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CHAPTER 19
TRANSPORT PROPERTIES

This chapter is concerned with properties that depend on
the movement of molecules in fluid phases (i.e. gases or
liquid phases).
19.1 Viscosity:

The viscosity of a fluid is a measure of the friction
resistance to its motion.

Assuming two parallel planes in a fluid that are separated
by a distance (dx) and having velocities of flow differing
by (d).
(See figure)

According to Newton's law of viscous flow, the frictional
force F, resisting the relative motion of two adjacent layers
in the liquid is proportional to the area A and to the
velocity gradient (dv/dx):
F
dv
dx
A
F  A
(velocity gradient)
(area)
dv
dx

The proportionality constant  is known as the
viscosity (or viscosity coefficient).

The SI unit for the viscosity  is Kg m-1 s-1 or N s m-2,
commonly used unit is the poise (g cm-1s-1).
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Measurements of Viscosity:

Viscosity is usually studied by allowing the fluid to flow
through a tube of circular cross section and measuring the
rate of flow. From this rate, and with the knowledge of the
pressure acting and the dimensions of the tube, the
coefficient of viscosity can be calculated on the basis of a
theory developed in 1844 by the French physiologist J. L.
Poiseuille.

The rate of the flow is given as:
dv ( P1  P2 )R 4

dv
8l
(Poiseuille equation)
where, R is the radius of the tube
l is the length of the tube
P1 and P2 are the pressure at each and of the tube
Based on Kinetic theory:
1)
Viscosities of gases:

We have seen that viscosity arises because in a fluid
there is a fictional force, or a drag between two
parallel plates moving with different velocities.

Theory of the viscosity of a gas is very different
from the theory of the viscosity of a liquid. In the
case of a gas, the origin of the frictional drag
between the two parallel planes is that molecules are
passing from one plane to the other.
2
(mkBT )1 / 2

 3 / 2d 2
where,
d is molecular diameter in nm
m
is molecular mass in gram
kB
Boltzmann constant
Example:
The viscosity of carbon dioxide at 25oC and 101.325k is
13.8x10-6 kgm-1s-1. Estimate the molecular diameter.
Answer:
(mkBT )1 / 2

 3 / 2d 2
MMgmol 1
44.01g
m

 7.308x1026 kg
23
1
23
3
6.022 x10 mol
6.022 x10 x10 g / kg
d
(mkBT )1/ 4
 3 / 4 1/ 2
[7.308 x1026 (kg) x1.381x1023 ( Jk 1 ) x298.15(k )]1 / 4
d
(3.14)3 / 4 [13.8 x106 (kgm1s 1 )]1 / 2
d  4.75 x10 10 m  0.475nm
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2)
Viscosities of liquids:

Based on Kinetic theory of gases the viscous
behavior of liquids is very different from that of
gases.

It is shown empirically that the fluidities of liquids
obey a law of the Arrhenius type:
  A fl e
 E fl / RT
(Fluidity)
Ф = 1/

Kinetic theory expression for viscosity of a liquid:
  Avise E
vis
/ RT
liq  e1/ T

(Viscosity)
Kinetic theory expression for viscosity of a gas:
(mkBT )1 / 2
  3/ 2 2
 d
 gas  T 1/ 2

Therefore, the viscosities of gases increase
with rising temperature, while the viscosities of
liquid decrease.
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3)
Viscosities of solutions:

It is difficult to work out a satisfactory treatment of
the viscosities of solutions based on fundamental
principles. As a result, it is usually more satisfactory
to proceed empirically.

When a solute is added to a liquid such as water the
viscosity is usually increased.

Suppose that the viscosity of a pure liquid is o and
that of a solution is .

