Question About Ideal Gas and Average Free Movement of Molecules

[Calstiel]

Hello,

I have a question regarding the ideal gas and the average free movement of a single molecule. For simplicity, let's consider a model where we have only one atom moving, and we assume all atoms have the same radius.

First, we calculate the cross-section σ using the formula σ=d², where d is the diameter of the atom. After obtaining the cross-section, we calculate the average free movement λ.

Next, we multiply these two values, σ⋅λ to find the volume associated with the average movement.

However, I wonder if this formula is correct. Shouldn't we also consider adding a half-sphere at the end of the cylinder to account for potential collisions? Would the complete expression then be σ*λ + (4 *π* d)/2?

Thank you for your insights!

 

[Borek]

If we are talking about ideal gas we assume it is not too dense, so most likely σ*λ >> (4 *π* d)/2 and the latter can be safely ignored.

Moving the thread to physics, I feel like it will fit there better.

 

[Calstiel]

If we are talking about ideal gas we assume it is not too dense, so most likely σ*λ >> (4 *π* d)/2 and the latter can be safely ignored.

Moving the thread to physics, I feel like it will fit there better.

Thanks for your reply, but i am interested in the exact formula, so this "σ*λ >> (4 *π* d)/2" is right, but my question is, if we need the (4 *π* d)/2 for die total correct formula.

regards
C.

 

[PeroK]

Thanks for your reply, but i am interested in the exact formula,

There's no exact formula, as this is already an idealisation.

 

[Calstiel]

... this is already an idealisation.

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.

 

[PeroK]

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.

Physics is one big approximation! If you really don't like it, transfer to pure mathematics.

 

[Calstiel]

Physics is one big approximation! If you really don't like it, transfer to pure mathematics.

I am trying to derive physics philosophically deductively, but unfortunately I cannot work with approximations. Btw the model can be viewed purely mathematically.

Regards

 

[PeroK]

I am trying to derive physics philosophically deductively, but unfortunately I cannot work with approximations.

That's bad luck for you.

 

[Calstiel]

That's bad luck for you.

I understand your point and yes, unfortunately I know the problem. But let's see how far we can go. :)

 

[Charles Link]

The cross section for collisions $\sigma=\pi d^2$, because the cylinder of collision assuming hard spheres, has radius $d$. You can do some very precise calculations assuming the atoms move in straight lines and collisions occur within this cylinder of radius $d$ which has cross-sectional area $\sigma$ and get that the mean free path $\lambda=1/(n \sigma )$, where $n$ is the density=number of atoms per unit volume.

Each atom, i.e. the atom you are studying, (the others can be assumed to be points in regard to this atom), basically makes a cylinder of volume $v= \sigma \lambda$, and when $n v=n \sigma \lambda=1$, that sort of defines $\lambda$.

(This formula for mean free path, $\lambda=1/(n \sigma )$, is widely used and well-established. I remember discussing it many years ago with a well-known plasma physicist, Professor Manfred Raether (RIP) at the U of Illinois in Champaign-Urbana).

 

[Calstiel]

The cross section for collisions $\sigma=\pi d^2$, because the cylinder of collision assuming hard spheres, has radius $d$. You can do some very precise calculations assuming the atoms move in straight lines and collisions occur within this cylinder of radius $d$ which has cross-sectional area $\sigma$ and get that the mean free path $\lambda=1/(n \sigma )$, where $n$ is the density=number of atoms per unit volume.

Thanks for your replay! So you think the first formular in the pic is right? My idea was die image under the "classic one". Because the atom can push an atom on the same "line they moved" to, or? An then this atom is not in the cylinder?

Regards

 

[Charles Link]

Please see the addition I made just before you posted it (in post 10), where the atom under study makes the cylinder of radius $d$, and the others are regarded as points.

The formula assumes low density, so that $\lambda >> d$ and you don't need to worry about the endcap.

 

[Borek]

I cannot work with approximations

Then you should already stopped earlier, at the assumption of atoms having a well defined diameter.

Basically you are saying "no" to approximations you don't like, but you say "yes" to approximations that you like. That's not a solid logic.

