back to the OP.

Let's assume the following:
Both gases are perfectly insulated from cylinder and pistons.
The pistons are quite heavy so the expansion and compression are close to quasi-static.
The collision between the pistons is completely inelastic so that they stay together after the collision.
The mechanism that holds the pistons in place in various positions can be retracted and inserted with almost no friction.
The process is stopped by inserting this mechanism exactly when the two pistons stop moving on the far right.

With these specifications I would say that the expansion of gas A and the compression of gas B are both adiabatic so there is no heat transfer between the gases and therefore their entropy is constant.

The heat produced during the collision of the pistons all goes into the pistons which leads to a temperature and an entropy increase in the pistons.
As far as I can see that is the only entropy change in this process.

Happy to receive feedback, especially if I've missed something.

 

[Squizzie]

And I think you can fit a thousand angels on the pin.
This is a total waste of time!

 

[Chestermiller]

Back to the OP.

Let's assume the following:
Both gases are perfectly insulated from cylinder and pistons.
The pistons are quite heavy so the expansion and compression are close to quasi-static.
The collision between the pistons is completely inelastic so that they stay together after the collision.
The mechanism that holds the pistons in place in various positions can be retracted and inserted with almost no friction.
The process is stopped by inserting this mechanism exactly when the two pistons stop moving on the far right.

With these specifications I would say that the expansion of gas A and the compression of gas B are both adiabatic so there is no heat transfer between the gases and therefore their entropy is constant.

The heat produced during the collision of the pistons all goes into the pistons which leads to a temperature and an entropy increase in the pistons.
As far as I can see that is the only entropy change in this process.

Happy to receive feedback, especially if I've missed something.

I have another version that I like. The heat capacities of the pistons are negligible so that all the kinetic energy of the piston ultimately ends up as internal energy of the gases. The only issue then is how it is distributed between the gases, assuming that the piston is non-conducting.

 

[berkeman]

As I said, that is the OP's responsibility.

The OP last visited PF in June of this year...

This is a total waste of time!

Then stop posting in this thread. Or I can thread ban you if you prefer...

 

[aliinuur]

I would need to know what is the nature of the mechanism ensuring "piston B can only move to the right"

there are a lot of ways to make piston move only in one direction. For instance, i think of mechanism that just locks piston B so that the piston is oscillates between gas B and the mechanism until gas B completley absorbs the kinetic energy.

 

[aliinuur]

Show us your calculation that supports yo

Not sure if this has been pointed out: In this experiment work has been done on the system!
Somehow you have to achieve the situation shown on the left. For example you can pull the two pistons apart to create a vacuum between them.
Your device uses external work to transport heat from cold to hot.

i get this as "if heat from hotter pbject flows to colder one that is because external work has been done on hotter object to heat it up". this argument makes the 2nd law meaningless.

 

[Philip Koeck]

Not sure if this has been pointed out: In this experiment work has been done on the system!
Somehow you have to achieve the situation shown on the left. For example you can pull the two pistons apart to create a vacuum between them.
Your device uses external work to transport heat from cold to hot.

You write: "i get this as "if heat from hotter pbject flows to colder one that is because external work has been done on hotter object to heat it up". this argument makes the 2nd law meaningless."

Could you please explain that a bit more? I'm not following.

 

[Standard]

Let's assume that TA=TB
Piston A will be pushed back into the vacuum.

Let's assume that the second piston (gas B) is held to prevent it from doing the same thing as piston A.

Now if piston A moves forward.
When piston A reaches piston B, the temperature of A has changed, so TA<TB, because gas A has expanded (it now occupies a larger volume).

So the pressure of piston B will prevent piston A from advancing and so piston B will not be pushed back, but on the contrary piston B will push piston A back on contact when piston B is released.

 

[Chestermiller]

What is the exact word-for-word statement of this problem?

 

[aliinuur]

What is the exact word-for-word statement of this problem?

the statement of the problem: Heat transfer on its own, without external work, what is impossible by the 2nd law.

 

[Chestermiller]

the statement of the problem: Heat transfer on its own, without external work, what is impossible by the 2nd law.

That is specified by the Clausius inequality. The entropy generated by a process in a closed system is positive or zero.

 

[aliinuur]

That is specified by the Clausius inequality. The entropy generated by a process in a closed system is positive or zero

does that mean definiton with 2 bodies should be abandoned in favor of entropy definiton?

 

[berkeman]

does that mean definiton with 2 bodies should be abandoned in favor of entropy definiton?

What in the world are you saying / asking?

 

[Demystifier]

@aliinuur A statement of the 2nd law relevant in your case is this: https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Clausius_statement
Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.

Here the crucial part is "without some other change ...". In your example there is also some other change involved, so your example does not violate the 2nd law as stated above.

 

[Philip Koeck]

@aliinuur A statement of the 2nd law relevant in your case is this: https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Clausius_statement
Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.

Here the crucial part is "without some other change ...". In your example there is also some other change involved, so your example does not violate the 2nd law as stated above.

Just to expand a bit: The other thing that changes is that the potential energy that's stored in the device due to the pressure difference between gas A and vacuum is used up when gas A is allowed to expand.

