Register to reply 
Free Expansion of gas 
Share this thread: 
#19
Apr2313, 06:10 PM

Mentor
P: 5,370

Chet 


#20
Apr2413, 03:14 PM

Sci Advisor
HW Helper
P: 6,683

AM 


#21
Apr2513, 02:41 AM

PF Gold
P: 1,168

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. 


#22
Apr2513, 09:29 AM

Mentor
P: 5,370

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 springmass 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 


#23
Apr2913, 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 nonspontaneous?!  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 