The specific viscosity is then defined as:
 o
Specific viscosity = 
o
= (increase in viscosity/viscosity of pure solve)

Dividing the specific viscosity by the mass concentration
 (g/dm3) of the solution gives what is known as
the reduced specific viscosity:
1   o
Reduced specific viscosity =

 o
In order to eliminate the effect of intermolecular
interaction between the solute and solvent molecules
the reduced specific viscosity is extrapolated to
infinite dilution, then we obtain what is known as
the intrinsic viscosity []:
 1   o 
[ ]  lim  o 



o


(Viscosity at infinite dilution)
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In general:
The intrinsic viscosity is proportional to the relative
molecular mass:
[ ]  M r
[ ]  kMr
where, k and  are constants and
Mr is relative molecular mass
By taking logarithms:
log [] = log k +  log Mr
Plotting log [] against log Mr gives straight line of
slope = 
(Very important for determining molecular weight of
polymer chain).
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Problem 19.5:
The viscosity of pure toluene is 5.90x10-4 kg m-1s-1 at 20oC.
Calculate the intrinsic viscosity of solution, containing 0.1
g/dm3 of polymer in toluene, having a viscosity of 5.95x10-4 kg
m-1s-1.
Answer:
o = 5.90x10-4 and
 = 5.95x10-4
 1   o 
 5.95x104  5.90 x104 
1
[ ]  lim po 


3 


5.90 x104

o

 0.1gdm 
 1  0.05  3 1
3
1
[ ]  

m kg  0.084m kg
 0.1  5.90 
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19.2 Diffusion:

Solutions of different concentrations are in contact with
each other, the solute molecules tend to flow from regions
of higher concentration to regions of lower concentration.

The driving force leading to diffusion is the Gibbs energy
difference between regions of different concentrations.

In 1855 Adalf Eugen Fick formulated two fundamental
laws of diffusion:
Fick's First Law:

According to the law the rate of diffusion dn/dt of a
solute across an area A is known as the diffusive flux
J is:
J
dn
C
  DA
dt
X
(Fick's first law)
where, (C / X ) is the concentration gradient of the
solute, and dn is the amount of solute crossing the
area A in time dt.
Fick's Second Law:
and
C
 2C
D
X
X 2
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(Fick's second law)
Brownian Movement:

Individual solute particles move in constant and irregular
motion as a result of collision with solvent molecules.

When a particle diffuses a distance X in any direction, in
time (t), the mean square distance is:
2
X  2 Dt
2
D
X
2t
Example 19.2:
The diffusion coefficient for carbon in α-Fe is 2.9x10-8
cm2s-1 at 500oC. How far would a carbon atom be
expected to diffuse in 1 year (3.156x107s)?
Answer:
2
X  2 Dt
= 2x2.9x10-8cm2s-1x3.156x107 = 1.83cm2
2
X  1.35cm

Such diffusion has important practical consequences.
For example, diffusion of carbon into the area of a
mechanical weld might lead to weakening of the
weld and to possible rupture.

Also, a fresh crystal surface becomes contaminated
by diffusion of impurities from the bulk.
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Frictional Force Ff:

When a driving force is applied to a molecule its
speed increases until the frictional force Ff acting on
it is equal to the driving force. The molecule has then
attained a limiting speed :
Ff  
F f  f
where, f is known as the frictional coefficient.
and
D
k BT
f
(Einstein equation)
This equation was first derived by Albert Einstein in
1905.
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Stokes's Law:

The diffusion coefficient depends on the ease with
which the solute molecules can move.

In aqueous solution, the diffusion coefficient of a
salute is a measure of how readily a solute molecule
can push aside its neighboring water molecules and
move into another position.

This is why as shown in Table 19.3, diffusion
coefficients decrease with increasing the molecular
size, since larger solute molecules has to push more
solvent molecules aside during diffusion. Therefore
the large solute molecules will move more slowly
than smaller molecules.

In 1851 the British physicist George G. Stokes
considered a simple situation in which the solute
molecules are so much larger than the solvent
molecules that the latter can be regarded as not
having molecular character.

For such a system, stokes deduced that the frictional
force Ff opposing the motion of a large spherical
particle of radius (r) moving at speed () through a
solvent of viscosity () is given by:
F f  6r  f
where, the frictional coefficient f is given as:
f  6r
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From Einstein equation into Stokes Equation we get
Stokes-Einstein equations:
k BT
D
6r

Measurement of D in a solvent of known viscosity
therefore permits a value of the radios r to be
calculated.
(See example 19.5)

However stokes law involves errors for solvents that
are neither very large nor spherical. In spite this
error, the law proved to be useful in providing
approximate values of molecular sizes.
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