Physics 213 Problem 5 Week 3
Diffusion
1. A few atoms of Argon (atomic weight = 40) are released from a storage tank into a room full
of air at 300 K. Assume that the Ar atoms have an average energy of (3/2)kT and a mean free
path of 0.1 micrometer. Determine the diffusion constant of the Ar gas and the average radial
displacement of the atoms from their point of origin at 1 second and at 1 hour after their release.
Plot a few points of rav(t) to get a feeling for the t1/2 function.
The Ar atoms have mass m = 0.040 kg/6.022x1023 = 6.64 x 10-26 kg,
v = (3kT/m)1/2 = (3 x 1.38 x 10-23 x 300/6.64 x 10-26) = 432.6 m/s
= 0.1 µm = 10-7 m
D = v / 3 = (433 m/s )(10-7)/3 = 1.44 x 10-5 m2/s
Diffusion distance in 1 sec = (6Dt)1/2 = (6 x 1.44 x 10-5 x )1/2 = 0.93 cm
Diffusion distance in 3600 sec = 0.93 cm x (3600)1/2 = 56 cm
2. Now calculate the average displacement in the x-direction at 1 second and 1 hour.
This basically the same problem, realizing that <r2> = <x2> + <y2> + <z2> ≈ 3<x2>
Therefore, <x2> = <r2>/3 xrms = rrms/√3.
xrms(1 s) = 0.93 cm/√3 = 0.54 cm
xrms(3600 s) = 56 cm/√3 = 32.3 cm
Note: You can get the same results by using the formula <x2> = 2Dt.
rav(cm)
0 1 2 3 4 5 6 t (s)
2.5
2.0
1.5
1.0
0.5
0.0
