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Assume there are two ideal gases. The first is made of a light particle, and the second is made of a heavy particle. The two are of the same amount, in the same volume container, and at the same temperature. Assuming the two gases are ideal, both are therefore at the same pressure.

Since the temperatures are the same, both gases have the same average KE. The lighter gas will therefore have a higher average speed than the heavier gas. Its collision frequency is therefore greater. Although the collisions are less frequent, the heavier gas has more momentum. Because of this, it strikes the sides of its container with more force, albeit less often. Supposedly, this makes up for the lower collision frequency in the heavier gas. This is why both gases have the same pressure. Is there any way to prove this in the ideal case?


Thought I would qualify some things. Assume that the heavy particle is "x" times more massive than the light particle. This would make the speed of the light particle $ \sqrt{x} $ times greater than that of the heavy particle. This would also mean that the heavy particle has momentum: $ \rho_{H} = {x} / \sqrt{x} $. The light particles would have to collide $ \sqrt{x}/x $ times more frequently, no?

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You seem to have stated the situation nicely. You are clearly starting to think about this like a physicist. Why not finish the work? Do the calculation in terms of $m$ and $M$ and see what you get. – dmckee Sep 21 '13 at 17:24
On another note: this might not be what you're looking for, but for an ideal gas the ideal gas law holds. Since the mass of the individual particles does not appear in it, it is clear that the pressure is the same for both gasses. – Danu Sep 21 '13 at 17:26
InRe: the edit. You are essentially done with the calculation at that point. BTW, we have the MathJax rendering engine running on the site which means that putting \sqrt{x}/x between dollar signs gets you $\sqrt{x}/x$. You might prefer \frac{\sqrt{x}}{x} which renders as $\frac{\sqrt{x}}{x}$. The language is essentially $\LaTeX$ math-mode. – dmckee Sep 21 '13 at 17:30
$ {l} $ = mean free path, which is a constant for both cases. For the light particle, the time $ {t} $ it takes for a collision is $ {l} = {v_L}*{t} $. Extrapolating this and solving the collision frequency of the light particle is $ {f_L} = \sqrt{x} * {f_H} $. In fact, my original statement of collision frequency didn't make sense because I was implying that the light particle was colliding less often than the heavy particle. – David Sep 21 '13 at 17:52
@David Was your question answered? If it wasn't I think I can provide some help. – Gotaquestion Sep 21 '13 at 23:16
up vote 1 down vote accepted

Ok, I believe I figured this out. The pressure of each gas is proportional to the force with which the collide and the frequency of each collision. The force with which they collide is also proportional to the momentum of the particles. Therefore, the pressure is proportional to the collision frequency and momentum. The frequency of collision for the light particles is: $ {f_L} = \sqrt{x} {f_H} $, and the ratio of the pressures is $ {P_L} / {P_H} = (\sqrt{x}/x * \rho_H * \sqrt{x}{f_H}) / \rho_H * {f_H} = 1/1 = 1 $, therefore the pressures must be equal.

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