 

[Calstiel]

Then you should already stopped earlier, at the assumption of atoms having a well defined diameter.

Basically you are saying "no" to approximations you don't like, but you say "yes" to approximations that you like. That's not a solid logic.

No, I don't have a problem with that, but I can't explain it so easily here. The fact that the electron moves in a probability cloud around the proton is actually not a problem for me here. Btw.
in philosophy, we do not speak of atoms as "things-in-themselves".

 

[Charles Link]

Basically you are saying "no" to approximations you don't like, but you say "yes" to approximations that you like

I think we should work harder at trying to encourage someone who is new to the forum. He does have some good ideas and seems like he is willing to learn. :)

 

[Charles Link]

and notice "on the average", even for non-uniform densities, (i.e. when you don't always encounter the same number of particles in your cylinder), you will have exactly one collision (on the average) for the atom under study when it travels the distance $\lambda$ when $n \sigma \lambda=1$.

 

[kuruman]

To @Calstiel:

Have you seen this hyperphysics calculation?

 

[Charles Link]

@Calstiel If you want to see a very precise probability theory type calculation on this, I think I can work it out:
(You may find this a little advanced, but this is how it is done in detail).

Probability $P$ it goes a distance $\Delta x$ without a collision is $P=1-n \sigma \Delta x$.
Probability it goes a distance $x=N \Delta x$ without a collision is $P(X>x)=(1-n \sigma \Delta x)^N =e^{-n \sigma x}$, where $X$ is the random variable that says where the first collision occurs at.

$P(X<=x)=1-e^{-n \sigma x}$, so that the probability density function for $X$ is $p(x)=dP(X<=x)/dx=n \sigma e^{-n \sigma x}$.
Finally the mean free path $\bar{X}=\int\limits_{0}^{\infty} x p(x) \, dx=1/(n \sigma)$

$P(X<=x)$ is called the probability distribution function for the random variable $X$.

(In the above, you take the limit for large $N$, and the function is for $x>0$).

also notice in the first line, for small $\Delta x$ that $n \sigma \Delta x$ will be the probability it has a collision in distance $\Delta x$.

 

[Charles Link]

additional comment or two on the previous post: The probably distribution function=$P(X \leq x)$ is most often designated as $F(x)$, and the probability density function $p(x)$ where $P(x<X \leq x+dx)=p(x) \, dx$ above, where $p(x)=dP(X \leq x)/dx$ is most often designated as $f(x)$ with $f(x)=F'(x)$.

Note: $F(x)=\int\limits_{-\infty}^{x} f(t) \, dt$.

@Calstiel If you have had some calculus, you can probably follow a good portion of the previous post=otherwise you might look at it again after you have had a calculus course or two.

 

[Calstiel]

@Charles Link

I had a quick look at it yesterday and I agree with the beginning of the derivation, it looks like a solid way to solve the problem, but I think I have one more comment to make, which I will probably comment on later. Is this your derivation or a standard solution?

 

[Charles Link]

Is this your derivation or a standard solution?

Both. I had a course in probability theory and the exponential probability distribution is well-known and was presented there. One case I think I remember seeing in that class (many years ago) is the case of light bulbs that can fail randomly. One other way to derive a very similar thing is to solve the differential equation $dN/dt=- \alpha N$, where time is the variable instead of distance.

(Edit: Note how long the light bulb lasts is mathematically analogous to how far does the atom go before it has a collision, and likewise, the mean lifetime of the light bulb is analogous to the mean free path of the atom.)

You may find it of interest that I learned the subject well enough ( I studied very hard in college) so that I didn't need to open up any textbooks or google the topic just now to do the derivation.

 

[Calstiel]

First i have try to get your first post, he is a little dense. I have started to loosen up a bit:
Thoughts on Ideal Gas (Mathematically)

-----------------
The probability that a particle travels a very short distance (Δ𝑥) without colliding with another particle decreases depending on the particle density and the size of the particles.

P = 1-n* σ * Δx

P = The probability that a particle travels a distance Δ𝑥 without colliding with another particle.

n = The number of particles with which the particle could collide.

σ = The cross-section of the particles.

