# Deriving the pressure exerted by classical particles on the walls of a cubical container (Ideal Gas)

Given that the probability to find a particle in a state $$|i\rangle$$ is $$p_i \propto \exp(-\beta E_i)$$ one can conclude that for classical particles (:=particles obeying the classical kinetic energy relation) with mass $$m$$ the velocity distribution is given by $$P(|\mathbf{v}|)=\sqrt{\frac{2}{\pi}} {(\beta m)}^{3/2}\exp\left(-\beta m \frac{|\mathbf{v}|^2}{2}\right)$$ Now consider a cubical box with Volume $$L^3$$. Which pressure exert $$N$$ classical particles confined in this box on the six identical walls of the confinement?

One can arrive at the answer $$p=\frac{N}{\beta V}$$ by assuming that

1. There are no interactions between particles
2. Collisions with the walls are elastic and that the impulse of the particle after a collision with a wall is $$\mathbf{p'}=\mathbf{p}-2\mathbf{p}\mathbf{n}$$. Where $$\mathbf{n}$$ is normal to the wall.
3. Every particle is traveling for a distance $$L$$ before it collides with a wall. Particles only travel in a straight path from Side $$S_i$$ to the corresponding opposite side $$S_i'$$. They never travel from one side to an adjecent side.

Now the logic is easy. Force exerted on one side of the cube by one particle with velocity $$v$$ is just $$F(v)=\frac{\Delta p}{\Delta t}=\frac{2mv}{2L/v}$$ Force exerted on the side by one particle on average $$\langle F\rangle=\frac{2m}{2L}\langle v^2\rangle=\frac{m}{L}\int_{\mathbb{R}^+}v^2P(v)dv=\frac{3}{\beta L}$$ Which leads to the total pressure exerted on this side $$p=\frac{N}{3}\frac{\langle F\rangle}{A}=\frac{N}{\beta V}$$ Which by symmetry is the same for all six sides.

Note that not $$N$$ but $$N/3$$ particles travel from one side to it's opposite side, since there are six sides in total.

Can someone arrive at the result without using the third assumption, that I made up? I would consider this as a purely geometric problem and can not think of a clever way to tackle it. I want a treatment that also allows particles to travel from one side to one of it's adjacent side.

We assumed that the particles are non-interacting. Let us pick a particle at random. This particle has with $$p_i$$ probability an energy of $$E_i$$. Assume this energy is also a given. Now, we have a substitute assumption for your third:

1. Any particle has equal probability of being headed to any direction. In $$\Bbb R^3$$, this means that probability that a particle is headed to a direction is $$dP(\theta+d\theta,\phi+d\phi)=\sin(\theta)\,d\theta\,d\phi\ /\ 4\pi$$.

This last assumption is not really an assumption than a symmetry argument. Now assume we have a particle with velocity $$\mathbf v$$. This particle's contribution to the total pressure on the cube's upper face (that is, $$(x,y,L)$$) is,

$$p_z = \frac{m (\mathbf v\cdot\hat z)^2}{L} = \frac{m}{L}\,(v\cos(\theta))^2$$

This follows since this particle has speed $$v_z=\mathbf v\cdot\hat z$$ on $$+\hat z$$ direction, and it goes through a total path (along $$\hat z$$) of $$L$$ in the process.

We calculated the probability of this event above.

$$p=\langle p_z \rangle=\int p_\, dP(\theta+d\theta,\phi+d\phi) = \int_0^\pi \frac{mv^2}{L}\,\cos^2(\theta)\,\sin(\theta)\,d\theta\ /\ 2$$ $$p=\frac{mv^2}{6L}\,(\cos^3(0)-cos^3(\pi)) = \frac{mv^2}{3L}$$

The deal is, assuming only $$N/3$$ particles are moving in $$\hat z$$, and the ones that do have velocities only on $$\hat z$$ is identical to the calculations made by assuming all particles ($$N$$) have an arbitrary velocity vector.

• Thank you, this is what I was looking for. But you say "This follows since this particle has speed $v_z=v⋅\hat{z}$ in $+\hat{z}$ direction, and it goes through a total path (along $\hat{z}$) of $L$ in the process." Why is the path along $\hat{z}$ equal to $L$? This assumes that the particle starts at the bottom of the box. On average this is not the case though. – TheoreticalMinimum Nov 17 '19 at 20:57
• I think I answered this last point for myself: Pick a random particle with speed $v$ and look at it after a collision where the upper wall received an impulse of $\Delta p=2p_z=2mv_z=2mv\cos(\theta)$. The totally elastic collision did not change the velocity $v$. How long does the particle need to again hit the upper wall? The answer is $\Delta t=\frac{2L}{v_z}=\frac{2L}{v*cos(\theta)}$, since the particle has to travel down to the wall on the floor and than back up again, a total $\hat{z}$-distance of $2L$. So the force on the upper wall is indeed $F=\frac{\Delta p}{\Delta t}=mv_z^2/L$ – TheoreticalMinimum Nov 17 '19 at 21:11