Problems and Solutions on Thermodynamics and Statistical Mechanics

Statistical Physics 345

Solution:

As molecular diffusion is a random processl we have

L~ = nL2 ,

where L is the total displacement of the molecule, 1 is its mean free path, n
is the number of collisions it suffers as it moves through the displacement
L. Therefore, the required time is

where we have taken 1 = 5 x lo-' m and v = 5 x lo2 m/s.

2157

You have just very gently exhaled a helium atom in this room. Cal-

culate how long (t in seconds) it will take to diffuse with a reasonable

probability to some point on a spherical surface of radius R = 1 meter

surrounding your head.

(UC, Berkeley)

Solution:

molecular collisions:

First we estimate the mean free path 1 and mean time interval 7 for

1 kT 1.38 x x 300

nu pa 1 x lo5 x 10-20

1 4 x lo-'

= 1.4 x lo-' s.

v 300

I=-=-= = 4 x lo-' m

r=-=

Since R2 = NL2, where N is the number of collisions it suffers in traversing

the displacement R, we have

R2

lv

t = Nr = (f) r = - = 8.6 x lo2 s NN 14 min.