# Derivation of pV = 1/3 Nmc^2

Homework Statement:
Question regarding derivation of pV = 1/3 Nmc^2
Relevant Equations:
$$\Delta t = 2L/v_{x}$$
This is such a small point but I just wondered if anyone could clarify.

It is easy to work out the time between successive collisions with one wall of the box, namely $$\Delta t = 2L/v_{x}$$However, if this time interval is for instance 1 second, then we could have either 1 or 2 collisions in this time interval depending on whereabouts the particle starts (i.e. could be at t=0 and t=1). These would result in different values for exerted force, but evidently this only is a problem at the beginning and end since double counting is definitely wrong.

My thinking was that when taking an average over a larger period of time - suppose n time intervals - the total time becomes $n\Delta t$ and if n becomes fairly large the calculated time interval for each collision $\frac{n\Delta t}{n}$ or $\frac{n\Delta t}{n+1}$ approach each other, so it doesn't matter if we count the first/last collisions.

Am I overcomplicating this?

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Homework Statement: Question regarding derivation of pV = 1/3 Nmc^2
Homework Equations: $$\Delta t = 2L/v_{x}$$

This is such a small point but I just wondered if anyone could clarify.

It is easy to work out the time between successive collisions with one wall of the box, namely $$\Delta t = 2L/v_{x}$$However, if this time interval is for instance 1 second, then we could have either 1 or 2 collisions in this time interval depending on whereabouts the particle starts (i.e. could be at t=0 and t=1). These obviously result in different values for exerted force.

My thinking was that when taking an average over a larger period of time - suppose n time intervals - the total time becomes $n\Delta t$ and if n becomes fairly large the calculated time interval for each collision $\frac{n\Delta t}{n}$ or $\frac{n\Delta t}{n+1}$ approach each other, so it doesn't matter if we count the first/last collisions.

Am I overcomplicating this?

That's one argument. Another is that if the particles are randomly distributed at ##t=0##, then the time to the first collision averages out. In any case, it's the large numbers that support a statistical approach.

• etotheipi
That's one argument. Another is that if the particles are randomly distributed at ##t=0##, then the time to the first collision averages out. In any case, it's the large numbers that support a statistical approach.

Ahh I hadn't thought of that, that's quite a nice way of thinking of it.