# Why is the time $\frac{2L}{v}$ in the derivation of $P=\frac{Nm\langle v^2\rangle}{3V}$

In the derivation of $$P = \frac{Nm\langle v^2\rangle}{3V}$$, why do we take $$\mathrm{d} t = \frac{2L}{v}$$ for $$F = \frac{\mathrm{d} p}{\mathrm{d}t}$$ when the particle collides with the wall of the container.. I understand that time for particle will be $$\frac{2L}{v}$$ for returning to that position. But, shouldn't $$\mathrm{d} t$$ be the time for the change in momentum of the particle? I also see many explanations saying its to get the average force, I can't understand why is average force needed as we need the force the particle exerts on the walls.

The derivation of which I am talking about is here.

Edit: Symbols:

$$P\rightarrow$$ Pressure

$$N\rightarrow$$ Number of Particles

$$m\rightarrow$$ Mass of each particle

$$\langle v^2\rangle \rightarrow$$ mean squared velocity of the particles.

$$F\rightarrow$$ Force on the wall of the containers

$$\mathrm{d}t$$ -> Change in Time

• Define your symbols. Is T time or temperature? Is P pressure or momentum? You've also made a mistake copying things over ($dT \neq 2l/T$). Please provide more information on where the derivation starts and what it's trying to show so that future readers can understand what's going on if the link breaks. Aug 6 '21 at 6:43
• Oh I am sorry, I was in a hurry while uploading this question. Will make sure to correct it in a moment. Aug 6 '21 at 7:23

The pressure on one wall is the force per unit area. The force is momentum per unit time. It receives momentum discretely due to individual particle arrivals, but the particles are so numerous and so frequently visiting it seems continuous. So the force is inversely proportional to how long it takes the particle to hit the wall again, and that's $$2L/v$$.

$$dT = \frac{2L}{v}$$ is the time between collisions of a molecule with the wall, as it has to travel both ways across the container before hitting the wall again.

The force comes from the change in momentum per second, so if you want the number of collisions with the wall, per second, it's $$\frac{1}{dT}$$.

You're right that if we wanted to find the average force during only one collision then $$\mathrm{d}t$$ would need to be the time for the collision to occur. We want to find the average force for all time, though. To do that it suffices to average the force over how long it takes the particle to hit the wall and come back to hit it again. That's why you use $$\mathrm{d}t = l / v_x$$.

The one-dimensional motion along each direction is periodic with a period $$T$$ corresponding to the time for returning at the same coordinate with the same velocity. Being a uniform motion such a period is related to the speed along a direction parallel to the box side of length $$L$$, say the $$x$$ direction, by $$T=2L/v_x$$.

Now, if one has a periodic function $$f(t)$$ of period $$T$$, the time average over long times can be obtained as: $$\langle f \rangle=\lim_{\tau\rightarrow \infty}\frac{1}{\tau}\int_0^{\tau}f(t)dt=\lim_{N\rightarrow \infty}\frac{1}{NT}\int_0^{NT}f(t)dt=\frac{1}{T}\int_0^{T}f(t)dt. \tag{1}$$ If $$f(t)$$ corresponds to a sudden change of momentum by $$\Delta P$$ at times $$t_0+n T$$, ($$n=0,1,2,\dots$$; where $$t_0$$ is an arbitrary starting time): $$f(t)= \Delta P \sum_n \delta(t-nT-t_0), \tag{2}$$ we get $$\langle f \rangle=\frac{\Delta P}{T}$$.

To summarize, the average we are interested in kinetic theory is the time average over long times, not the average over the microscopic collision time.