NPTEL – Chemical – Mass Transfer Operation 1
MODULE 2: DIFFUSION
LECTURE NO. 7
2.7 Diffusivity in Solids and its Applications
In many processes such as drying, adsorption and membrane separations
require the contact of gases or liquids with solids. Diffusion occurs in these cases
in the solid phase and the diffusion mechanism is not as simple as in the case of
gases or liquids. However, it is possible to describe by the Fick‟s law used in the
case of fluids. If the diffusivity is independent of concentration and there is no
bulk flow, the steady state molar flux (NA) in the Z direction is given by Fick‟s law
as follows:
N A D A
dCA
(2.76)
dz
where DA is the diffusivity of A through the solid. Integration of the above
Equation gives diffusion through a flat slab of thickness z:
NA 
D A (C A1  C A2 )
z
(2.77)
where CA1 and CA2 are the concentrations at two opposite sides of the slab. This
is similar expression for diffusion obtained for fluids under identical situation. For
other solid shapes the general Equation for rate of diffusion (w) is
w  N A Sav 
D A Sav (C A1  C A2 )
z
(2.78)
where Sav is the average cross section for diffusion.
2.7.1 Mechanism of diffusion in solids and its application
The diffusion of solutes through the solids plays an important role in many
processes such as heterogeneous catalytic reactions. The structure of solid and
interaction with the solutes are important for the rate of diffusion.
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(a) Diffusion in porous solids
The solid sometimes may act as porous barrier or as porous catalyst pellets and
is normally surrounded by a single body of fluid. The inward or outward
movement of the solutes through the pores of the solid is mainly by diffusion.
This movement may occur inside the pore or at the surface of the adsorbed
solute.
(b) Diffusion inside pore
At low pressure the mean free path of the molecules may be larger than the
diameter of the passage when the diffusion occurs inside the fine pores of the
solid. The collision with wall becomes important compared to collision among
molecules. The diffusion of this kind is known as „Knudsen diffusion‟. To quantify
Kundsen diffusivity a simple Equation based on kinetic theory of gases was
proposed as follows:
2 
D K    rp v
3
(2.79)
where rp radius of passage and v is the average velocity of the molecules due to
their thermal energy which is defined as
1
 8RT  2
v

 M 
(2.80)
Here T is the temperature in K and M is the molecular weight. The flux due to
Knudsen diffusion is similar to Fick‟s law:
J K DK
dCA
dz

DK d p A
RT
dz
(2.81)
(c) Surface diffusion
The diffusion of adsorbed molecules on the surface due to concentration gradient
is kwon as surface diffusion. If the fractional coverage of the surface is less than
unity then the some of the active sites remain empty. Adsorbed molecule having
energy greater than the energy barrier tends to migrate to an adjacent vacant
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NPTEL – Chemical – Mass Transfer Operation 1
site. This migration is visualized to occur by „hoping‟ mechanism. The flux due to
surface diffusion may be written similar to Fick‟s law:
J s Ds
dCs
(2.82)
dz
where Ds is the surface diffusion coefficient (m2/s) and Cs is the surface
concentration of the adsorbed molecules (kmol/m 2). Js is the number of moles
transported across unit distance on the surface normal to the direction of
transport (kmol/m.s).
(d) Diffusion through polymers
The diffusion of solutes through polymeric solids is more like diffusion in liquids,
particularly for the permanent gases. The gas dissolves into the solid exposed to
the gas and usually described by Henry‟s law. The gas then diffuses from high to
low pressure side. Hence, high pressure is applied to increase gas
concentrations, thus increasing respective solubility (difference in solubility is the
key feature) of gases. The polymeric chains are in a state of constant thermal
motion and diffusing molecule move from one location to the adjacent location
due to the potential barrier. An Arrhenius type Equation may be applied for the
temperature dependency of the diffusion coefficient in polymers.
DA  D0 e H D / RT
(2.83)
where HD is the energy of activation and D0 is a constant. For the permanent
gases the typical diffusivity value is in the order of 10-10 m2/s.
Example Problem 2.1: A test tube, 1.5 cm in diameter and 18 cm long, has 0.4
gm camphor (C10H16O) in it. How long will it take for camphor to disappear? The
pressure is atmospheric and temperature is 200C. The sublimation pressure of
camphor at this temperature is 97.5 mm Hg; diffusivity of camphor can be
estimated by using Fuller‟s Equation:
DAB 
 1
1 


2 
1
M
M
B
 ( ) B3   A

1.0133  10 7 T 1.75
P ( ) A3

1
1
2
m2/s;
where, T in K; P in bar, MA, MB are molecular weights of A and B, respectively.
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NPTEL – Chemical – Mass Transfer Operation 1

