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Does Gradient of Fugacity Create Entropy?

by Darwin123
Tags: entropy, fugacity, gradient
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DrDu
#19
Jan11-13, 02:05 AM
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The forces and fluxes of linear thermodynamics are identified using the fundamental law
##dS=dU/T+p/TdV-\sum_i\mu_i/T dN_i ##.
So entropy production due to a flux of energy U is due to the gradient of 1/T, due to volume change to grad p/T and due to particle number flux due to grad μ/T.
See, e.g.,
http://en.wikipedia.org/wiki/Onsager...ocal_relations
Chestermiller
#20
Jan11-13, 09:50 AM
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Quote Quote by Rap View Post
Thanks for posting those equations, I read them using googlebooks and once you read them, you cannot re-read them.

I'm still thinking about those equations, but the statements you made above I cannot see. If I understand your scenario, you have a cylinder with a gas at equilibrium, and you move the piston to the right with a velocity much larger than the speed of sound. This is Joule expansion, which I think of as having a cylinder with a boundary in the middle. No particles go thru, no motion. On the left is the equilibrium gas, on the right, a vacuum. Then you remove the boundary instantaneously. I think the two scenarios are effectively the same. Anyway, looking at it intuitively from a particle point of view, I see the really hot particles in the region, where the boundary was, heading immediately into the vacuum with few collisions, the warm particles doing the same, but more slowly and with more collisions, etc. and the region around the boundary cooling as a result of losing these hot particles. Mathematically, since the Maxwell-Boltzmann distribution admits particles with arbitrarily high energy, the rate of propagation to the right will be instantaneous. Also, the effect of the boundary removal moves in the other direction, to the left, into the gas, at the speed of sound, dropping its temperature. The slower rate will be due to the high density and consequent collisions. I see a non-maxwellian distribution occurring to the right, and if you define a pseudo-temperature as the average kinetic energy per particle divided by 3k/2, then the pseudo-temperature profile will go down to the right of the dividing point, then going up towards the new right face of the cylinder. A short time after the boundary removal, the pseudo-temperature at the right face will be extremely high (corresponding to the very high energy tail of the formerly Maxwell-Boltzmann distribution) but will then drop rapidly as the whole volume equilibrates. I don't think there will be a narrow transition region during the equilibration process. The transition region will be much more spread out. This is not to say that the mathematical equations above will not predict this, only to say that if they do, then Joule expansion is outside their ability to accurately calculate.

During the main part of the expansion, I think the region to the right of the dividing line will be seriously out of equilibrium, not describable by a Maxwell-Boltzmann distribution, maybe not really having a well defined temperature or entropy. I don't think it can be described using the above equations. On an S-T diagram, I see the gas being a point, then the boundary is removed, the point disappears, then appears at the new equilibrium point at a later time, same T (and N), larger S. Maybe the final approach to equilibrium can be roughly estimated, always plotted on that constant T line.
I think we are having trouble communicating because of differences in our backgrounds. Yours is from a molecular background, and mine is from a continuum mechanics background.

In any event, I didn't mean to imply that the piston is moving at a speed exceeding the speed of sound, but only at the speed necessary to drop the pressure at the piston face to a somewhat lower value (say 50% of the original pressure) and hold it there. I also didn't mean to imply that I am creating anything close to a pure vacuum in the right hand portion of the cylinder. In both regions of the gas, the pressure and density are high enough for the continuum gas dynamics equations to apply. The mean free path of the molecules will still be very short, as will the time between collisions. The Maxwell-Boltzmann distribution should not be seriously disturbed.

Gas dynamics problems of this type can nowadays be routinely solved using computational fluid dynamics (CFD) software. Although I have not followed the literature in great detail, I am confident that the results of such calculations have been extensively verified experimentally over the years. If you are interested in seeing how simple problems are analyzed using gas dynamics, I refer you to our old friend BSL, Transport Phenomena. A worked example, presenting all the assumptions used, is presented in section 11.4, p. 350, Example 11.4-7 One-Dimensional Compressible Flow: Velocity, Temperature, and Pressure Profiles in a Stationary Shock Wave (If I remember correctly, you were interested in learning how all this applies to shock waves). The system they are analyzing is an open system (flow system) rather than a closed (fixed mass) system that I discussed in my post, but I'm sure you will get the connection.

