# Molecular Velocity

1. Mar 9, 2012

### roam

1. The problem statement, all variables and given/known data

http://img191.imageshack.us/img191/3277/59596364.jpg [Broken]

3. The attempt at a solution

I used the Maxwell-Boltzmann distribution equation:

$KE_{avg}=\frac{1}{2}mv^2 = \frac{3}{2}k T$

$\implies v = \sqrt{\frac{3kT}{m}}$

The most probable velocity for argon would be:

$v = \sqrt{\frac{3\times (1.38 \times 10^{-23}) \times 293.15 \ K}{40}} = 3.034 \times 10^{-22}\ m/s$

And for water vapour at (not sure if I need to modify the equation):

$v = \sqrt{\frac{3\times (1.38 \times 10^{-23}) \times 293.15 \ K}{18}} = 6.74 \times 10^{-22} \ m/s$

Is this correct? How do we get to use the info we were given about the molecular radius and the pressure (at 1 atm)?

And do we expect them to have collision rates? Personally I guess water vapour molecules have a higher collision rate because they have a higher velocity and thus collide more often. How do we make the actual calculation?

I appreciate any help and suggestions.

Last edited by a moderator: May 5, 2017
2. Mar 9, 2012

### cepheid

Staff Emeritus
No, your answers are not correct at all. If you actually stop and think about it, you'll realize that 10-22 m/s is a ridiculously slow velocity, and is not at all consistent with the picture of molecules in a gas zipping around.

Your error comes from the fact that you haven't converted all of your quantities to SI units. Your masses need to be in kilograms, but they are currently in units of "molecular weight."

When doing a physics computation, always stop and ask yourself whether the result makes sense. It's never obviously right, but sometimes it can be quite obviously wrong.

Also check to see if you're using the right equation. The most probable, mean, and rms speeds are all different for this distribution:

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

You use it in the second part where it asks about collision rates.

You can think about this in terms of the mean free path -- how far a molecule can get, on average, before colliding with another one. There is a simple way to figure this out from the number density.

The second part of the problem is, what is the condition for a "collision" to occur? Hint: the mean free path you calculate will be the average spacing between the molecules' centres.

3. Mar 10, 2012

### roam

Thank you for the link, what the question is asking for is the most probable velocity, I believe it's given by

$v_p = \sqrt{\frac{2kT}{m}}$

But how can I convert from "molecular weight" to Kgs? I'm a bit confused.

1 atm = 1.01325 × 105, so number density of molecules is given by

$n = \frac{p}{kT} = \frac{1.01325 \times 10^5}{(1.38 \times 10^{-23}) \times 293.15 \ K} = 2.504 \times 10^{25}$

so that the mean separation between molecules is given by

$d = n^{-1/3} = (92.504 \times 10^{25})^{-1/3} = 3.417 \times 10^{-9} \approx 3.4 \ nm$

Mean free path between collisions is

$\lambda = \frac{1}{n \pi d^2}$

The collision rate per unit volume is given by

$n^2 \pi d^2 \bar{v}$

where the mean velocity is given by

$\bar{v} = \sqrt{\frac{8kT}{\pi m}}$

Is that the right equation for comparing the collision rates?

Last edited: Mar 10, 2012
4. Mar 11, 2012

### roam

To find the most probable velocity I must use the equation

$v_p = \sqrt{\frac{2kT}{m}}=\sqrt{\frac{2RT}{M}}$

But I need to either have the mass m or the molar mass M. The only thing they say is "molecular weights of 40 and 18". How do I get this to either Kgs or molar mass? Are these numbers in unified atomic mass units, and I have to multiply by the conversion factor 1.66 ×10−27 to get to Kg?

5. Mar 11, 2012

### roam

And is that the right equation for the rate of collisions in my second post?

6. Mar 11, 2012

### cepheid

Staff Emeritus
I find it hard to believe that you would be given molecular weights in a problem if you hadn't already been exposed to the definition of molecular weight in your class. So, this might just be a question of looking in your notes or in your book.

But even if your teacher didn't define what it was, you could always just look it up.

http://en.wikipedia.org/wiki/Molecular_mass

This article explains that molecular mass is the mass of the molecule in unified atomic mass units (u), and that molecular weight, also called relative molecular mass, is simply the molecular mass divided by 1 u (in other words, it is the dimensionless version of the molecular mass).

7. Mar 11, 2012

### cepheid

Staff Emeritus
I haven't checked your arithmetic, but I agree with your method. Now, naively, I would just assume that the mean free path is equal to this mean spacing d that you have calculated above. But I guess that's only true for the case where the molecules themselves do not take up any space.

