Question About Ideal Gas and Average Free Movement of Molecules

[Charles Link]

@Calstiel I can read German reasonably well=I studied it a lot in both high school and college. I looked through the German article somewhat carefully, and they seem to be doing a similar derivation that Hyperphysics does, but they explicitly use $v_{rel}^2$, perhaps because using $v_{rel}$ as Hyperphysics does gives the problem that I pointed out.

In the German paper method of doing this calculation, they work with $\bar{ v_{rel}^2}$, in which case you get a cosine term that is not nested in an expression in a root sign, so it will then average to zero. In that case, you then get $\sqrt{2}$ instead of the 4/3 I got for the final result, (in their case after converting from $v^2$ to $v$).

I'm willing to go along with the $\sqrt{ 2}$ as being a reasonably accurate result. In the case I have, with the cosine expression inside the root sign, I think it is necessary to assume that the speeds are all the same, and just in arbitrary directions. Without the root sign, the speed distribution doesn't matter, because the cosine term does indeed average to zero, without any correction like I found.

Thank you very much for posting the "link". :)

 

[Calstiel]

The approach on tec-science.com reminds me of Feynman's path integral method. This is typically how I approach these things, and I hope this method is more precise than it may initially seem to you. I'm curious to see how PeroK will respond. The "German solution" (tec-science) seems solid to me, but we'll see what other users think. I didn't examine the HyperPhysics link closely, as I had already taken a more in-depth look at the content on tec-science.com. I'll do that, but first I want to see what the others have to say about it. :)

I can read German reasonably well=I studied it a lot in both high school and college

Gut zu wissen,
Grüße aus Österreich. :)

 

[Charles Link]

and a follow-on: Note that the average for the terms $v^2$ is simple regardless of the distribution of speeds, but once they are placed as a sum of a couple of terms and put inside a square root sign, as Hyperphysics does, they become very difficult to evaluate. (I believe the result is then in general dependent on the velocity distribution).

The approximation the German paper linked above does is to work with $v^2$ terms, along with a term involving the cosine, and solve for the average of $v_{rel}^2$. Once that is obtained, they then assume that $\bar{v}_{rel}$ is approximately the square root of this number.

That's at least my analysis of it, and their approximation to get the $\sqrt{2}$ factor is indeed a pretty good one.

One additional thought or two I had on it though is that the problem is more mathematically correct if you evaluate it for $\bar{v}_{rel}$ if it were possible. That's where using 4/3 (@PeroK please check my math of post 49 if you can) as a factor is really also reasonably good, and is in fact the precise factor you get if all the speeds of the atoms are identical, but simply with a random direction.

Consider the problem of riding a motorcycle in the rain, where you have velocity $\vec{v}_1$ and where the rain has velocity $\vec{v}_2$. The amount of rain that reaches you is proportional to $|\vec{v}_{rel}| =|\vec{v}_1-\vec{v}_2|$. That's the way Hyperphysics proceeds and that is the more correct way. You then average over the entire distribution of all $\vec{v}_2$ vectors, but the result is only simple, and gives the 4/3 result, if all the $\vec{v}_2$ have the same amplitude but random direction. (It does not give the $\sqrt{2}$ result that Hyperphysics claims. The cosine term does not average to zero when it is inside the expression.)

 

[Calstiel]

The approximation the German paper linked above does is to work with $v^2$ terms, along with a term involving the cosine, and solve for the average of $v_{rel}^2$. Once that is obtained, they then assume that $\bar{v}_{rel}$ is approximately the square root of this number.

That's at least my analysis of it, and their approximation to get the $\sqrt{2}$ factor is indeed a pretty good one.

I would have done it the same way. My model that I mentioned earlier was based on particles at rest, so I hadn't thought about the relative speeds. I would have solved it the same way. I must mention, however, that I am assuming hard determinism, which only appears to be random because we lack the exact initial condition.

 

[Charles Link]

Yes, you first solve it for the other particles at rest, and then figure out what the correction factor is to get the average relative velocity, which is important and may be better understood by working the problem of riding a motorcycle in the rain that I presented above (post 53). It was only when I worked with that scenario, that I was completely sure that the average $|\vec{v}_{rel} |$ is indeed what is needed.