Considering the formula, we can note that: The probability P is always 100% if there are no particles, or if the cross-section σ is 0, or if the distance Δ𝑥 is 0. If we assume one particle with a cross-section of 1 and a distance of 1, then the sum of the number of particles, the cross-section, and the distance is 1, resulting in 1−1=0 , making the probability equal to 1. If we take 2 particles and assign a cross-section of 0.5, we also get a probability of 0, or conversely, if we take half a particle with a cross-section of 2, or half a particle with a cross-section of 1 and a distance of 2.

We can describe the product n⋅σ as the potential collision area (pca), which leads to:

P = 1-pca *Δx

If the potential collision area is equal to the potential change in the particle's position Δ𝑥, then the probability that the change occurs is 0. This case is also fulfilled when pca 0,5 and Δ𝑥 = 2 etc.

We can note that the probability is 1 when we have no distance, or when pca is 0, meaning either the number of particles is 0 or the particles have no cross-section.

The probability that a particle travels a total distance x = N*Δx is now:

P(X>x) =(1-pca *Δx)^N

What does the formula describe? First, we derived the probability that a particle achieves a change in position x with probability P for Δx. To calculate the probability as a function of the previous position, we must raise the probability to the power of N, thus:

(1-pK *Δx)^N

We can call this P′, so that we can also write:

P'= (P)^N

If we have a probability of P for example as 0.5, then for N = 1⇒ P = 0,5 means the particle travels the distance. Thus, we must take 0,5 ² resulting in a probability that the particle again travels the distance Δx of 0.25. If it successfully covers this distance again, we must now take 0,5³, which gives us 0.125, or 12.5%.

Now we calculate the probability that Δx changes further by determining the negative proportionality to P'; we will call this P'’.
P(X≤x) =1-P'= P''

If the probability falls exponentially as our particle has moved further, then the probability that it does not move anymore increases inversely proportional to that. So, we have initially for P 0,5 for Δx at N=1, and for P’ at N = 2 we have 0.25. For P'' , at N = 1 with P = 0.5 , we also have 0.5 , whereas at N = 2 we have 0.75 , and so on.

So we can write:

P'' = ∝ 1/P’

-----------------

What you think about it? (As you see I’m not done yet.)

 

[PeroK]

First i have try to get your first post, he is a little dense. I have started to loosen up a bit:
Thoughts on Ideal Gas (Mathematically)

-----------------
The probability that a particle travels a very short distance (Δ𝑥) without colliding with another particle decreases depending on the particle density and the size of the particles.

P = 1-n* σ * Δx

P = The probability that a particle travels a distance Δ𝑥 without colliding with another particle.

n = The number of particles with which the particle could collide.

σ = The cross-section of the particles.

In that equation, $P$ could be negative if $n$ is large enough.

 

[Calstiel]

In that equation, $P$ could be negative if $n$ is large enough.

Yes there is a maximum for n. If n is high enought, then we had P = 0, because we have so many atoms, that no atom can move. But yes your are right, we need to add the case to (but there are some more additions to make).

So:

Px = 1-n* σ * Δx

If Px ≥ 0 ⟶ P = Px
If Px < 0 ⟶ P = 0

 

[PeroK]

Yes there is a maximum for n. If n is high enought, then we had P = 0, because we have so many atoms, that no atom can move. But yes your are right, we need to add the case to (but there are some more additions to make).

So:

Px = 1-n* σ * Δx

If Px ≥ 0 ⟶ P = Px
If Px < 0 ⟶ P = 0

That's not the way probabilities work. If $\sigma \Delta x$ is small, then you have approximately a linear formula for small $n$. That formula is only an approximation as long as $P$ is close to $1$.

The other problem is that $n\sigma \Delta x$ is a volume, not a number. It's numerical value depends on the units. More important, $P$, in that formula, is independent of the volume. Whereas, it must depend on the total volume available to the $n$ particles.

 

[Calstiel]

Whereas, it must depend on the total volume available to the $n$ particles.

Finally the mean free path $\bar{X}=\int\limits_{0}^{\infty} x p(x) \, dx=1/(n \sigma)$

I think we need to change here the number 1 with the volumina and in the calculation above we need to add the units. As fair as i understand the way its an absolute reference without the units.

