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

In summary, the conversation discusses the derivation of the mean free path formula for a gas, which relates the mean free path between collisions to the mean distance between nearest neighbors. The formula is derived by considering the volume swept by the moving molecule and the number of collisions it makes in that volume. The formula is not mathematically rigorous and may not apply to all cases, but it is generally accepted in the physics community. The conversation also mentions the physical justification for setting the volumes of a cube and cylinder to be equal in the derivation.
  • #1
FallenLeibniz
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1

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?
 
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  • #2
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.
 
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  • #3
Charles Link said:
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. :)
 
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Related to Question about the relation b/w mean free path and other variables

1. How is the mean free path related to temperature?

The mean free path is inversely proportional to temperature. As temperature increases, the average kinetic energy of particles in a gas also increases, leading to more frequent collisions and a shorter mean free path.

2. What is the relationship between mean free path and pressure?

The mean free path is inversely proportional to pressure. As pressure increases, the density of particles in a gas also increases, resulting in more frequent collisions and a shorter mean free path.

3. How does the mean free path change with gas density?

The mean free path is inversely proportional to gas density. As gas density increases, the number of particles in a given volume increases, leading to more frequent collisions and a shorter mean free path.

4. Is there a relationship between mean free path and molecular mass?

Yes, the mean free path is inversely proportional to the square root of molecular mass. This means that gases with heavier molecules will have a shorter mean free path compared to gases with lighter molecules at the same temperature and pressure.

5. How does mean free path affect the thermal conductivity of a gas?

The mean free path is directly proportional to the thermal conductivity of a gas. A longer mean free path means that there is a lower chance for collisions between particles, allowing heat to transfer more easily through the gas.

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