Question About Ideal Gas and Average Free Movement of Molecules

[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/

 

[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-Boltzmann 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.