PHYSICAL CHEMISTRY IN BRIEF

CHAP. 14: DISPERSION SYSTEMS [CONTENTS] 461

Solution

By substituting into the second relation (14.4) we obtain

v=

2 × 9. 81 ×(5× 10 −^6 )^2

9 × 0. 001

(2. 6 −1)× 103 = 8. 72 × 10 −^5 m s−^1 = 31. 4 cm h−^1 ,

The rate ofsedimentation in the centrifugal fieldin an ultracentrifuge rotating with the

angular velocityωis given by the relation

v=

dh

dτ

=

m ω^2 h

6 π η r

(

1 −

ρ 0

ρ

)

=

2 ω^2 r^2 h

9 η

(ρ−ρ 0 ), (14.5)

wherehis the distance from the axis of rotation. In this case the rate of sedimentation is
proportional to the distance from the axis of rotationh.

Example

The sedimentation rate of quartz particles (ρ= 2.. 6 g cm−^3 ) of the radiusr= 1× 10 −^7 m in water

(η

.

= 1mPa s,ρ 0

.

= 1g cm−^3 ) is 1 cm in about 80 hours. What must be the angular velocity and

the number of rotations in an ultracentrifuge for the particles to sediment at a rate of 1 cm min−^1

at a distance of 10 cm from the axis of rotation?

Solution

From relation (14.5) we obtain

w^2 =v

9 η

2 r^2 h(ρ−ρ 0 )

=

0. 01

60

9 × 0. 001

2 × 10 −^14 ×(2. 6 −1)× 1000

= 4. 6875 × 105 s−^2 ,

w= 684. 6 s−^1 , n=

ω

2 π

= 108. 96 s−^1 = 6538min−^1.

After a certain time the rate of sedimentation starts to decrease because the diffusion flow of
particles caused by their different concentrations begins to assert itself against the gravitational
or centrifugal force. After a sufficiently long time the rate drops to zero and asedimenta-
tion equilibriumis attained. For sedimentation in the gravitational field this equilibrium is