1. Mean free path and diffusion
Thermal & Kinetic
Lecture 8
Mean free path and diffusion
LECTURE 8 OVERVIEW
Recap… (and refresher on approximations)
Mean free path
Random walk
Coefficient of diffusion for a gas

2. Last time….
Blackbody radiation – classical and quantum descriptions (Rayleigh-Jeans vs Planck)
<E> for collection of simple harmonic oscillators (quantum vs classical)

3. ? Approximations Maclaurin series expansion: If f(x)=ex, f’(x) =
(a) xex, (b) ex/x, (c) ex ?
ANS: (c)
Take f(x)= ex. We can get an approximation to ex:
ex = e0 + e0x + (e0x2/2!) + (e0x3/3!) +…….
ex = 1 + x + x2/2! + x3/3! + ……….

4. Mean free path of gas molecules
The average distance a molecule travels betweeen collisions is called the
mean free path (
<v> d 2d
In one second, a molecule travels on average a distance <v> and collides with all molecules within a cylinder of radius d (the molecular diameter). ?
Write down an expression for the number of molecules within the cylinder
(i.e. the number of collisions the molecule undergoes in 1 second).
ANS: n’pd2<v> ?
Write down an expression for the average distance between collisions.
ANS:

5. Mean free path of gas molecules
The expression for l on the previous slide is over-simplified. Number of big
assumptions made:
Other molecules are stationary.
Collision depends only on molecular diameter.
U(r) r
Need to consider mean relative speed of molecules, <vr>/<v> = 2 (CW 4)
Due to interactions!
Hence, a more accurate expression for the mean free path, l, is:
s0 is the effective collision cross section.

6. Diffusion and random walks
<x2> varies linearly with time. Hence, the rms displacement t. ?
How does this compare with the displacement of a single free molecule that doesn’t undergo collisions? Is the random walker faster or slower?
ANS: Slower

7. Coefficient of diffusion of a gas
Take a container of gas in thermal equilibrium and introduce a small amount of extra gas into a small region within the container.
n’ for gas within that region is greater than n’ elsewhere in container.
Molecules will diffuse through container until equilibrium is reached.
The rate at which equilibrium is reached depends on the coefficient of diffusion.
NB We assume that there are no convective currents. x A B
Let no. of molecules per unit volume (n’) at time t vary only in the x direction with gradient:
Let net number of molecules per second crossing unit area of plane AB be J(x,t).
Coefficient of diffusion, D, is defined as:
Fick’s law

8. ? Coefficient of diffusion of a gas
Why is there a minus sign in the equation for Fick’s law?
ANS: Diffusion is in +x direction but number density increases in –x direction
Equilibrium is reached by molecules moving randomly but with more moving on
average in the +x direction.
Transport of molecules across unit area of the plane AB. x A B
x + l
x - l
Consider molecules diffusing from x + l and x - l
Make the (rather large) assumption that all molecules move in either positive or negative x, y, and z directions with the mean speed <v>. (Turns out that a more detailed mathematical analysis - taking speed and velocity distributions into consideration – gives same result).
Therefore, 1/6 of the molecules travel in the +x direction.

9. Coefficient of diffusion of a gas
x - l
x + l
At time t, molecules per second arrive at unit
area of the plane AB having left the layer at x + l.
Similarly, molecules/s arrive at unit area of AB
having left the layer at x – l. B x
Therefore, net number of molecules crossing unit area of AB in the +x
direction is…? ?
As l is very small, by taking l = dx we can approximate n’(x ± lt):

10. Coefficient of diffusion of a gas
Substituting this expression into the equation for J(x,t) we get:
Compare this with the expression for Fick’s law:
The diffusion coefficient for a gas is therefore: