# Relation between particle numbers and collisions between gas molecules inside a closed container

I'm not a physicist, nor studying physics, so this may be a dumb or a very hard question, I don’t know. I'm not sure if I used correct tags, feel free to correct them if you feel like it is necesary. Forgive me about my English aswell, it’s not my mother language.

I'd like to know if there is a known relation between the average collisions per particle with other particles (not with walls) and the total amount of particles in a given volume and temperature. I guess its somehow a stochastic mechanic though, but how high is the typical deviation?. Will proximity to walls affect the number of collisions or wall collisions will compensate particle collisions?

My real interest is not to apply this to real particles, but with agents behaving like gas particles in a closed container, so every particle will be moving at the same speed (which if I’m not wrong solely depends on temperature in nature), and perfect elasticity can be simulated as well. I'm not taking gravity into consideration either. The "perfect simple answer" here is a formula relating volume, number of collisions per particle, temperature and amount of molecules (density?).

I’m always happy to learn, so if after a “simple answer” you want to refer me to additional documentation I’ll be happy to look further.

Thank you all.

Let's track a single spherical particle moving in an ideal gas of identical particles: Assuming our particle of interest has a diameter $d$ and a constant average velocity $\bar v_x$ because the gas is at constant temperature, then after time $t$ it will have swept out a cylindrical path of collision with volume $V$:

$$V\ =\ (\text{base of cylinder})(\text{height of cylinder})\ =\ (\pi d^2)(\bar v_x t)$$

To count the number of particles our particle of interest encountered i.e. the number of collisions it had in this time, we just need to multiply by the density:

$$\small \text{number of collisions}\ =\ \text{(volume swept out by our particle)} \times \text{(the number of particles per unit volume)}$$

Substituting in the result from the first equation above and terms from the ideal gas law:

$$\large \text{number of collisions}\ =\ (\pi d^2)(\bar v_x t)\ \times\ \left( \frac{n}{V} \right)$$

Dividing both sides by $t$ and substituting the expression for the average velocity of a gas particle, $\bar v_x = \sqrt{8k_B T/ \pi m}$ results in our final equation:

$$\large \text{number of collisions per second}\ =\ (\pi d^2) \left( \sqrt{8k_B T/ \pi m}\right) \times\ \left( \frac{n}{V} \right)$$

Or written in terms of pressure, by using the ideal gas law:

$$\large \text{number of collisions per second}\ =\ (\pi d^2) \left( \sqrt{8k_B T/ \pi m}\right) \times\ \left( \frac{P}{RT} \right)$$

Source notes:

You may ask, "Why is the base of the swept-out cylinder $\pi d^2$?" It is the effective collision area taking into account the size of the "target" particles. See the below image from the HyperPhysics article on "Mean Free Path" (C.R. Nave, Georgia State University): • Thanks, this is EXACTLY what I was looking for. I think that for my model I can simplify it, as every particle is "inteligent" and able to measure its own speed (which will be exactly the same for all, kind of a spherical chicken in void) and relative position, so I think I'll be able to get rid of physical constants. In the first experiments, particles (agents) will have perfect sensors and engines, so what do you think, will I have something like a gaussian distribution or something much more randon when checking individual particle collisions rather than the average collisions per particle? – DGoiko Nov 8 '17 at 14:38
• Unless you are gonna actually simulate this all out, I would say: number of collisions is proportional to particle speed right? So if you pick a distribution of speeds, there's your distribution of number of collisions. – pentane Nov 9 '17 at 0:54

The formula for Mean Free Path for gas particles depends not only on density, but on the "size" of the particles. As the size decreases, the chance of collision goes down.

If you have the mean free path and the average velocity, then path divided by speed will give you mean free time for collisions per particle.

• Hello. I cant upvote your answer, however I accepted it. Just to make things straigh, if I can make my agents (which are just dots in space) think they "crash" with another agent and then repel each other, then the diameter where this fenomena occurs is called the kinetic diameter of my gent? – DGoiko Nov 8 '17 at 5:24
• That's one way to think of it. Real molecules don't have a sharp boundary between zero interaction and hard collision, so the distance is slightly fuzzy. But for a model, that's where you'll want to start. – BowlOfRed Nov 8 '17 at 5:27
• Thank you very much! I got just what I needed. I'll give some feedback if something interesting comes out from my algorithm. – DGoiko Nov 8 '17 at 5:35

For an ideal gas, PV = nRT.

Pressure is proportional to the number of collisions each gas molecule sees-- not just the collisions with walls, but the collisions with other gas molecules as well. Hence how you can have different air pressures at different altitudes even though there are no "walls."

• Hi. Sorry I didnt make my question clear, I know that formula, what I want to know is the relation between pressure and number of collisions. Is there a formula that gives me the average number of collisions a particle will have over a certain amount of time? – DGoiko Nov 8 '17 at 4:47
• @DGoiko gotcha. Hopefully my new answer helps :) – pentane Nov 8 '17 at 7:03