# Some questions about Hard-sphere Maxwellian Gas Collisions

1. Aug 3, 2013

### erielb

Consider a unit volume (rigid walled container of surface area S) containing N molecules with diameter d, having a maxwellian speed distribution with a mean time between collisions t*.

Allowing for a stable (i.e constant) equilibrium mono-layer distribution of some number of these molecules on the wall (say W); what are the answers to the following questions for the balance of molecules distributed throughout the containing volume, assuming only binary collisions.

1) For any time freeze instant, what fraction of the (N-W) translating molecules

(a) are in physical contact with one other molecule?
(b) have a nearest neighbor distance x such that d< x <= 2d ?
(c) on average, what fraction of t* will a typical molecule be separated from its next, or last collision partner by x, where x ranges from d to 2d

2) For the above type gas it is commonly asserted that the "effective" exclusion volume per molecule is 4 times the actual molecular volume. Can anyone falsify this?

2. Aug 4, 2013

### Philip Wood

Where did this come from, I wonder. I can't understand it - which doesn't mean it's wrong. Are we to treat the molecules as spheres? Is x the separation between the CENTRES of two spheres?

The usual simple treatment of the problem is to imagine one molecule (diameter d) as sweeping out a volume of $\frac{4}{3}\pi d^3$, that is 8 times its actual volume, but to represent the other molecules by dots at their centres. No dot can come within the sphere of radius d if the spheres are impenetrable (because when 2 spheres of radius r touch, their centres are 2r = d apart). So the excluded volume for two colliding molecules is $\frac{4}{3}\pi d^3$, that is 8 times the volume of a molecule. But this excluded volume is excluded to BOTH colliding molecules, so the excluded volume per molecule is 4 times the volume of a molecule.

Last edited: Aug 4, 2013
3. Aug 4, 2013

### erielb

Thanks Phillip: orthodox explanation too neat to be true

The total excluded volume of an isolated molecule (under the hard sphere assumption) is as you say 8 times the volume of the molecule itself.

In a collision pair, the actual total excluded volume (and the bounding surface area) attending the paired molecules is less than that of two separated molecules whenever the centers of the paired molecules are less than 2 molecular diameters apart. For molecules in direct contact, the total excluded volume for the pair is some 17% less than that of the fully separated molecules; this reduction monotonically decreases to zero when the separation is greater than or equal to 2d.

Moreover Van der Waal's "b" (the difference between the total containing volume and that portion accessible to the centers of the contained molecules) attributed to the total exclusion volume of the molecules collectively, is commonly asserted to be only 4 times (not 8X) the total collective molecular volume.

Given the reduction of exclusion volumes attending collisions of the molecules among themselves and with the walls, the statement that "b" should precisely equal half the exclusion volume of an isolated molecule is for me simply too neat to be true.

Also the classic argument that collisions of point-particles (molecular centers with no exclusion volume) with the bounding surface of the excluded volume of one finite size molecule can be treated with equal weight in apportioning the excluded volume of the pair to each collision member neglects the fact that in actual collisions of finite hard-sphere molecules, only 1/16 of the bounding surface of each colliding molecule is effective in excluding the other center from entering its excluded zone, and that portions of the total exclusion volume of the pair are due to the "materiality" of the molecules themselves, in addition to the constraints of relative spatial configurations

Just thought others may have similar concerns.

Again, thanks for your feedback Phillip.

Last edited: Aug 4, 2013
4. Aug 5, 2013

### Philip Wood

You are quite right to be dissatisfied with the simple argument I gave. It's the division of 8Vm by 2 which I find hard to justify. I want to say that we do it because the excluded volume comes into play only when molecules collide, so it is only ever relevant when shared by pairs of molecules - but I'm not sure that this is any better than hand-waving.

D Tabor in Gases, Liquids and Solids, Third edition (1991) gives brief discussions of four approaches to estimating the excluded volume.

5. Aug 15, 2013

### erielb

Thanks for the reference, I'll check it out.

My own intuitive approach to this issue is as follows:

if the original empty containing volume were filled with an incompressible fluid, and then supposing that each molecule added to the volume could/would displace an equal volume of fluid out of the container, the totality of the fluid remainder (total volume less volume of molecules collectively) is necessarily interstitial and clearly a fixed and definite multiple of the total molecular volume, no matter how the spatial configurations of the molecules within the volume vary.

However those portions of the interstitial fluid (at any specified instant) inaccessible to the centers of any of the contained molecules [viz. the excluded volume] constitute a variable fraction of the interstitial fluid, dependent upon the relative spatial configurations of the molecules among themselves and the walls.

Given the particle density, molecular diameters, and temperature parameters, the distribution of times for successive collisions can be determined exactly. Binary collision mechanics, allow for a valid "time average" value for the total excluded volume (as a multiple of the total molecular volume) if one uses a rigorous analysis of the relative frequency of collisions of molecular centers with the stationary and translating bounding surfaces of the excluded zones within the volume.

I'm still working to complete and check my calculations.

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