This gas work violates 2nd law of thermodynamics?

  • Context: Undergrad 
  • Thread starter Thread starter aliinuur
  • Start date Start date
  • Tags Tags
    Gas Work
Click For Summary

Discussion Overview

The discussion revolves around the implications of a gas performing negative work, specifically in the context of a piston system, and whether this scenario violates the second law of thermodynamics. Participants explore theoretical aspects, calculations, and the mechanics involved in heat transfer and energy conversion within the system.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants argue that the scenario described does not violate the second law of thermodynamics, questioning the reasoning behind the claim.
  • Others assert that a piston can transfer heat between gases at equal temperatures, suggesting that mechanical work is sufficient for energy transfer.
  • A participant emphasizes that heat is defined as energy transferred due to temperature differences, which may not apply in the described scenario.
  • There is a discussion about the nature of the pistons, with some suggesting that if the pistons are massless, the analysis would differ significantly.
  • Concerns are raised about the collision between pistons being elastic or inelastic, which could affect the outcomes of the energy transfer.
  • One participant proposes a specific scenario where both gases are insulated and the pistons are heavy, leading to adiabatic processes and constant entropy for the gases.
  • Another participant expresses reluctance to engage in detailed calculations without clear specifications of the mechanism involved.

Areas of Agreement / Disagreement

Participants generally disagree on whether the described scenario violates the second law of thermodynamics, with multiple competing views and interpretations of the mechanics involved. The discussion remains unresolved, with no consensus reached on the implications of the gas work performed by the piston system.

Contextual Notes

Limitations include assumptions about the nature of the pistons, the specifics of the mechanism ensuring piston movement, and the conditions under which heat transfer occurs. The discussion also highlights the complexity of analyzing energy transfer in thermodynamic systems.

  • #61
Philip Koeck said:
I've tried to simplify the problem, hopefully retaining its essence, and do an analysis, but I ran into some difficulties. Maybe someone can tell me how to proceed.

If I have a closed cylinder as in the sketch but only 1 piston between gas A and gas B, so no vacuum, then I don't need to think about collisions between pistons.
Initially I have the following situation:
Gas A occupies 1/3 of the cylinder on the left, gas B occupies 2/3 on the right and they are separated by a piston. The number of moles (n) is the same for both and remains constant.

For the starting values of volume, pressure and temperature we have:
VB = 2 VA
PA = 2 PB
TA = TB = T

There is no heat transfer between a gas and anything else (piston, cylinder or the other gas).
The piston is extremely heavy so that expansion and compression will be quasi-static.

If I release the piston gas A will expand very slowly and gas B will be compressed. The sum of volumes is constant.

This process should stop when the pressures of gas A and B are the same.
Gas A will have cooled off and gas B will have heated up.

Here's my problem:
If I treat the expansion of A and the compression of B as reversible adiabatic, then the whole process will be reversible since nothing happens in the surroundings.
(I checked that analytically, just to be sure, and if I use expressions for reversible adiabatic processes ( P Vγ = constant, for example), I get ΔS = 0 for both gases (with some rounding errors).

However, by just looking at it, I would swear the process must be irreversible.

How would you calculate the final temperatures and volumes of the gases and the entropy-changes?
There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.
 
  • Like
Likes   Reactions: Philip Koeck
Science news on Phys.org
  • #62
Chestermiller said:
There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.
Chestermiller said:
There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.
I think gas A and gas B volumes oscillate between 1/3V and 2/3V, so yeah, this is reversible process.
 
  • Sad
Likes   Reactions: weirdoguy
  • #63
This is incorrect. Gas viscosity damps the motion so that, in the final state, the pistons are no longer moving.
 
  • #64
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!!!!
There is no transfer of heat in this scenario. There is work being done, and temperatures are changing, but neither of those things need be associated with a transfer of heat.

Heat is a transfer of energy due to a temperature difference. Just because there's a temperature change you can't assume there's heat energy being transferred. There could instead be (and indeed there is) work done. That is the essence of the 1st Law of Thermo.
 
  • Like
Likes   Reactions: aliinuur and Philip Koeck
  • #65
Chestermiller said:
This is incorrect. Gas viscosity damps the motion so that, in the final state, the pistons are no longer moving.
I assume even an ideal gas would have a certain viscosity due to random collisions between molecules.
Do you agree?
 
  • #66
Philip Koeck said:
I assume even an ideal gas would have a certain viscosity due to random collisions between molecules.
Do you agree?
Definitely. See chapter 1, Transport Phenomena by Bird el al.
 
  • Like
Likes   Reactions: Philip Koeck

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
1K
  • · Replies 60 ·
3
Replies
60
Views
10K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 69 ·
3
Replies
69
Views
7K
  • · Replies 14 ·
Replies
14
Views
2K