Transcript Last Lecture:

Last Lecture:
• The thermodynamics of polymer phase separation is
similar to that of simple liquids, with consideration
given to the number of repeat units, N.
• For polymers, the critical point occurs at cN=2, with
the result that most polymers are immiscible.
• As cN decreases toward 2, the interfacial width of
polymers becomes broader.
• The Stokes’ drag force on a colloidal particle is
Fs=6phav.
• Colloids undergo Brownian motion, which can be
described by random walk statistics: <R2>1/2 = n1/2 ,
where  is the step-size and n is the number of steps.
• The Stokes-Einstein diffusion coefficient of a colloidal
particle is given by D = kT/(6pha).
Colloids under Shear and
van der Waals’ Forces
3SCMP
2 March, 2006
Lecture 7
See Jones’ Soft Condensed Matter, Chapt. 4 and Israelachvili, Ch. 10 &11
Flow of Dilute Colloidal Dispersions
The flow of a dilute colloidal dispersion is Newtonian (i.e.
shear strain rate and shear stress are related by a
constant h).
In a dispersion with a volume fraction of particles of f in a
continuous liquid with viscosity ho, the dispersion’s h is
given by a series expression proposed by Einstein:
h = ho (1 + bf + b f + ...)
2 2
A typical value for the constant b is 2.5; the series can
usually be truncated after the first two or three terms,
since f must be << 1 for the equation to hold.
Flow of Concentrated
Colloidal Dispersions
At higher f, h is a function of the shear strain rate,  ,
and the flow is non-Newtonian. Why?
Shear stress influences the arrangement of colloidal particles.
At high shear-strain rates, particles re-arrange under the
applied shear stress. They form layers or strings along the
the shear plane to minimise dissipated energy. Viscosity is
lower.
At low shear-stain rates, Brownian diffusive motion is able to
randomise particle arrangement and destroy any ordering
due to shear stress. Viscosity is higher.
Effects of Shear Stress
on Colloidal Dispersions
With no shear
Under a shear stress
Confocal Microscope Images
MRS Bulletin, Feb 2004, p. 88
The Characteristic Time for
Shear Ordering
Both the shear strain rate and the Brownian diffusion are
associated with a characteristic time, t.
A
Dx

y
A
v
F
F
s =
A
d
 =
dt
The characteristic time for the shear strain, ts, is simply:
Slower shear strain rates thus have longer characteristic
shear times. One can think of ts as the time over which
the particles are re-distributed under the shear stress.
1

Characteristic Time for
Brownian Diffusion, tD
The characteristic time for Brownian diffusion, tD, can be
defined as the time required for a particle to diffuse the
distance of its radius, a.
DSE 

So,
tD
 2
(R )
t
=
a2
tD
a
a
a2

DSE
Substituting in an expression for the Stokes-Einstein
diffusion coefficient, DSE:
6pho a3
tD =
kT
Competition between Shear Ordering and
Brownian Diffusion: Peclet Number, Pe
To determine the relative importance of diffusion and
shear strain in influencing the structure of colloidal
dispersions, we can compare their characteristic times
through a Peclet number:
t D (a unitless parameter)
Pe =
tS
Substituting in values for each characteristic time:
6pho a 3 6ph o a 3
Pe =
=
1
kT
kT ( )

Thus, when Pe >1, diffusion is slow (tD is long) relative to
the time of shear strain (tS). Hence, the shear stress can
order the particles and lower the h. Shear thinning is
observed!
A “Universal” Dependence of h on Pe
Data for different
colloids of differing
size and type
Shear thinning
region
Small a; Low 
Large a; High 
When Pe <1, tD is short in comparison to tS, and the
particles are not ordered because Brownian diffusion
randomises them.
van der Waals Forces
between Particles
The van der Waals attraction between isolated molecules is
quite weak.
However, because of the additivity of forces, there can be
significant forces between colloidal particles.
Recall the London result for the interaction energy
between pairs of non-polar molecules:
3  o 2 h
C
w (r ) =
(
) 6 = 6
4 4p o r
r
The total interaction energy between colloidal particles is
found by summing up w(r) for the number of pairs at each
distance r.
Interaction Energy between a Molecule
and a Ring of the Same Substance
x

r is the molecular density
in the condensed state.
Israelachvili,
p. 156
Interaction Energy between a Molecule
and a Ring of the Same Substance
The cross-sectional area of the ring is dxdz.
The volume of the ring is thus V = 2pxdxdz.
If the substance contains r molecules per unit volume
in the condensed phase, then the number of molecules
in the ring is N = rV = 2prxdxdz.
The distance, r, from the molecule to the ring is:
r = ( x 2 + z 2 )1 / 2
The total interaction energy between the molecule and N
molecules in the ring can be written as:
W = Nw (r ) = N (
C
r
6
)=
2prxdxdzC
(( x 2 + z 2 )1 / 2 )6
Interaction Energy between a Molecule
and a Slab of the Same Substance
x
Semi-
slab

