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How many of the molecules will collide within a given area A of the wall normal to the x-direction in some fixed time interval Δt ? On average half the molecules have a positive x-component of velocity. Therefore half the molecules contained inside a cylindrical volume of cross sectional area A and length vΔt will strike the wall within a given area A during the time interval Δt.

So, the number of collision with A during $\Delta{t}$ is

$\frac{1}{2}\frac{NAv_x\Delta{t}}{V}$

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So my question is how can we assume that half of the molecules are moving one way and half the other? I know that we are focusing on one co-ordinate i.e. x-coordinate, thus, we are not focusing in other directions like up and down etc. Nevertheless,I wondered, if I were the physicist coming up with this derivation, why would I divide the number of collisions by 2 saying that half of the particles go one way and half the other way. How can it be just half not one third any other probability.

Thanks for the time and help.

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  • $\begingroup$ Another way an equivalent relation is derived is to evaluate the time taken for one molecule to go from one wall to the other wall and dark to the original wall whilst moving parallel to the x-axis $\frac{2L}{v_{\rm x}}$ where $L$ is the separation of the walls. $\endgroup$
    – Farcher
    Commented Oct 18, 2018 at 6:10
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    $\begingroup$ These approaches to deriving the theory have always bugged me because they only work for perfectly rectangular boxes where the gas is so dilute that particle collisions can be entirely ignored. A more robust approach requires you to consider the volumetric density of material and write a distribution function for the possible angular trajectories of the particles toward the wall. By integrating these out you get the number of particles hitting the wall and from this you can calculate a force and thus a pressure, which gives the ideal gas law. $\endgroup$
    – Matt Hanson
    Commented Apr 8 at 3:55

2 Answers 2

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A simple way of thinking about this is to think about the motion of particles through an imaginary plane somewhere inside the fluid. The number of particles crossing the plane from one side to the other must be the same as the number of particles crossing in the other direction. This must be the case because first of all the gas is homogeneous and it has the same properties (e.g. average kinetic energy, particle velocity, etc.) everywhere. If the particles moved from one side to the other more than in the other direction, you'd have a build-up of particles on one side, violating the second law of thermodynamics.

For particles near a wall, they don't 'know' they are near the wall until they hit it. So the same rule must apply to particles near a wall.

Viewing it from the perspective of the 2nd law might be the easiest way of thinking about it, but you can also view it purely from the basic assumptions of an ideal gas: a bunch of non-interacting particles. If you had just a single particle in the box, the same law would apply. On average, over long enough time, the particle would find itself moving toward a wall the same number of times it would find itself moving away from it.

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  • $\begingroup$ So, basically it's the second law of thermodynamics that governs the fact that the particles in average move in both the directions. Otherwise it defies the spontaneous nature of air molecules i.e. to diffuse. Did I get it right? $\endgroup$
    – suiz
    Commented Oct 18, 2018 at 4:59
  • $\begingroup$ Look at my edit. $\endgroup$
    – A Nejati
    Commented Oct 19, 2018 at 4:57
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Projections of the velocities of gas molecules follow Maxwell distribution (i.e., Gaussian/normal distribution): $$ f(v_x)dv_x=\sqrt{\frac{m}{2\pi k_B T}}e^{-\frac{mv_x^2}{2k_BT}}dv_x, $$ that is exactly half of them have positive velocities, moving in the direction of the wall, while the other half move in the opposite direction, and will never collide with the wall.

Related:
One may ask, what will happen to the molecules that move away from the wall, after they collide with the other side of the container and return. The assumption here is that we calculate the pressure average over a period of time that is sufficiently long for many molecules to hit the wall (and neglect the shot noise), but still too short, to account for the molecules that go to the other side of the container and return.

This time is also shorter than the mean free time between molecular collisions, which is not appearing explicitly in Maxwell-Boltzmann theory, but which must be kept in mind, whenever we talk about thermodynamic equilibrium.

The finite averaging time also poses another mathematical issue - of the molecules with positive but very small velocities, which never reach the wall of the container during this time...

A while ago, for my own amusement I carried out a calculation of a pressure due to the molecules being scattered back and forth between two walls: Proof of pressure of ideal gas from first principles.

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