Chemistry Reference
In-Depth Information
realized that it is impossible to model the bombardments of all solvent molecules
explicitly (see Example 6-3). Therefore, he used an average force randomly acting
on the Brownian particle to represent all such collisions and the force is governed
by the Newtonian dynamics.
_
m
v
52γ
v
(6-31)
_
In the above equation, m is the mass of the Brownian particle; v and
v are its
γ
velocity and acceleration, respectively, and
is the friction coefficient that sig-
nifies the average viscous drag force experienced by the particle. It is assumed
that the drag force is governed by Stokes law (the particle is large enough that the
solvent can be treated as a continuum medium). Accordingly, the friction coeffi-
cient for a macroscopic spherical particle in solvent is:
γ 5
6
πη
r
(6-32)
where r is the particle's radius and
is the solvent viscosity. Since the viscous
drag force would lead to a decay of the particle's velocity, Langevin introduced a
fluctuating force
η
(t) to describe the erratic motion of the particle resulting from
random, uncompensated impacts of the solvent molecules as shown in
Figure 6.5
.
He assumed that the fluctuating forces have a zero mean and that they are inde-
pendent of location in the solution. It is obvious that without such random forces,
the Brownian particle would come to rest due to viscous drag forces. The final
equation of motion of the particle is given by:
ξ
mv
52γ
v
1 ξð
t
Þ
(6-33)
The above equation is the basis of the Langevin dynamics or the fluctuation
dissipation theorem.
EXAMPLE 6-3
Consider a 0.00001 molar solution of poly(vinyl alcohol) with a degree of polymerization of
200 in water. To analyze the dynamics of one polymer chain in such a solution, how many
water molecules should be considered explicitly? Note that the density and molecular weight
of water are 1 kg/L and 18.02 kg/kmol, respectively.
Solution
First of all, let us find out the total number of atoms (each water molecule contains two
hydrogen and one oxygen atoms) in 1 liter of water.
1kg
1L
3
1 kmol
18
1000 mol
1 kmol
3
N
a
water molecules
mol
3 atoms
water molecule
5166:5 N
a
1L3
02 kg
3
3
:
where N
a
is Avogadro's number. On the other hand, the total number of atoms of the polymer
chains in 1 liter of the solution is calculated as follows:
0:00001
mol
N
a
water molecules
mol
L
3
0
@
1
A
198
monomer
atom
monomer
1
atom
end monomers
3
chain
3
7
2
3
8
50:01402 N
a
