Question

at what pressuredoes the mean free path of argonat 25 degree Celsius become comparable to the size of a 1L vessel that contains it? take sigma= 0.36nm^2. (answer is in mPa)

Answer 1

I gather you want an answer from first principles, i.e. kinetic theory. So...

Consider a thin (or low density) slab of the gas of thickness z and area A. The total cross-sectional area occupied by the argon molecules will be N*s, where s is the cross-sectional area of 1 argon atom and N is the number of argon atoms in the slab. (You can see why I specify a thin or low density slab -- so that no argon atoms are overlapping, when see along the z axis.) N = p*A*z, where p is the density of the argon atoms.

So the probability that any argon atom traveling through this slab will collide with another argon atom in the slab is the fraction of the cross sectional area of the slab occupied by argon atoms, which is (p*A*z*s)/A = p*s*z. For a rough estimate of the mean free path l, just ask when this crude measure of the probability of collision through a slab of length z reaches 1:

1 = p*s*l solve for l gives l = (p*s)^-1.

Now what is asked is when l = V^(1/3) for V = 1 L:

V^(1/3) = 1/(p*s)

Solve for the density:

p = 1/(s*V^(1/3))

At this point I'm stuck without using some kind of other theory, because you want the pressure, not the density. I'm forced to use, for example, the ideal gas equation of state,P = p k T:

P/(kT) = 1/(s*V^(1/3))

P = k*T/(s*V^(1/3))

If I put in T=25C, s=0.36 nm^2, V = 1 L, I get P = 1. x 10^-6 atm, or 0.0009 torr, which seems pretty reasonable. You need a very good vacuum to have mean free paths on the order of decimeters.