Ah, I see. So you already had a canned formula for mean free path. Let me see if it makes any sense to me. Suppose a molecule of the gas in question has radius r. Then, as it travels, it presents a circular "cross-section" of area $\pi r^2$ for collision with other molecules. If it travels a distance λ, then you can consider it to have "swept out" a cylindrical volume. The volume of this cylinder is $\pi r^2 \lambda$. So, the number of collisions to have occurred over this distance is simply equal to the number (N) of other molecules contained within this cylindrical volume. This number N is just given by the number density multiplied by the volume:$$N = n\pi r^2 \lambda$$However, as soon as the molecule collides with one other molecule, it will be sent off in some other direction. So, we want to figure out the cylindrical volume that contains only one other molecule, so that we can figure out how far our travelling molecule gets before its first collision. In others, we want to find the volume such that N = 1:$$1 = n\pi r^2 \lambda$$Solving for λ, we get$$\lambda = \frac{1}{n\pi r^2}$$

So, your equation makes sense to me provided that the "d" in your equation is in fact the radius of the molecule. So be careful with your notation. You have the letter "d" doing double duty as both the molecular radius and the mean separation. In any case, it looks like we didn't need the mean separation after all. Using the argument above, I've shown that the molecule, on average, will travel a distance λ given by the expression above before colliding with another molecule.

Well, I have no idea if it's the right equation for the collision rate. But I can do what I did above for the mean free path, and use physical reasoning to try and derive an equation for the collision rate. This is a skill you should develop: physics is not about memorizing formulae and then plugging numbers into them. It is about understanding physical concepts and then applying them to different situations in order to understand how things work.

I imagine that the collision rate for a molecule would be given by the reciprocal of the average time $\bar{t}$between collisions. For example, if the average time between collisions was 1/10 th of a second, then you'd expect a collision rate of 10 collisions per second. Now, since distance = speed*time, it follows that time = distance/speed. Therefore, you'd expect the average time between collisions to be given by$$\bar{t} = \frac{\lambda}{\bar{v}}$$which means that the collision rate is given by$$\frac{1}{\bar{t}} = \frac{\bar{v}}{\lambda} = n\pi r^2 \bar{v}$$But this is just the collision rate for a single molecule. There are n molecules in a unit of volume. Therefore, we'd expect the number of collisions per unit volume and per unit time to be "n" times the rate for a single molecule:$$\frac{n}{\bar{t}} = n^2 \pi r^2 \bar{v}$$So, as far as I can tell just using simple reasoning, your equations seem to be right.

8. Mar 12, 2012

### roam

But "d" is not the radius, it is the mean separation between molecules according to my notes. And my note clearly define and use "d" not "r":

$d = n^{-1/3} = (KT/p)^{1/3}$

Mean free path between collisions is λ=1/nπd2. And collisions are given by

$n^2 \pi d^2 \bar{v} = c \frac{p^2 \sigma}{\sqrt{m(kT)^3}}$

Is this a typo in my notes?

Unfortunately I was never exposed to that before and I don't have it in my notes that's why I'm very confused. So, does this mean I can simply substitute 40 an 18 as m into the equation for most probable velocity?

Since 40 = m x (1.66×10-27)/1x(1.66×10-27) => m = 40

(Because 1 u = 1.66×10-27 Kg according to the Wikipedia article).

Last edited: Mar 12, 2012
9. Mar 13, 2012

### cepheid

Staff Emeritus
I was curious to see if there was a gap in my understanding, so I looked up in a first-year physics textbook (University Physics by Young & Freedman, 10th Edition). It turns out that I was essentially right, I just made one minor conceptual error. Anywhere that you see "r" in my equations, you need to replace it with "2r". The reason is because if two molecules' centres are separated by a distance 2r, then their edges are touching, and a collision has occurred. So the radius of the "cylinder" that I referred to before has to be expanded to 2r.

As for your notes, you can't possibly expect me to know what's going on with them. Ask your teacher for clarification on what "d" means in that context.

I should note that my result is for a very primitive model that simply treats the molecules as hard spheres of radius r. Maybe you are using a more sophisticated model in your class. I don't know. You need to ask the person who is teaching your class.

No, no, you completely misinterpreted what I (and the article) meant when I said that the molecular weight is the molecular mass "divided by 1 u". What I meant was that, if the mass of a molecule was 40 u, then its molecular weight would be 40 u / 1 u = 40. In other words, the molecular weight is a dimensionless number that tells you how many "u's" your mass is.

Working backwards, if the molecular weight is 40, then the mass of the molecule is 40 u, and you need to convert this into kg using the conversion factor.

ALSO: Echoing what I said in my first reply to you: PLEASE employ a little bit of common sense. You are trying to figure out what the mass of a single molecule in this gas is, in kilograms. There is no way that the answer can be 40 in units of kg. That would be almost the same mass as a human being, which is a ridiculous result. So I don't know what made you think that "40" could possibly be reasonable thing to substitute in.

10. Mar 13, 2012

### roam

In the equation for the mean free path, and the one for the collision rates, "d" stands for diameter or as you said 2r.

Looking at the equation n2πd2vmean, since argon has the bigger radius, so I would expect it to have higher collision rates. But on the other hand water vapour molecules have higher mean velocities...

Thank you for the explanation. I've realized that shortly after I made that post, so I multiplied by the conversion factor and I got a reasonable answer.