 

[PeroK]

I think we need to change here the number 1 with the volumina and in the calculation above we need to add the units. As fair as i understand the way its an absolute reference without the units.

Okay, I'll leave it to @Charles Link when he's back on line.

 

[PeroK]

@Calstiel If you want to see a very precise probability theory type calculation on this, I think I can work it out:
(You may find this a little advanced, but this is how it is done in detail).

Probability $P$ it goes a distance $\Delta x$ without a collision is $P=1-n \sigma \Delta x$.

So, $n$ is molecules per unit volume?

 

[kuruman]

$P=1-n\sigma\Delta x$ doesn't look right to me either.

You have a total probability on the left hand side and an increment $\Delta x$ on the right. If you want to do this "philosophically deductively", I think you need to start with an ansatz that makes physical sense and write the incremental probability of not undergoing a collision when the distance traveled is $x$ as $$dP=-\alpha(x) dx$$ where $\alpha(x)$ is some function of distance that has units of inverse length and is in accordance with your model. Then integrate to find $P(x).$

 

[Charles Link]

@Calstiel Some of the math might be a little advanced for you, but the first part I think you should be able to understand. I used the letter $N$ later in the derivation to take $x=N \Delta x$, but let's use $N$ right now to represent the number of particles an let $V$ be the volume. Then we have particle density $n=N/V$. For small distances $\Delta x$ , the probability $P$ that there will be a particle in the cylinder of volume $v=\sigma \Delta x$ will be $P= v N/V=n \sigma \Delta x$. This $P$ will be very small for small $\Delta x$ , and it will be nearly exact for very small $\Delta x$. This $P$ only holds for very very small $\Delta x$ and is said to be exact in the limit that $\Delta x$ goes to zero. We see that for $\Delta x > 1/(n \sigma )$ , the computed $P$ is greater than $1$. We can not use this formula for finite $\Delta x$. It only works for infinitesimally small $\Delta x$.

We will instead compute something else for finite $x$: The probability that there is no particle in the volume $\sigma x$. The probability there is no particle in the volume $\sigma \Delta x$ is $P_s=1-n \sigma \Delta x$. This is also exact in the limit $\Delta x$ goes to zero. We then take a whole bunch of these in a row to compute the same thing, but for finite volume $\sigma x$. You really need calculus to know how to compute this number though=it involves taking $x=M \Delta x$ where there are $M$ number of volumes $\sigma \Delta x$ pieced together to make a finite volume $\sigma x$. The probability of no particle in the volume $\sigma x$ is $(1-\sigma \Delta x)^M$.

Note: I do think the mathematics of this derivation might be a little advanced for you. In any case the result as $M$ gets very large to make $x$ finite is $e^{-n \sigma x }$. Thereby the probability that the distance $X$ for the first collision is greater than $x$ is $P(X>x )=e^{-n \sigma x }$.

 

[Charles Link]

To add to the above the limit of $(1-\frac{n \sigma x}{M})^M$ as $M$ gets large is $e^{-n \sigma x }$. This is a well known calculus result, but if you have taken a course in calculus, you might have seen it.

It would perhaps be a whole lot simpler if the derivation is a little too advanced, to be able to follow that the mean free path is $\lambda= 1/(n \sigma )$, and is found by the $\lambda$ where $n \sigma \lambda = 1$, i.e. where the product of the volume $v$ that the cylinder makes and the density of particles is equal to $1$.

 

[Charles Link]

@kuruman 's method of post 29 will also work for this and get the same result. I did mention this alternative method in post 21 above, and perhaps it is a simpler route.

 

[Charles Link]

The math of the derivation of post 18 might be a little advanced for @Calstiel , but @PeroK and @kuruman should be able to follow it. I really can't claim credit for it, even if I know all the steps involved without resorting to a textbook. It is really a very standard type derivation of the exponential distribution in probability theory.

 

[PeroK]

The math of the derivation of post 18 might be a little advanced for @Calstiel , but @PeroK and @kuruman should be able to follow it. I really can't claim credit for it, even if I know all the steps involved without resorting to a textbook. It is really a very standard type derivation of the exponential distribution in probability theory.

