Physics of Diffusivity: Brownian Motion & Local Composition

[Clausius2]

While calculating the diffusion coefficient D of a binary gaseous mixture as Hirschfelder do so based in Lennard Jones potential in his book about "Molecular Theory of Gases and Liquids", I have realized that D does not depend on the local composition of the mixture (i.e. D does not depend on the mass fraction of each substance). On the contrary, the coefficient of viscosity depends on such local composition. I am trying to figure out the reason of this disparity, and I think that it has to do with the Brownian motion involved in diffusion mechanism. Have you got any explanation of why D does not depend on local composition?.

Thanks.

 

[alien308]

From general considerations this can not be.
Assume that one of the components have molecules of big radius and big mass.
Accompaniment of such component will bring about reducing D.
Regrettably beside I am absent a book for the detailed answer.

 

[Clausius2]

From general considerations this can not be.
Assume that one of the components have molecules of big radius and big mass.
Accompaniment of such component will bring about reducing D.
Regrettably beside I am absent a book for the detailed answer.

radius and mass are included in Hirschfelder formula:

D=constant\frac{\sqrt{T^3(W_1+W_2)/2W_1W_2}}{P\sigma^2\Omega}

where \sigma is the average between the two radius. Molecule mass is included into molecular weights W1 and W2.

I haven't found any expression of D as a function of local mass fraction of each component.
And I want to know why there is no such dependence.

 

[Gokul43201]

Clausius, that is indeed a most bizarre looking and highly counterintuitive result. There must be some underlying assumptions/conditions related to the system. Are you absolutely sure that the quoted equation is not specifically for an equimolar mixture ?

 

[marlon]

Hey Clausius,

to be honest i must admit that i do not know the answer to your question here but it strikes me as very strange as to how diffusion can be independent of mass. I mean isn't it the mass-gradient that causes the diffusion process in this case ?

regards
marlon

 

[Clausius2]

Clausius, that is indeed a most bizarre looking and highly counterintuitive result. There must be some underlying assumptions/conditions related to the system. Are you absolutely sure that the quoted equation is not specifically for an equimolar mixture ?

Hi Gokul! Thanks!

I have the book "Molecular Theory of Gases and Liquids" of Hirschfelder, Curtiss and Bird, just in front of me. In page 539, there is the "Coefficient of Diffusion in a Binary Mixture". And the formula which I have posted in my last post, where \Omega is a parameter which measures the deviation of the spheroidal molecular model. Moreover, few lines below, Hirschfelder says, D can be corrected by a function f for giving an effective coefficient of diffusion Df. The f function is a function "of molecular weights, mole local fractions and viscosities of both components", and Hirschfelder adds: "An explicit expression for this function which varies only slightly from unity is given in Appendix A of this chapter. The dependence of the diffusion coefficient on the composition of a mixture of gases is hence only slight".

So that I am not inventing anything. My question is why?. As it can be thought at first sight, it would seem to be such dependence.

to be honest i must admit that i do not know the answer to your question here but it strikes me as very strange as to how diffusion can be independent of mass. I mean isn't it the mass-gradient that causes the diffusion process in this case ?

You are confusing the mechanism of diffusion, which can be represented by the Fick's law Y_iV_i=-D_{ij}\nabla Y_i where Y_i is the mass fraction of component (i), V_i is the diffusion velocity vector, and D_{ij} is the coefficient of diffusion. You are confusing how this expression works with the proper coefficient of diffusion. For example, viscosity produces a diffusion of momentum due to velocity gradients, but the coefficient of viscosity is not a function of velocity. By the same reasonement, a difference of composition produces a diffusion of mass, but the coefficient of diffusion has slight dependence on mass fractions. The main difference is that coefficient of viscosity does depend on the local composition, but the binary coefficient of diffusion doesn't make so. It seems like an "average" across all the fluid flow. But I am thinking about the Brownian motion, which is linked to this coefficient, ¡sn't it?. Is not the brownian motion something what have to be averaged also to be quantified?.

I don't know, what do you think?

 

[quetzalcoatl9]

Hi Gokul! Thanks!

I have the book "Molecular Theory of Gases and Liquids" of Hirschfelder, Curtiss and Bird, just in front of me. In page 539, there is the "Coefficient of Diffusion in a Binary Mixture". And the formula which I have posted in my last post, where \Omega is a parameter which measures the deviation of the spheroidal molecular model. Moreover, few lines below, Hirschfelder says, D can be corrected by a function f for giving an effective coefficient of diffusion Df. The f function is a function "of molecular weights, mole local fractions and viscosities of both components", and Hirschfelder adds: "An explicit expression for this function which varies only slightly from unity is given in Appendix A of this chapter. The dependence of the diffusion coefficient on the composition of a mixture of gases is hence only slight".

So that I am not inventing anything. My question is why?. As it can be thought at first sight, it would seem to be such dependence.



You are confusing the mechanism of diffusion, which can be represented by the Fick's law Y_iV_i=-D_{ij}\nabla Y_i where Y_i is the mass fraction of component (i), V_i is the diffusion velocity vector, and D_{ij} is the coefficient of diffusion. You are confusing how this expression works with the proper coefficient of diffusion. For example, viscosity produces a diffusion of momentum due to velocity gradients, but the coefficient of viscosity is not a function of velocity. By the same reasonement, a difference of composition produces a diffusion of mass, but the coefficient of diffusion has slight dependence on mass fractions. The main difference is that coefficient of viscosity does depend on the local composition, but the binary coefficient of diffusion doesn't make so. It seems like an "average" across all the fluid flow. But I am thinking about the Brownian motion, which is linked to this coefficient, ¡sn't it?. Is not the brownian motion something what have to be averaged also to be quantified?.

I don't know, what do you think?

i would think that the diffusivity would definitely depend upon the mole fraction. your velocity autocorrelation and mean-square displacements would be different due to the masses, and also the radial distribution function would be altered.

for a binary mixture in lennard jones, it is the arithmetic mean of the sigma. however, i would guess that this would no longer be the case with varying mole fractions, but would then be a weighted average to get the correct sigma for the ensemble.

 

[Clausius2]

i would think that the diffusivity would definitely depend upon the mole fraction. your velocity autocorrelation and mean-square displacements would be different due to the masses, and also the radial distribution function would be altered.

for a binary mixture in lennard jones, it is the arithmetic mean of the sigma. however, i would guess that this would no longer be the case with varying mole fractions, but would then be a weighted average to get the correct sigma for the ensemble.

Thanks, but Hirschfelder does not mention any modification of sigma.

 

[quetzalcoatl9]

Thanks, but Hirschfelder does not mention any modification of sigma.

the lorentz mixing rules prescribe averaging the sigma (hard-sphere radius) of each atom.

however, it is my understanding that this is for equal mixtures. i have not found anything yet that describes varying mole fractions - and yet clearly the mole fraction will affect the diffusivity, wouldn't you think?

 

[Astronuc]

I would have to agree with Gokul.

I would image the diffusion coefficient like yield strength is measured as a bulk property and does not really apply on the microscopic or local level. Surely local diffusion as well as local strain at the grain level (10 micron, or < 100 micron) cannot be accurately described by the coarse models.

This is a problem when looking at local boiling, particularly where the chemical species have a very complex relationship with the two or more phases, and local temperatures, enthalpies, and electro-chemical potentials.

I am not familiar with Hirschfelder et al, so I cannot comment on their treatment of the matter.