Physical Chemistry Third Edition

450 10 Transport Processes

2 1.5 2 1.0 2 0.5

c 0

0 0.5 1.0 1.5

z/cm

Concentration

t 5 4 hours

t 5 2 hours

t 5 1 hour

t 50

Figure 10.3 Concentration as a Function of Position in a Diffusing System Accord-

ing to Eq. (10.2-14).

ofD 2 equal to 1. 0 × 10 −^9 m^2 s−^1 , a typical value for ordinary liquids. This figure

illustrates the fact that ordinary liquids require several hours to diffuse 1 cm.

Exercise 10.3

Show by substitution that the function of Eq. (10.2-14) satisfies Eq. (10.2-12). You will need the

identity

d

dx

(∫x

0

f(u)du

)

f(x) (10.2-15)

Another case that can be mathematically analyzed is approximated when the bottom

half of a cell is filled with pure solvent and a very thin layer of solute is layered carefully

on it, followed by more pure solvent to fill the cell. An idealized representation of this

initial condition is

c 2 (z,0)

{

∞ if z 0

0ifz 0

For an infinitely long cell, a solution of Eq. (10.2-12) corresponding to this initial

condition is

c 2 (z,t)

n 0

2

√

πD 2 t

e−z

(^2) / 4 D 2 t
(10.2-16)
wheren 0 is the total amount of substance 2 initially present per unit cross-sectional
area. This function is aGaussian function,orGaussian distribution, as introduced in
Chapter 9. It is shown in Figure 10.4 for a value ofD 2 equal to 1. 0 × 10 −^9 m^2 s−^1 and
for three values of the time.
Exercise 10.4
a.Show by substitution that the function in Eq. (10.2-16) satisfies Eq. (10.2-12).
b.Show that the same amount of substance 2 is present at any time by showing that
∫∞
−∞
c 2 (z,t)dzn 0