The only thing I would add is that we should take into account that the average relative speed between molecules is $\sqrt 2$ greater than the average speed of an individual molecule. See the derivation on the Hyperphysics page. This gives the probability density function:
$$P(X \le x) = 1 - e^{-\sqrt 2 n_v \sigma x}$$And the mean free path:
$$\bar X = \frac 1 {\sqrt 2 n_v \sigma}$$

 

[Charles Link]

@PeroK I'm going to need to study this latest result. I don't know that the average speed should make a difference, but I gave it a "like" anyway.

 

[PeroK]

@PeroK I'm going to need to study this latest result. I don't know that the average speed should make a difference, but I gave it a "like" anyway.

If we take the rest frame of a particle, then the other particles are moving with average speed $\bar v$. That's used to calculate the distribution:
$$P = e^{-n_v\sigma \bar v \Delta t}$$To convert that to $\Delta x$, where $x$ is the distance the particle travels in the lab frame, we need $\Delta x = v\Delta t = \frac{\bar v \Delta t}{\sqrt 2}$.

The average speed itself doesn't make a difference: it's the difference between average speed between molecules and the average speed of a individual molecule that introduces the factor of $\sqrt 2$ - as per the Hyperphysics page.

 

[Calstiel]

@Calstiel Then we have particle density $n=N/V$.

As I wrote, I took a standard volume of 1, N/1 is always N, so n = N. This gives me the relationship I described above, which ultimately leads to p1*V1 = p2*V2 and that is also stated in all chemistry books, so i think this is also a way to go.

 

[Charles Link]

As I wrote, I took a standard volume of 1, N/1 is always N, so n = N. This gives me the relationship I described above, which ultimately leads to p1*V1 = p2*V2 and that is also stated in all chemistry books, so i think this is also a way to go.

That will not be consistent with my nomenclature and my formulas. I think the math may be a little advanced for you anyway, but $n=N/V$ in many textbooks, and that is what I use. Chemistry books often use $n$ as the number of moles. It pays to be flexible in this case though. Throughout your education you will need to adapt to whatever is being used.

Meanwhile @PeroK , I gave it a little more thought, and I think I strongly disagree with Hyperphysics with their root 2 in the formula. I do think you should be able to follow my derivation of post 18 of the mean free path. (Oh, I didn't see your second reply yet above=let me study it)... It looks to me like it may result from every atom being in motion, and thereby the volume that is traveled through throughout the whole container is increased, but that is going to take me a little extra work to resolve...

 

[Calstiel]

@Charles Link I do not speak about n in chemistry books, i speak about the relationship between pressure and volume (p1*V1=p2*V2). If I increase the volume by double, the pressure decreases by half, and vice versa.I think you won't doubt that? It should be easy to see how one arrives at the ratio using my method. I think you may not have read the post and are only comparing your and my use of n. I have described exactly how I use n*. If you think my derivation is wrong, then you would also have to show where, otherwise it is not a valid argument ("its not an argumentum ad rem"). Or you doubt p1*v1 =p2*v2, which I don't think so.

* Of course we can use another variable here, but that is not important now.

 

[Calstiel]

P'' ∝ 1/P’

Did you understand why this (In my context) leads to p ∝1/V (p = pressure) ?

If the connection is not clear I will plot it in the next days step by step with Jupyter in python.

 

[Charles Link]

@Calstiel My chemistry background is fairly good, especially when it involves $PV=nRT$. If you keep $n$ and $T$ constant, then of course you will have $P_1 V_1 =P_2 V_2$.

What is keeping me busy at the moment is @PeroK 's post of how you treat the case of atoms that are in relative motion w.r.t. one another. That one has me stuck pretty good at the moment, but you would still do well to try to follow whatever you can of my derivation, and not get stuck on the choice of symbols that are used. When the definitions are clearly presented in one way or another, you need to work with what is presented, and there really isn't the need to ask that it be translated to some other units.