Interaction Energy between a Molecule
and a Slab of the Same Substance
For a ring of
radius, x:
 2prCxdxdz
W  Nw 
(( x 2  z 2 )1 / 2 )6
Let the molecule be a distance z = D from a semi- slab.
The total interaction energy between the molecule and slab
is found by integrating over all depths into the surface. A slab
can be described by a series of rings of increasing size.
z  x 
xdx
W ( D)  2pCr  dz 
2
2 3
(
x

z
)
x 0
zD
2pCr
W ( D) 
4
z 

dz
z D z
4

 pCr
6D3
Attractive Force between a Molecule
and a Slab of the Same Substance
W (D ) =
•
pCr
6D 3

D
Force is obtained from the derivative of energy with
respect to distance:
dW (D ) pCr
F=
=
dD
2D 4
Interaction Energy between a Particle
and a Slab of the Same Substance

x
z=0
z=2R
R
x
D
D+z
z
2R-z
dz
Slice Thickness = dz
z
R = particle radius
Interaction Energy between a Particle
and a Slab of the Same Substance
For a slice of thickness dz and radius x, the volume is px2dz.
Each slice contains N = rV= rpx2dz molecules.
For a single molecule in the particle at a distance z+D,
pCr
the interaction energy with the slab is:
W=
6(D + z )3
To calculate the total interaction energy between a particle
and the slab, we need to add up the interactions between
every slice (with N molecules) and the slab.
W ( D)   Nw  - rp
rpC z2 R x 2 dz
6

3
(
D

z
)
z D0
Interaction Energy between a Particle
and a Slab of the Same Substance
W ( D)   Nw  - rp
rpC z2 R x 2 dz

3
(
D

z
)
z D
0
6
For a sphere with a radius of R, the chord theorem tells us
that x2 = (2R - z)z. Substituting in for x2:
W ( D)  
r 2p 2C
6
z 2R

z 0
(2 R - z ) zdz
( D  z)
3
But if D<<R, which is the case for close approach when
vdW forces are active, only small values of z contribute
significantly to the integral, and so integrating up to z =
will not introduce much error. We can also neglect z in the
numerator as z <<R when forces are large.
 r 2p 2C
W ( D) 
6
z 
 r 2p 2CR


3
6D
z D0 ( D  z )
2 Rzdz
Attractive Force between a Particle and
a Slab of the Same Substance
2 2

r
p C
Integrating:W ( D) 
6
z 
 r 2p 2CR


3
6D
z D0 ( D  z )
2 Rzdz
It is conventional to write a Hamaker constant as
A = r2p2C. Then,
r 2p 2CR
AR
W (D ) =
=
6D
6D
Note that although van der Waals interactions vary with
molecular separation as r -6, particle/surface interaction
energy varies as D -1.
The force between the particle and a slab is found from
the derivative of W(D):
dW (D ) AR
F=
=
2
dD
6D
1 2 6
Units of A:
( 3 ) Jm = J
m
Hamaker Constants for Identical
Substances Acting Across a Vacuum
A = p2Cr2
Substance
Hydrocarbon
C (10-79 Jm6)
r (1028 m-3) A (10-19 J)
50
3.3
0.5
CCl4
1500
0.6
0.5
H2O
140
3.3
1.5
“A” tends to be about 10-19 J for all substances. Why?
If v = molecular volume, we know that r  1/v and o r3  v
Recall the definition of the London constant: C  o2
So, roughly we see: A Cr2 o2r2  v2/v2 = a constant!
Surface-Surface Interaction Energies
The attractive energy between two semi- planar slabs is !
Can consider the energy between a unit area (A) of
surface and a semi- slab.
dz
D
z
z=0 Unit area
In a slice of thickness dz, there are N =rAdz molecules. In
a unit area, A = 1, and N = rdz.
rpC
We recall that for a single molecule: W (D ) =
6D 3
Surface-Surface Interaction Energies
dz
W (D ) =
z
D
rpC
6D 3
z
z=0
Unit area
To find the total interaction energy per unit area, we integrate
over all distances for all molecules:
rpC z  rdz
W ( D)   Nw  
 3
6
zD z
z=
W (D ) =
r 2pC
6
1
( z
2
2
)
=
z=D
r 2pC
6
[
1
1