Once the gas dynamics equations have been solved (say for the closed system situation that I discussed), you can take the results and calculate the local transient rate of entropy generation from temperature gradients and velocity gradients, and plug these rates into the entropy change equation I gave (which applies to both the reversible and irreversible process in which diffusional mass transport is not occurring). You can thereby calculate the change in entropy for the system, between the two equilibrium states as well as at all points along the reversible path. All this is valid as long as the assumption of continuum behavior for all portions of the system is valid (short mean free path and short time between collisions).

Another interesting feature that the equation I provided for the entropy change tells us is that, while the dQ/T at the boundary is less for an irreversible path than for a reversible path, the entropy generation from temperature and velocity gradients for the irreversible process compensates for this so that the total entropy change between the starting and final equilibrium states is the same for both paths.
Studiot
#21
Jan12-13, 04:25 AM
P: 5,462
First let me observe that this thread has veered way off topic but I assume the mods are allowing it to continue for interest's sake.

Dr Du has posted the link between Rap's analysis and that of ChesterMiller.

For those (like me) who are not deeply into this part of subject I found pages 307 to 320 of Callen (Thermodynamics and an Introduction to Thermostatistics) highly illuminating, offering a tight but readable introduction to Onsager.
In particular, Rap, you will find the answer to your question in post#18 there.
One caution, however, like so many workers in This subject Callen has some differences in terminology and sign conventions. In relation to this thread his use of the words closed or closure should be checked.
Rap
#22
Jan12-13, 11:11 AM
P: 789
Quote Quote by Studiot View Post
First let me observe that this thread has veered way off topic but I assume the mods are allowing it to continue for interest's sake.

Dr Du has posted the link between Rap's analysis and that of ChesterMiller.

For those (like me) who are not deeply into this part of subject I found pages 307 to 320 of Callen (Thermodynamics and an Introduction to Thermostatistics) highly illuminating, offering a tight but readable introduction to Onsager.
In particular, Rap, you will find the answer to your question in post#18 there.
One caution, however, like so many workers in This subject Callen has some differences in terminology and sign conventions. In relation to this thread his use of the words closed or closure should be checked.
Yes, Dr. Du's Onsager link made me realize that this is very important. I have never studied the Onsager relationships, and it's very interesting. Not sure I get it yet. I will check out Callen. I think the link is still on topic (in my mind anyway) its just that the fugacity fits into a larger picture, and we have to get the larger picture first.
Rap
#23
Jan23-13, 11:05 PM
P: 789
Quote Quote by DrDu View Post
The forces and fluxes of linear thermodynamics are identified using the fundamental law
##dS=dU/T+p/TdV-\sum_i\mu_i/T dN_i ##.
So entropy production due to a flux of energy U is due to the gradient of 1/T, due to volume change to grad p/T and due to particle number flux due to grad μ/T.
See, e.g.,
http://en.wikipedia.org/wiki/Onsager...ocal_relations
Ok, hopefully I understand Onsager a bit better. It seems to me that you cannot talk about "does a gradient in x produce entropy" without specifying what other variables are held constant. In the example above, a gradient in /T produces entropy when U and V are held constant. Is this what we wish to be held constant when answering the question of whether a fugacity gradient creates entropy? Also, I am curious to find out what the entropy creation is for a specified fugacity gradient. If U and V are constant (no gradients), then the rate of entropy creation is [itex]dS_c/dt = L_{NN} (\nabla (\mu/T))^2[/itex] where [itex]L_{NN}[/itex] is one of the elements of Onsager's matrix elements, directly related to Fick's law and the coefficient of diffusion, and [itex]S_c[/itex] is the density of created entropy. The problem is that [itex]\nabla (\mu/T)[/itex] expands out to give a gradient in fugacity plus terms involving a gradient in temperature. We have specified that there is no gradient in U and V, but not that there is no gradient in T. So its not that simple.

I feel sure that a gradient in fugacity will produce entropy in general, but there appear to be cases where it does not, depending on what is held constant. I think I have learned that the statement that differences (or gradients) in X cause changes (or fluxes) in Y means nothing until you have specified which other state variables have been held constant.


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