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.