Chemistry Reference
In-Depth Information
λ
2
λ
λ
1
λ
3
FIGURE 6.9
Spatial distribution of molecules in a liquid with
λ
signifying the distance between the two
equilibrium positions.
du
dy
f
5η
(6-55)
where u is the velocity in the direction of the force (i.e., x direction);
η
is the
coefficient of viscosity.
Rearranging
Eq. (6-55)
gives:
f
λ
1
Δ
η5
(6-56)
u
where
u denotes the difference between the velocity of the two layers of mole-
cules with a distance of
Δ
y) apart. Consider that a hole is created on the
right-hand side (energy is required to move a molecule to create the hole). As a
result, the molecules in the upper layer can move to the right and settle in a new
equilibrium state. Let
λ
1
(i.e.,
Δ
be the distance between two equilibrium positions in the
direction of motion (we do not know the exact value of
λ
but it should be at the
molecular length scale). The distances between molecules in the two other direc-
tions are
λ
λ
3
. Owing to the intermolecular interactions between neighboring
molecules, there exists an energy barrier
λ
2
and
ε
0
for the upper layer of molecules to
move, as shown in
Fig. 6.10
.
Based upon the absolute reaction rate theory
[18]
, viscosity is given by:
η5
λ
1
k
B
T
k
(6-57)
2
λ
λ
2
λ
3
where k is a constant related to the partition function of the liquid. Obviously, it
is necessary to know the partition function of the liquid to calculate the absolute
value of viscosity. Nonetheless, this is not of concern here.
At the molecular level, the process of diffusion is similar to that of viscous
flow. In order to diffuse in a liquid, two molecules of liquid are required to slip
past each other. Suppose the distance between two successive equilibrium posi-
tions is
used in the viscous
flow. The change of the standard free energy as a function of distance can then
be represented by the curve shown in
Fig. 6.11
.
λ
, which can have a different magnitude than the
λ
