Thermodynamics and Statistical
Mechanics
Transport Processes
Thermo & Stat Mech Spring 2006 Class 15
1
Mean Free Path
The average distance that a particles travels
between collisions is called the mean free path,
l. In order to have a collision, the centers of
two particles of radius R must approach to
within a distance of 2R = d. The picture on the
next slide illustrates this point.
Thermo & Stat Mech - Spring 2006
Class 15
2
Collision Cross Section
s = p d2
Thermo & Stat Mech - Spring 2006
Class 15
3
Mean Free Path
1
ns vrel = 1, so  =
ns vrel
1
l = v =
ns (vrel / v )
Thermo & Stat Mech - Spring 2006
Class 15
4
Mean Free Paths(?)
3 1
1
l=
= 0.75
4 ns
ns
3p 1
1
l=
= 0.77
4 ns
ns
1 1
1
l=
= 0.71
ns
2 ns
Thermo & Stat Mech - Spring 2006
Class 15
5
Mean Free Path
1
ns vrel = 1, so  =
ns vrel
1
1
l = vm =
=
ns (vrel / vm ) ns ( 2v / vm )
l=
1
1
= 0.63
ns
8
ns
vm
fc = =
 l
1
p
Thermo & Stat Mech - Spring 2006
Class 15
6
Some Molecular Speeds
v =

0
8kT
vf (v)dv =
pm

3kT
v =  v f (v)dv =
0
m
3kT
2
vrms = v =
m
2
vmax
2
2kT
=
m
Most Probable
Thermo & Stat Mech - Spring 2006
Class 15
7
Effusion
If there is a hole in a container of gas,
molecules will escape through it. The rate at
which they pass through the hole is equal to the
flux of particles hitting the hole times the area
of the hole, FA. Then the rate of change of the
number of molecules in the container is:
dN
= F A
dt
Thermo & Stat Mech - Spring 2006
Class 15
8
Effusion Limit
If the hole is small equilibrium conditions
prevail, and F is known.
If
p D2
 4p l 2
4
i.e. D 2  0.16l 2
or
or
D 2  16l 2
D  0. 4 l
dN
Then, F = nv and
= FA =  14 nv A
dt
1
4
Thermo & Stat Mech - Spring 2006
Class 15
9
Transport Processes
1. Molecular diffusion. The movement of the
particles is of interest. This is particle transport.
2. Thermal conductivity. Particles carry energy from
a high temperature region to a lower temperature
region. This is energy transport.
3. Viscosity. Drag is created when particles move to a
region which is moving at a different speed. This is
momentum transport.
Thermo & Stat Mech - Spring 2006
Class 15
10
Transport Processes
1. F y =  D
dn
dy
1 dQ
dT
2.
= 
A dt
dy
Fx
dv x
3.
=
A
dy
Thermo & Stat Mech - Spring 2006
Class 15
11
Viscosity
Consider a plate moving parallel to another
plate with a layer of gas between them. The gas
in contact with either plate is at rest relative to
the plate, so there is a velocity gradient in the
gas. A force F is applied to each plate to
maintain the motion, and the area of the upper
plate is A.
Thermo & Stat Mech - Spring 2006
Class 15
12
Viscosity
Thermo & Stat Mech - Spring 2006
Class 15
13
Viscosity
Gas diffuses up and down, transferring x
momentum from layer to layer. The gas that
moves in the + y direction is going slower than
the layer it moves into, and the gas that moves
in the – y direction is going faster than the layer
it moves into.
Thermo & Stat Mech - Spring 2006
Class 15
14
Viscosity
Thermo & Stat Mech - Spring 2006
Class 15
15
Viscosity
First we need to know how far particles travel
between collisions in the y direction. We must
find the average value of y = l cos q, averaged
over the flux of molecules.
 n
dF = 
 4p

[vf (v)dv] sin q cos qdqd

Thermo & Stat Mech - Spring 2006
Class 15
16
Viscosity
l cos qdF

y=
F
ln
y=
4p 

0
vf (v)dv 
p 2
0
2p
sin q cos qdq  d
2
0
1
nv
4
l
2
1
y=
(v ) (2p ) = l
p v  3
3
Thermo & Stat Mech - Spring 2006
Class 15
17
Viscosity


2 dv x

G = n v m v x  3 l
dy 



1
2 dv x

G = 4 n v m v x  3 l
dy 

1
4
 4 dv x  1
dv x
 = 3 n v ml
G = G  G = n v m 3 l
dy 
dy

Fx 1
dv x
G=
= 3 n v ml
 = 13 n v ml (poise)
A
dy
1
4
Thermo & Stat Mech - Spring 2006
Class 15
18
Viscosity
 = n v ml
1
3
8kT
v=
pm
l=
1
8
ns
p
8kT
= n
m
pm
1
3
1
8
ns
p
mkT
mkT
=
=
2
3s
3p d
Thermo & Stat Mech - Spring 2006
Class 15
19
Diffusion
dn 

2

G = v  n( y )  3 l
dy 

dn 

1 
2

G = 4 v  n( y )  3 l
dy 

1
4
dn
 4 dn 
1
 =  3 v l
F = G  G =  v  3 l
dy
 dy 
D = 13 v l
1
4
Thermo & Stat Mech - Spring 2006
Class 15
20
Thermal Conductivity


2 d
G = n v   ( y )  3 l 
dy 



1
2 d
G = 4 n v   ( y )  3 l 
dy 

1
4
 4 d 
1 dQ
d
1
1
= G  G =  4 n v  3 l  =  3 n v l
A dt
dy
 dy 
cvT
=
NA
Thermo & Stat Mech - Spring 2006
Class 15
21
Thermal Conductivity
cvT
=
NA
cv dT
d d dT
=
=
dy dT dy N A dy
cv dT
d
1 dQ
1
1
=  3 n vl
=  3 n vl
N A dy
dy
A dt
cv
 = n vl
NA
1
3
Thermo & Stat Mech - Spring 2006
Class 15
22
Transport Processes
1. F y =  D
dn
dy
D = 13 v l
2.
1 dQ
dT
= 
A dt
dy
cv
 = n vl
NA
3.
Fx
dv x
=
A
dy
 = 13 n v ml
Thermo & Stat Mech - Spring 2006
Class 15
1
3
23