Actually it's not even clear that any heat is transferred between the gases in the process. The expansion of A and the compression of B could both be adiabatic.

 

[Chestermiller]

does that mean definiton with 2 bodies should be abandoned in favor of entropy definiton?

I don't understand what you are asking.

 

[Demystifier]

does that mean definiton with 2 bodies should be abandoned in favor of entropy definiton?

Unlike others, I understood that you are talking about different formulations of the 2nd law. Some formulations don't explicitly talk about entropy, but about heat, which are equivalent because $dQ=TdS$. But AFAIK no formulation talks about two bodies.

 

[Chestermiller]

Unlike others, I understood that you are talking about different formulations of the 2nd law. Some formulations don't explicitly talk about entropy, but about heat, which are equivalent because $dQ=TdS$.

dQ=TdS only for a reversible path.

 

[Demystifier]

dQ=TdS only for a reversible path.

It doesn't make sense. For a reversible path $dS=0$ (if by $S$ one means entropy of the whole system), hence $dQ=0$, hence the content of $dQ=TdS$ is trivial for a reversible path. For $dQ=TdS$ to be non-trivial, it has to be applicable for non-reversible processes as well.

 

[Chestermiller]

It doesn't make sense. For a reversible path $dS=0$ (if by $S$ one means entropy of the whole system), hence $dQ=0$, hence the content of $dQ=TdS$ is trivial for a reversible path. For $dQ=TdS$ to be non-trivial, it has to be applicable for non-reversible processes as well.

dS is not generally zero for a reversible path. Consider reversible isothermal expansion.

 

[Chestermiller]

There have been many statements of the Second Law over the years, couched in a complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800s. (The following is a somewhat fictionalized account, designed to minimize the historical discussion, and focus more intently on the scientific findings.) Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
$$I=\int_{T_i}^{T_f}{\frac{dQ}{T_{int}}}$$where $T_{int}$ is the temperature at the interface with the surroundings. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing: For any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) is not arbitrary; instead, there is a unique upper bound to the value of the integral. Clausius also found that this observation is consistent with all the “word definitions” of the Second Law.

Reference: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

 

[Philip Koeck]

TL;DR Summary: Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

View attachment 328132

I'd really like to see some more details from the OP.
For example, why is gas A loosing heat and where is there a transport of heat in this process?

 

[Demystifier]

dS is not generally zero for a reversible path. Consider reversible isothermal expansion.

I don't get it. Reversible means the final state is identical to the initial one, right? The identical states have the same $S$, hence the overall change of entropy during the whole path is $\Delta S=0$. Hence either (i) $dS=0$ at all parts of the path, or (ii) $dS>0$ at some parts of the path and $dS<0$ at other parts. But $dS<0$ contradicts the 2nd law, which rules out (ii). What remains is (i) $dS=0$ during the whole path. Do I miss something?

 

[aliinuur]

I'd really like to see some more details from the OP.
For example, why is gas A loosing heat and where is there a transport of heat in this process?

heat of gas A goes to kinetic energy of the piston A, heat transfered via pistons, but there is no external work or something like that

 

[Philip Koeck]

heat of gas A goes to kinetic energy of the piston A, heat transfered via pistons, but there is no external work or something like that

Gas A is not necessarily giving off any heat. If the expansion is adiabatic then Q = 0, for example. However the temperature decreases!

If the expansion is somewhere between adiabatic and isothermal then heat actually flows into gas A. Even so gas A would get colder.

If the expansion is isothermal then heat flows into gas A to balance the work done during expansion and the temperature is constant.

You could find an expansion where heat leaves the gas, but that would have to be steeper than an adiabatic curve in a PV-diagram.

Maybe this helps: For an ideal gas inner energy is proportional to temperature. So if the temperature of gas A drops this just means some of its inner energy has been used to do expansion work. It doesn't necessarily mean there was a flow of heat.

Concerning work that's put in: Some work was obviously needed to get the device into the state shown on the left.

 

[Chestermiller]

I don't get it. Reversible means the final state is identical to the initial one, right? The identical states have the same $S$, hence the overall change of entropy during the whole path is $\Delta S=0$. Hence either (i) $dS=0$ at all parts of the path, or (ii) $dS>0$ at some parts of the path and $dS<0$ at other parts. But $dS<0$ contradicts the 2nd law, which rules out (ii). What remains is (i) $dS=0$ during the whole path. Do I miss something?

Reversible CYCLE means that the final state of the system is identical to the initial one. Reversible path between state A and state B of the system does not require that states A and B of the system are the same.

 

[Demystifier]

Reversible CYCLE means that the final state of the system is identical to the initial one. Reversible path between state A and state B of the system does not require that states A and B of the system are the same.

Fine, the states $A$ and $B$ are not the same. But do they have the same entropy, namely, is $S(B)=S(A)$? You seem to be saying that one entropy can be larger, e.g. $S(B)>S(A)$. But then in the path $B\to A$ we have $dS<0$, which contradicts the 2nd law. So how can such a path exist?