Edit: Note that the collision rate will increase (when the motion of the other atoms is considered as well) by this factor of $\bar{v}_{rel}/\bar{v}_1$, so the mean free path will be shortened by precisely this factor.

 

[Calstiel]

My visualization was a bit different, I first tried to insert it into a "different system" and that doesn't work without relative speed. I had done it like that before (as mentioned in the link above) (in a different area), but somehow I didn't see it right away (I'm a bit tired). Then I imagined the balls in motion. Normally I always draw the arrows of motion, like Feynman in QED. But everyone has their own approach, of course you can also imagine larger objects. It would be interesting to discuss the difference between real coincidence and random occurrence, I wanted to address that point later. I have another critical comment for a other point, but I want to try to solve the problem that I still see myself, I had already started to do that, when it is more "tangible" I will address it. In my opinion the thread is developing very constructively, great forum!

 

[Herman Trivilino]

Yes, but assuming the idealized model there must be a correct formula. I don't want an approximation, but rather the exact calculation assuming the ideal model.

AFAIK in the ideal gas the volume of each particle is zero. We are using the particle model here so the size of each molecule is negligible compared to the distance between the molecules.

 

[Charles Link]

AFAIK in the ideal gas the volume of each particle is zero. We are using the particle model here so the size of each molecule is negligible compared to the distance between the molecules.

Keep reading. This one gets much better. We have an almost exact formula, that the mean free path $\lambda=1/(n \sigma)$. Then @PeroK observes in a Hyperphysics article that there is a slight correction to this due to the relative motion of the particles of $1/\sqrt{2}$ . More careful inspection of this is that it is somewhat handwaving=4/3 instead of $\sqrt{2}$ is what I computed if all the particles have the same speed.

In addition the OP @Calstiel finds an article in a German website where they don't have the same handwaving as Hyperphysics, but choose to work with $v^2$ instead of $v$ , perhaps to try to avoid the handwaving, and being more accurate with their mathematics, but making an approximation in the process.

I welcome your additional feedback. :)

 

[PeroK]

@PeroK I think you might find this part very interesting=please double-check my work=I need to do the same myself. If my calculations are correct, the Hyperphysics derivation is almost right, but their $\sqrt{2}$ factor needs to be 4/3 instead. Consider taking the $\cos(\theta)$ factor in their expression and integrating/averaging it in spherical coordinates over the $4 \pi$ steradians using $\sin(\theta) \, d \theta \, d \phi$. I'll write out more of the details momentarily, but I get for the $\theta$ integral $\int\limits_{-1}^{+1} (1-x)^{1/2} \, dx=(4/3) \sqrt{2}$, so that the complete result with all the factors included is 4/3 rather than root 2.

The $\phi$ integral gives $2 \pi$, and there is another $\sqrt{2}$ that factors out of there expression for $\bar{v}_{rel}$.

I only did it for the case where the speeds were all the same. To do for a distribution of velocities would prove to be very difficult.

The Hyperphysics derivation, IMO, does not generally hold. What is true is that, for a velocities in random direction, the cross term in the dot product vanishes on average. This gives the relationship between the average (expected) squares of the velocities:
$$E(v_r^2) = 2E(v^2)$$The relationship between the average speeds then depends on the distribution, and the variance. To do an accurate calculation, you would have to use the Maxwell-Boltzman distribution as your starting point for the distribution of particle speeds. The Wikipedia page does this and, after a more involved calculation, gets the same answer as Hyperphysics of $E(v_r) = \sqrt 2 E(v)$. This must be a property of the specific Gaussian distribution:

https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution#Typical_speeds

 

[Charles Link]

@PeroK Your response above is reasonably good, but I think this one is worth a second look. Please see my post 53 and some of the details there including the analogous problem I mentioned of riding a motorcycle in the rain.

I did the detailed calculation in post 49 for the Hyperphysics method for identical speeds but random directions and I did get 4/3, (instead of $\sqrt{2}$). The Hyperphysics method of using $|\vec{v}_{rel}|=|\vec{v}_1-\vec{v}_2|$ can be seen to be the correct way of working this problem with the motorcycle in the rain analogy, but I found their final result ($\sqrt{2}$) to only be approximately accurate.

