Molecular diffusion, often simply called diffusion, is the thermal motion of all (liquid or gas) particles at temperatures above absolute zero. The rate of this movement is a function of temperature, viscosity of the fluid and the size (mass) of the particles. This ppt contains everything about molecular diffusion which is one of the important topic in mass transfer operations. References - K.A Gavhane, Treybal. Covers some topics for molecular diffusion with numericals. For notes/doubts contact me ! Mail : chauhanjay657@gmail.com Notes and presentation for other subjects like Chemical Reaction Engineering, Instrumentation of process control, Heat Transfer, Fluid Flow Operations will also be provided with Question bank and important examples/Numericals.

2. Molecular diffusion is concerned with the movement of
individual molecules through a substance by virtue of
their thermal energy. As per the kinetic theory of gases
the molecule is imagined to travel in a straight line
unless gets obstructed by another molecule. Then its
velocity changes both in magnitude and direction. The
average distance the molecules travels between
collisions is its mean free path.
Thus the molecule travels highly zigzag path, the net
distance in one direction travelled in given time is
called as rate of diffusion.

4. Diffusion or Molecular diffusion
1.5 m
0.75 m
0.75 m
Salt Solution
Pure Water 87.5 % in 10 years
99 % in 28 years
22 r/min in 60 sec.

5. Rate of movement of molecules is defined as molar flux
Moles/area . Time = moles/cm2.sec.
Two types of flux
N and J flux
N: The flux relative to fixed location in space.
J: The flux of a constituent relative to the average molar velocity
of all constituents

6. The diffusivity or diffusion coefficient DAB which is a
measure of diffusive mobility is defined by Fick’s law of
diffusion
Z
δ
x
δ
D
Z
δ
C
δ
Z
δ
x
δ
D
Z
δ
C
δ
D
J
B
BA
B
A
AB
A
AB
A
C
D
J
C
BA
B 








7. Net Flux
+ve
- ve

13. 1 CA1 2 CA2
Z1=0
Z2= Z
Z2 - Z1 = Z

14. Rearranging and integrating
Let Z2 - Z1 = Z
General equation of flux

15. C
T
R
p
T
R
p
v
n
T
R
n
Pv
y
P
P
C
C
t
A
t
A
A






Molecular Diffusion in gases
Following general equation becomes

17. Evaluation of NA & NB
NA = CH4 & NB = H2
2
1
B
A N
N



18. Steady state diffusion of A through non-diffusing B

19. PA1 + PB1 = Pt Pt - PA1 = PB1
PA2 + PB2 = Pt Pt - PA2 = PB2
PA1 – PA2 = PB2 – PB1

21. NH3 + Air
1
2
PA1 PB1
NH3
PA2 PB2
Water
Water
2

22. Steady state Equimolal Counter diffusion
Distillation NA = - NB then Indeterminate
Above Equations not applicable

23. A
t
A
A
y
P
P
C
C


C
T
R
p
T
R
p
v
n t



dZ
dP
T
R
D
dZ
dP
pt
T
R
Pt
D
dZ
dP
Pt
C
D
dZ
dC
D
A
AB
A
AB
A
AB
A
AB







1

28. Steady state diffusion of A through non-diffusing B
1
2
1
2
1
2
1
1
ln
1
1
ln
1
1
ln
A
A
Avg
AB
A
A
A
AB
A
A
A
AB
A
x
x
M
z
D
N
x
x
z
C
D
N
C
C
C
C
z
C
D
N

















29. XA2 + XB2 = 1  1 – XA2 = XB2
XA1 + XB1 = 1  1 – XA1 = XB1
XB2 - XB1 = XA1 - XA2
1
2
1
2
2
1
1
2
1
2
2
1
1
2
ln
ln
1
1
ln
B
B
B
B
A
A
Avg
AB
A
B
B
B
B
A
A
Avg
AB
A
A
A
Avg
AB
A
x
x
x
x
x
x
M
z
D
N
x
x
x
x
x
x
M
z
D
N
x
x
M
z
D
N































31. Steady state Equimolal Counter diffusion
 
 
2
1
2
1
2
1
2
1
2
1
A
A
Avg
AB
A
A
A
AB
A
A
A
AB
A
C
C
A
A
AB
z
z
A
AB
A
x
x
M
z
D
N
C
C
C
C
C
z
D
N
C
C
z
D
N
dC
N
D
dz
dZ
dC
D
N
A
A


























32. At 9% Density of solution ρ = 1012 kg/m3
At 3% Density of solution ρ = 1003 kg/m3

39. Density of CCl4 = 1.59 gm/cc
Partial Pressure at location 1 PA1 = 33 mm Hg
Molecular weight CCl4 = 154 gm/gmol

40. A volatile organic compound Benzene costing Rs. 5/Kg is
stored in tank of 10 m diameter and open at top. A stagnant
air film 10 mm thick is covering the surface of the compound
beyond which the compound is absent. If atm temperature
is 250C and vapor pressure of compound is 150 mm Hg and
DAB = 0.02 m2/s. Calculate loss in Rs/day.