A
 202.16 m/s;

B
 20.1 m/s.
Solution2.1:
MA=152.128,
MB=28.9,
T=293K,
P=1.013 bar (1atm)
DAB 
1.0133 107 (293)1.75
1.013 (202.16)

1
3
1
1 


2

1
152.128 28.9 
 (20.1) 3  

1
2
m2/s
DAB=5.644×10-6 m2/s
pAo = 97.5 mmHg =0.1299 bar
pA1 = 0 bar
l=12 cm=0.12 m
NA 
DAB P ( P  p A1 ) 5.644  106  1.013
(1.013  0)
=
ln
ln
RTl
( P  p A0 ) 0.08314  293  0.12 (1.013  0.1299)
=2.685×10-7 k mol/m2.s
Area of tube,
NA  a t 
t
a
d 2 = 3.143  (0.015) 2 m 2 =1.768×10-4 m2
4
4
0.4  10 3
kmol
152.128
0.4 10 3
s
152.128  2.685 10 7 1.768 10 4
=5.539×104s
t =15.39 hr
Example Problem 2.2: A mixture of He and N2 gas is contained in a pipe at
250C and 1 atm. total pressure which is constant throughout. At one end of the
pipe, partial pressure of He is 0.6 atm. and at the other end (0.2 m) it is 0.2 atm.
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Calculate the steady state flux of He if DAB of He-N2 mixture can be estimated by
using Fuller‟s Equation with

A
 2.88 m/s; and

B
 17.9 m/s.
Solution 2.2:
D AB 
1.0133  10 7 (298)1.75
1.013(2.88)

1
3
1 
 1

2 
1
4.003 28 
 (17.9) 3  

1
2
m2/s
DAB=7.035×10-5 m2/s
pA0 = 0.6 atm
pA1 = 0.2 atm
l=0.2 m
T=298K
NA 
DAB
7.035  105
( p A0  p Al ) =
(0.6  0.2)
0.08314  298  0.2
RTl
=5.68×10-6 k mol/m2.s
Example Problem 2.3: MeOH (A) is separated from aqueous solution by
distillation. At a section of column, vapor phase contains 0.76 mole fraction
MeOH and liquid phase has 0.6 mole fraction. Temperature of the section is
71.20C and total pressure is 1 atm. throughout 1 mm thick vapor film. If molar
latent heat of vaporization of MeOH is 8787 K Cal/K mol and that of water (B) is
10039 K Cal/K mol at the given temperature. Calculate MeOH and water vapor
flux. Given: If mole fraction of MeOH in liquid is 0.6, equilibrium vapor will be
0.825. Vapor phase diffusivity of MeOH, DAB=1.816×10-5 m2/s.
Solution 2.3:
NA∆HA = - NB∆HB
NA×8787 = -NB×10039
yAl=0.76
A
yA0=0.825
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B
1 mm
xA0=0.6
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NPTEL – Chemical – Mass Transfer Operation 1
NB= -NA×
8787
10039
NB=-0.8753NA
Given data,
DAB=1.816 ×10-5 m2/s,
P=1 atm,
R=0.0823 m3atm/kmol
T=344.2 K,
l=1×10-3 m,
yA0=0.825,
yA1=0.76
We have,
N A  ( N A  N B ) y A  CD AB
 ( N A  0.8753N A ) y A 
 0.1247 N A y A 
l
y Al
0
y A0
N A  dZ  
NA×l =
P
RT

dy A
dZ
dy
P
DAB A
RT
dZ
dy
P
DAB A
RT
dZ
DAB Pdy A
RT (1  0.1247 y A )
DABln 1  0.01247 yA1
1  0.1247 yA0
NA 
P
1  0.01247 yA1
ln
RTl 1  0.1247 yA0
NA 
1.1816 105  1
1  0.01247  0.76
k mol/m2.s
ln
3
0.821 344.2  1 10
1  0.1247  0.825
=4.65×10-5 k mol/m2.s
NB=-0.8753×4.65×10-5 k mol/m2.s
=-4.07×10-5 k mol/m2.s
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NPTEL – Chemical – Mass Transfer Operation 1
References:
1. Treybal, R. E., “ Mass-Transfer Operations”, 3rd Eddition, McGraw-Hill,
1981
2. Geankoplis, C.J., “Transport Processes and Separation Process
Principles”. 4th Edition, Prentice-Hall of India, New Delhi, 2005.
3. Dutta, B.K., “Principles of Mass transfer and Separation Processes”.
Prentice-Hall of India, New Delhi, 2007.
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