Free Expansion of gas


by kulkajinkya
Tags: free expansion, joule expansion
Chestermiller
Chestermiller is online now
#19
Apr23-13, 06:10 PM
Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,505
Quote Quote by Jano L. View Post
The distinction between the internal pressure and external pressure is important only for macroscopic portion of gas, like all the gas inside the valve. What I meant is that the formula pV^γ applies to a portion of gas of volume V small enough that the gradient of the pressure is negligible over it. Then there is one pressure P and locally pV^γ = const. In mathematical treatments this is more often written as dS = 0 for any such volume, i.e. the flow is assumed isentropic. Of course, this is just a simplification, and the flow and other transport processes may very well result in a different behaviour, but the adiabatic process is the closest simple thing.
I agree with this assessment. I've done some analysis of gas dynamics involving the removal of a partition between a high pressure chamber and a low pressure chamber, and this is consistent with what my analysis led to. It is worth mentioning that the pressure will not be uniform in each chamber, and that pressure waves can travel down each chamber (at roughly the speed of sound) forming a boundary between a low pressure region and an higher pressure region. All the adiabatic reversible compression will take place at the wave front. This, at least, is the idealized situation that prevails in the absence of viscous dissipation and heat conduction.

Chet
Andrew Mason
Andrew Mason is offline
#20
Apr24-13, 03:14 PM
Sci Advisor
HW Helper
P: 6,591
Quote Quote by Jano L. View Post
In mathematical treatments this is more often written as dS = 0 for any such volume, i.e. the flow is assumed isentropic. Of course, this is just a simplification, and the flow and other transport processes may very well result in a different behaviour, but the adiabatic process is the closest simple thing.
And my point is that the gas flow is NOT isentropic. It is adiabatic, sure. But not isentropic. As the pressure in the two sides get close to being equal, the flow may approach isentropic flow.

AM
Jano L.
Jano L. is offline
#21
Apr25-13, 02:41 AM
P: 1,030
Well, it depends on how fast is the expansion. If it is very fast, then the state of the gas will be very far from equilibrium and it may be even impossible to assign it entropy.

If the expansion is slower, but a lot of friction and dissipation occurs, then the formula pV^γ does not apply; perhaps some other exponent could do better.

But if the expansion is slow and friction is negligible, even for high pressure difference, the flow may be modelled approximately as quasistatic process. Then the expansion is isentropic. Of course, in the end, the entropy has to increase, since the volume increased, but this is more properly assigned to subsequent conduction of heat that will equilibrate the temperature throughout the whole system. This process is much slower than the expansion, so within good approximation, it can be separated from the the latter.

Of course, in reality the expansion will not be exactly isentropic, but if we want to estimate e.g. temperatures right after the expansion, it seems that isentropic process is the simplest and quite reasonable approximation.
Chestermiller
Chestermiller is online now
#22
Apr25-13, 09:29 AM
Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,505
This is a very interesting discussion, and I'd like to add to it. In my posting #19, I alluded to a gas dynamics analysis I did to try to get a better understanding of what is going on in situations like these. I'd like to elaborate a little. In the analysis, I deliberately set the viscosity and thermal conductivity of the gas equal to zero to see what happens in the limit without heat conduction and viscous dissipation. I assumed a perfectly insulated cylinder, with a frictionless massless, insulating piston initially separating two chambers of equal volume. There were equal number of moles in the two chambers, but the temperature (and pressure) in one of the chambers was initially higher than in the other chamber. Prior to time zero, the piston was held in place, but, at time zero, the piston was released. Here are some of the results of the analysis:

1. Entropy was conserved for every parcel of mass within the system.
2. Mechanical energy was conserved even though the deformations were very rapid.
3. The system could never attain thermodynamic equilibrium because there were no dissipative processes operating.
4. The deformation and flow continued forever. The system behavior was analogous to a spring-mass system. The massless piston oscillated back and forth without ever coming to an equilibrium position.
5. Compression and expansion waves traveled up and down each of the cylinders at roughly the speed of sound (at the average temperature).
6. The kinetic energy of the gas was significant and oscillatory.

If anyone is interested in seeing a write up on the analysis, please contact me via a private message.

Chet
vivesdn
vivesdn is offline
#23
Apr29-13, 02:07 PM
P: 104
Reversibility is not about speed but about small differences. Of course, small differences account for slow processes. But an expansion against the vacuum, even when it is slow (small nozzle), it is irreversible as pressure difference is not differentially small between the parts. So the change is not isentropic (entropy will increase), and being adiabatic, the temperature will decrease.


Register to reply

Related Discussions
Free Expansion is non-spontaneous?! General Physics 6
free expansion of BEC Atomic, Solid State, Comp. Physics 0
Free expansion of an ideal gas. Classical Physics 4
free expansion Classical Physics 8
Free expansion of an ideal gas Classical Physics 5