This gas work violates 2nd law of thermodynamics?

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In summary, the gas by doing negative work(losing heat), say accelerating a piston violates the 2nd law of thermo because kinetic energy of the piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter.
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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!!!!
2ndLawT.png
 
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  • #2
Show us your calculation that supports your claim.
 
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  • #3
aliinuur said:
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.
 
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  • #4
russ_watters said:
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.
 
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  • #5
In case you didn't notice:

weirdoguy said:
Show us your calculation that supports your claim.
 
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  • #6
weirdoguy said:
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
 
  • #7
aliinuur said:
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.
 
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  • #8
There is no heat transfer.

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

https://en.wikipedia.org/wiki/Heat said:
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.
 
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  • #9
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.
 
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  • #10
aliinuur said:
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.
 
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  • #11
aliinuur said:
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.
 
  • #12
Philip Koeck said:
Maybe if the pistons are mass- and frictionless ...
The OP implied a "massy" piston as it acquired KE during the expansion.
 
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  • #13
Squizzie said:
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.
 
  • #14
Squizzie said:
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?
 
  • #15
Philip Koeck said:
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!
 
  • #16
Squizzie said:
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.
 
  • #17
Chestermiller said:
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"
 
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  • #18
Squizzie said:
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?
 
  • #19
Chestermiller said:
How would you envision this could be done?
I think that question should be addressed to the OP, @aliinuur
 
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  • #20
Squizzie said:
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?
 
  • #21
Chestermiller said:
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.
 
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  • #22
Squizzie said:
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.
 
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  • #23
Chestermiller said:
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.
 
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  • #24
aliinuur said:
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.
 
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  • #25
And I think you can fit a thousand angels on the pin.
This is a total waste of time!
 
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  • #26
Philip Koeck said:
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.
 
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  • #27
Squizzie said:
As I said, that is the OP's responsibility.
The OP last visited PF in June of this year...

Squizzie said:
This is a total waste of time!
Then stop posting in this thread. Or I can thread ban you if you prefer...
 
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1. Does this gas work really violate the 2nd law of thermodynamics?

No, the gas work does not violate the 2nd law of thermodynamics. The 2nd law states that the total entropy of an isolated system can never decrease over time. Gas work can still be performed in accordance with this law.

2. How can gas work be consistent with the 2nd law of thermodynamics?

Gas work can be consistent with the 2nd law of thermodynamics by considering the overall change in entropy of the system and its surroundings. As long as the total entropy of the system and its surroundings increases or remains constant, gas work can be performed without violating the 2nd law.

3. What are some examples of gas work that do not violate the 2nd law of thermodynamics?

Examples of gas work that do not violate the 2nd law of thermodynamics include expansion work done by a gas in a piston-cylinder system and compression work done on a gas by an external force. In both cases, the total entropy of the system and its surroundings increases or remains constant.

4. Can gas work violate the 2nd law of thermodynamics under any circumstances?

Gas work can potentially violate the 2nd law of thermodynamics if it leads to a decrease in the total entropy of an isolated system over time. However, in practice, gas work is typically performed in a way that is consistent with the 2nd law by ensuring that the overall entropy of the system and its surroundings increases or remains constant.

5. How can we ensure that gas work does not violate the 2nd law of thermodynamics?

To ensure that gas work does not violate the 2nd law of thermodynamics, it is important to carefully consider the entropy changes associated with the work being performed. By accounting for the overall change in entropy of the system and its surroundings, gas work can be carried out in a way that is consistent with the 2nd law.

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