The discussion centers on the interpretation of the Second Law of Thermodynamics in relation to gas behavior during piston movement. Participants argue that the process described does not violate the Second Law, as heat transfer occurs only under specific conditions, and mechanical work is involved in the energy transfer between gases. The key takeaway is that while one gas may cool as another heats up, this process adheres to thermodynamic principles, particularly regarding entropy changes and energy conservation.
PREREQUISITESStudents of thermodynamics, mechanical engineers, physicists, and anyone interested in understanding the principles governing energy transfer and thermodynamic systems.
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.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?
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.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.
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.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!!!!
I assume even an ideal gas would have a certain viscosity due to random collisions between molecules.Chestermiller said:This is incorrect. Gas viscosity damps the motion so that, in the final state, the pistons are no longer moving.
Definitely. See chapter 1, Transport Phenomena by Bird el al.Philip Koeck said:I assume even an ideal gas would have a certain viscosity due to random collisions between molecules.
Do you agree?