Using $v^2_{rel}$ the dot product term will vanish, and the computation is straightforward, ( and you do get $\sqrt{2}$ when you take the root of the $v^2_{rel}$), but I don't think the $v_{rel}$ case (the correct way to do it) has been computed in these articles properly for any distribution. I don't think anyone has worked the $v_{rel}$ case with the Maxwell-Boltzmann distribution.

 

[PeroK]

Using $v^2_{rel}$ the dot product term will vanish, and the computation is straightforward, but I don't think the $v_{rel}$ case (the correct way to do it) has been computed in these articles properly for any distribution. I don't think anyone has worked the $v_{rel}$ case with the Maxwell-Boltzmann distribution.

I'm just going by the result in Wikipedia. It says you can check the integral, which I admit I haven't done. I don't see that post #49 can compare with the calculation on Wikipedia.

 

[Charles Link]

I'm just going by the result in Wikipedia

Wiki =actually Hyperphysics does the general case, (they reference Maxwell-Boltzmann but work only the general case) and they do that somewhat incorrectly by writing $\bar{v_{rel}}=( \bar{v^2} +\bar{v^2}_2-2 \bar{\vec{v}_1 \cdot \vec{v}_2})^{1/2}$. This expression for the averages is only approximate and not exact, but Hyperphysics implies it is exact.

The bars for the average didn't come out properly in my Latex. Please go to the Hyperphysics article to see it more clearly: http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/menfre.html#c5

 

[Calstiel]

This expression for the averages is only approximate and not exact, but Wiki implies it is exact.

I'm currently working on a computer simulation of the topic here, so I'm a bit off the thread, but in general its often important to check the sources in Wiki, maybe someone here is doing that.

 

[PeroK]

Here are two more (longer) derivations that come to the same result. I suspect this result is well known:

https://chem346.class.uic.edu/wp-content/uploads/sites/521/2019/07/MB_ave_relative.pdf

https://chem.libretexts.org/Bookshe...ty_Density_Function_for_the_Relative_Velocity

 

[Charles Link]

@PeroK That makes me want to go back and check my result of post 49=I did get 4/3 for the $v_{rel}$ case where all speeds are the same. Thank you=this looks like two very good links that you gave us. :)

 

[PeroK]

@PeroK That makes me want to go back and check my result of post 49=I did get 4/3 for the $v_{rel}$ case where all speeds are the same. Thank you=this looks like two very good links that you gave us. :)

That may well be correct. All speeds the same is different from a M-B distribution.

 

[Charles Link]

That may well be correct. All speeds the same is different from a M-B distribution.

I'm almost a little surprised by the two different results, $\sqrt{2}$ and 4/3, especially when Hyperphysics does some incorrect handwaving to get the first result for what they seem to be showing as the case in general, regardless of the distribution. I have looked over my calculations of post 49 though, and I believe that I computed the case of all the same speed correctly. Thanks very much for your inputs. :)

 

[renormalize]

I have looked over my calculations of post 49 though, and I believe that I computed the case of all the same speed correctly.

So you've said repeatedly. But how is your calculation physically relevant? Can you cite any physical situation for which a gas is expected to have a mono-speed distribution (a Dirac delta function)?

 

[Charles Link]

So you've said repeatedly. But how is your calculation physically relevant? Can you cite any physical situation for which a gas is expected to have a mono-speed distribution (a Dirac delta function)?

This is more of an interest to me simply from a mathematical sense, because Hyperphysics seems to imply in the "link" of post 62 that the $\sqrt{2}$ is the result for the general case. The simple case of post 49 if my calculations are correct would show that this is not the case.

One reason I chose this case (of all speeds being the same) is that it is simple enough to solve. I still need to study the "links" that @PeroK provided in post 64. The mathematics for the M-B case looks somewhat complex from a first look at it.

 

[PeroK]

This is more of an interest to me simply from a mathematical sense, because Hyperphysics seems to imply in the "link" of post 62 that the $\sqrt{2}$ is the result for the general case.

The Hyperphysics page is clearly wrong in this respect. In general $E(X) \ne \sqrt{E(X^2)}$. But, if you do the full calculation for a Gaussian distribution of velocities, then $E(v_r) = \sqrt 2 E(v)$.