 

[PeroK]

What is keeping me busy at the moment is @PeroK 's post of how you treat the case of atoms that are in relative motion w.r.t. one another.

Look at it this way. Suppose you have two particles. One at rest and one bouncing around at some speed $v$. There will be an expected time $\Delta t$ until a collision. The expected distance that the moving particle will travel until a collision is $v\Delta t$.

Now consider that both the particles are travelling with speed $u$. In the rest frame of one particle, the other will be travelling with speed $v = \sqrt 2 u$ (using the statistical average). In this frame, the distance that the moving particle travels until a collision is $v\Delta t$. But, in the original frame, each particle has moved only $u\Delta t = \frac{v\Delta t}{\sqrt 2}$ until the collision.

 

[Charles Link]

In the rest frame of one particle, the other will be travelling with speed

It's the distribution of the various speeds/velocities, working in 3 dimensions, that I am struggling with, even if all particles have the same speed v, but in random directions.

I now have also worked it for the case where the test particle stays stationary and the other particles produce a flux from effusion onto the sphere from the test particle with the diameter corresponding to the cross section. The effusion rate is the number of particles in time t per unit area is $R=n \bar{v}/4$. This gives the same result of $\lambda =1/(n \sigma)$ that you get when you put the test atom in motion and keep the others stationary.

There may be a simple way to see whee the root 2 comes from=I don't doubt now that there will be some factor like that when both are in motion, but I don't see a simple way to derive it=Perhaps you have spotted something clever that I'm not seeing. :)

 

[Calstiel]

When the definitions are clearly presented in one way or another, you need to work with what is presented, and there really isn't the need to ask that it be translated to some other units.

Of course that's correct, but in the course of the thread I was only concerned with whether the derivation via the "other way" also works. The background is that I had followed an approach there that I had already worked on and I specifically put the book down to see whether I could derive it myself. That's not always the best way if you want to work with a formula, partly because of the units and because it can take longer or you can make mistakes, but it helps me learn better.

I had hastily assumed that you had thought the same thing here, without going into the different uses of n in more detail. A careless error on my part. In any case, it's not my intention to "harp on" about the units, I just wanted to present a slightly different approach.

I can't say anything about the other topic right now, I know the argument about skimming through the books on the subject, but I haven't thought about it in more detail yet.

 

[PeroK]

It's the distribution of the various speeds/velocities, working in 3 dimensions, that I am struggling with, even if all particles have the same speed v, but in random directions.

I now have also worked it for the case where the test particle stays stationary and the other particles produce a flux from effusion onto the sphere from the test particle with the diameter corresponding to the cross section. The effusion rate is the number of particles in time t per unit area is $R=n \bar{v}/4$. This gives the same result of $\lambda =1/(n \sigma)$ that you get when you put the test atom in motion and keep the others stationary.

There may be a simple way to see whee the root 2 comes from=I don't doubt now that there will be some factor like that when both are in motion, but I don't see a simple way to derive it=Perhaps you have spotted something clever that I'm not seeing. :)

A derivation is on the Hyperphysics page.

 

[Charles Link]

A derivation is on the Hyperphysics page.

Which was given in @kuruman 's post 17. Thank you. I'll need to study it a little, but yes, when you click on "show " they do give what looks like a very good derivation. Thanks. :)

 

[Charles Link]

@Calstiel Very good. Hope you find some of this mathematics interesting. :)

@PeroK From what I can tell of the dot product term, where the Hyperphysics says it correlates to zero, my guess is that is a reasonably good approximation, but I think it is likely the root 2 factor is not exact out to 6 decimal places and more. In any case, thank you=I didn't even see the root 2 part when @kuruman posted the link, or the derivation that you could click on to see. :)

 

[Calstiel]

@Calstiel Very good. Hope you find some of this mathematics interesting. :)

Yes, i do. :)

but I think it is likely the root 2 factor is not exact out to 6 decimal places and more

This would be an very interesting and important point, but i can not comment on that at the moment, i will follow your discussion about it.

 

[Charles Link]

@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.

 

[Calstiel]

@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.

At the moment i think √2 is correct. https://www.tec-science.com/thermod...-of-gases/mean-free-path-collision-frequency/