Question about the relation b/w mean free path and other variables

[FallenLeibniz]

Homework Statement

The problem that I am having stems from a problem given in the following way:[/B]

"a)Show that for a gas, the mean free path $\lambda$ between collisions is related to the mean distance between nearest neighbors $r$ by the approximate relation $\lambda \approx r\frac{r^2}{\sigma}$ where $sigma$ is the collision cross section."​

Homework Equations

a)Equation relating mean free path to molecule density and cross sectional area:

$\lambda \approx \frac{1}{n\sigma}$
---n is the number of molecules/number of collisions in a volume V that's
swept by the molecule during its travel
---$\sigma$ is the collision cross section defined as $\pi(2R)^2$ where R is the radius of the molecule modeled as a sphere.​

The Attempt at a Solution

[/B]After hours of working with it, I finally found that I can get the author's solution if I set the volume swept by the molecule approximately equal to the N cubes of length r...

$V_{swept_cyl} \approx V_{cubes}$
$(2\pi)(\sigma)(N)(\lambda) \approx N(r^3)$

However, I fail to see how this is totally justified in a physical context. Is it that we estimate that the cylinder is filled with those r length cubes as on average they (the neighbors) will be "about" that r distance away from the molecules the traveling one hits?

 

[Charles Link]

When $n$ is the density of atoms (targets), the number of collisions $N$, (assuming the target atoms stay stationary), when another atom travels a distance $s$ is given by $N=n s \sigma$ because $s \sigma$ is the volume that is traversed. (The cross section $\sigma$ is the projected area of the atom that is moving that is able to cause a collision, with the other atoms considered as points. $\sigma$ could also be viewed as the area of overlap.) The $s$ corresponding to $N=1$ is the mean free path $\lambda$. This gives $\lambda=\frac{1}{n \sigma}$. For atoms with density $n$, if they are in a cubic array, each atom occupies a volume $v=R^3$, where $R$ is the nearest neighbor distance, so that density $n=\frac{1}{R^3}$, which gives $\frac{1}{n}=R^3$ . $\\$ As a homework helper, the rules say I am not supposed to give you the answer, but you put sufficient effort into the problem, and it really shouldn't have been so difficult=the instructor could perhaps have given some more coaching if you got stuck. $\\$ And also note: The mean free path formula $\lambda=\frac{1}{n \sigma}$ is somewhat loosely derived with the conditions set above. I believe this is the generally accepted way of deriving it, and it isn't supposed to be tremendously mathematically rigorous and necessarily applying to the general case with all of the atoms in motion.

 

[FallenLeibniz]

When $n$ is the density of atoms (targets), the number of collisions $N$, (assuming the target atoms stay stationary), when another atom travels a distance $s$ is given by $N=n s \sigma$ because $s \sigma$ is the volume that is traversed. (The cross section $\sigma$ is the projected area of the atom that is moving that is able to cause a collision, with the other atoms considered as points. $\sigma$ could also be viewed as the area of overlap.) The $s$ corresponding to $N=1$ is the mean free path $\lambda$. This gives $\lambda=\frac{1}{n \sigma}$. For atoms with density $n$, if they are in a cubic array, each atom occupies a volume $v=R^3$, where $R$ is the nearest neighbor distance, so that density $n=\frac{1}{R^3}$, which gives $\frac{1}{n}=R^3$ . $\\$ As a homework helper, the rules say I am not supposed to give you the answer, but you put sufficient effort into the problem, and it really shouldn't have been so difficult=the instructor could perhaps have given some more coaching if you got stuck. $\\$ And also note: The mean free path formula $\lambda=\frac{1}{n \sigma}$ is somewhat loosely derived with the conditions set above. I believe this is the generally accepted way of deriving it, and it isn't supposed to be tremendously mathematically rigorous and necessarily applying to the general case with all of the atoms in motion.

@Charles Link: I do thank you for your reply. It's much appreciated. I wasn't looking so much for the answer as the fact that setting the volumes equal could be just written down on paper enough. More or less, I was wondering the physical justification from behind the derivation. But, I do get it now with regard to the density. If you assume the density is uniform throughout, a cube and a cylinder in the same space are going to have about the same density at the mean, so you can treat them as about equal.

To be honest, I'm actually doing self-study (prep for a computational phys course I'm doing this Fall), so in this case, my instructor was just about as clueluess as I was. :)