# Why does gas particle velocity affect rate of effusion?

I understand why smaller particles have more velocity, but I don't understand what velocity has to do with rate of effusion:

My reasoning is thus:

1. Pressure is the number of impacts of particles in a given period of time.

2. If He and Ar are both in balloons at, say, 1.5 atm, both gasses have the same average Kinetic energy but He is moving faster (because it's smaller). Simple enough.

3. If both gasses are subject to the same pressure, they have the same number of impacts over a given area over a given period in time.

4. This should mean that each gas has an equal number of particles approaching a hole over a given period of time.

5. If a hole is big enough for both particles to fit through, the rate of effusion should be the same, since the same number of particles are approaching the exit hole.

For an analogy to explain my thinking ... if you have two lanes of cars going through a checkpoint, one has 30 cars per minute at 50 miles per hour and one has 30 cars per minute at 30 miles per hour, the resulting number of cars through the checkpoint should still be 30 cars per minute in each lane, regardless of their velocity.

My book (and professor, and Wikipedia...) say that it is the higher velocity of He particles which cause the faster rate of effusion; however, from my reasoning, that doesn't make sense except with small enough holes where the size of the particles would have an effect, i.e. the holes are very small in size and He could fit through but Ar could not, or there are simply more holes that He could fit through.

But Graham's law is referring to pinholes, which should be large enough for either atom to fit through.

So why does velocity affect rate of effusion?

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Your point 1 is mistaken (or incomplete) $$P=\frac{F}{A}=\frac1A\frac{\Delta p}{\Delta t}=\frac1A\times\frac{2nmv}{\Delta t}$$
So, pressure is proportional to number of impacts, mass, and velocity. Pressure is not simply "number of impacts", rather it is a combination of number of impacts with other stuff. Since the energies are equal, we can say that $v\propto m^{-\frac12}$, and we get $P\propto n m^{\frac12}$. P is also the same, so $n\propto\frac1{\sqrt m}\propto v$.