2D
2
2] =
r 2pC
12D 2
Summary of Macroscopic Interaction Energies
Israelachvili,
p. 177
What Makes Adhesives Stick to a
Variety of Surfaces?
Soft polymers can obtain
close contact with any
surface - D is very small.
Then van der Waals
interactions are significant.
Significance of W(D) for Planar Surfaces
Per unit area:
A
W (D ) =
12pD 2
Typically for hydrocarbons, A = 10-19 J. Typical intermolecular
distances at “contact” are D = 0.2 nm = 2 x 10-9 m.
W (D ) =
10
19
J
12p (0.2 x10 9 m)2
= 66 x10
3
J
m2
To create a new surface by slicing an  slab in half would
therefore require -1/2 W(D) of energy per unit area of new
surface.
Hence, a typical surface energy, , for a hydrocarbon is 30
mJm-2.
Adhesion Force for Planar Surfaces
As we’ve seen before, the force between two objects is
F = dW/dD, so for two planar surfaces we find:
dW (D )
A
F=
=
dD
6pD 3
As W is per unit area, the force is likewise per unit area.
Thus, it is a pressure, P = F/A.
Using typical values for A and assuming molecular contact:
F=
10
19
J
6p (0.2 x10 9 m )3
= 6.6 x108 N
m2
This pressure corresponds to nearly 7000 atmospheres! But
it requires very close contact.
Attractive van der Waals’ Forces
Hence, when polymer fibers
make close contact to
surfaces, they adhere strongly.
But van der Waals’ forces
also cause attraction
between the fibers!
Ordering of Colloidal Particles
Numerous types of interactions can operate on colloids:
electrostatic, steric, van der Waals, etc.
Control of these forces during drying a colloidal dispersion
can create “colloidal crystals” in which the particles are
highly ordered.
MRS Bulletin,
Feb 2004, p. 86
Electrostatic Double Layer Forces
Colloidal particles are often charged. But, colloidal liquids
don’t have a net charge, because counter-ions in the
liquid balance the particle charge.
-
+
+
-
+
+
+
-
+
+
-
+
+
+
-
+
-
-
+
+
+
-
+
+
+ -
+
+
-
-
+
-
+
-
+
+
The charge on the particles is “screened”.
The Debye Screening Length, k
+
- +
+
+ +
+
+
+
+
+ + +
+ - - + - - +
+ - + + +
+
y
y = y o exp( kx )
k-1
It can be shown that k depends on the ionic (salt)
concentration and on the valency, z, as well as  for the
liquid ( = 85 for water).
2e 2no z 2
k=
 o kT
For one mole/L of salt in water, k-1 = 0.3 nm. Thus the
effects of charge on particle surfaces do not extend very far.
x
Packing of Colloidal Particles
When mono-sized, spherical particles are packed into an FCC
arrangement, they fill a volume fraction f of 0.74 of free space.
e
4r
r
f=
Sphere V
Occupied V
=
4 3
4( pr )
3
e3
Can you prove to yourself that this is true?
When randomly-packed, f is typically 0.6 for spherical
particles. Interestingly, oblique spheroid particles (e.g.
peanut M&Ms) fill a greater fraction of space when
randomly packed.
The Debye screening length can contribute to the effective
particle radius and prevent dense particle packing in the
presence of an aqueous solution.
Electrokinetic Effects
If particles have a charge, q, they can be moved by an
electric field.
FS
q
a
FE
E
At equilibrium, the force from the applied electric field,
FE, will equal the Stokes’ drag force, FS.
FE = Eq = FS = 6phav
The mobility, m, of a particle is then obtained as:
v
q
m= =
E 6pha
Problem Set 4
1. The glass transition temperature of poly(styrene) is 100 C. At a temperature of 140
C, the zero-shear-rate viscosity of a poly(styrene) melt is measured to be 7 x 109 Pas.
Using a reasonable value for To in the Vogel-Fulcher equation, and an estimate for the
viscosity at Tg, predict the viscosity of the melt at 120 C.
2. A polymer particle with a diameter of 300 nm is dispersed in water at a temperature
of 20 C. The density of the polymer is 1050 kg m-3, and the density of water is 1000
kgm-3. The viscosity of water is 1.00 x 10-3 Pa s.
Calculate (a) the terminal velocity of the particle under gravity, (b) the Stokes-Einstein
diffusion coefficient (DSE), (c) the time for the particle to diffuse 10 particle diameters,
and (d) the time for the particle to diffuse one meter.
Explain why DSE will be affected by the presence of an adsorbed layer on the particles.
Explain the ways in which the temperature of the dispersion will also affect DSE.
3. A water-based dispersion of the particle described in Question 2 can be used to
deposit a clear coating on a surface. A 200 mm thick layer is cast on a wall using a
brush. Estimate how fast the brush must move in spreading the layer in order to have a
significant amount of shear thinning. (Note that with a low shear rate, such as under
gravity, there is less flow, which is desired in this application.)
The Boltzmann Equation
yo +
y
+
Charged
+
surface
+
x
+
Ions (both + and -) have a concentration at a distance x
from a surface that is determined by the electrostatic
potential y(x) there, as given by the Boltzmann Equation:
zey ( x )
n+ = no exp(
)
kT
+ zey ( x )
n = no exp(
)
kT
Here e is the charge on an electron, and z is an integer value.
n
no
-
“Bulk”
concentration
+
x
The Poisson Equation
In turn, the spatial distribution of the electrostatic potential is
described by the Poisson equation: d 2y
r (z)
2 = 
dx
o
The net charge density, r, (in the simple case in which
there are only ions to counter-balance the surface charge)
is r = ze(n+ n )
But n+ and n- can be given by the Boltzmann equation,
and then the Poisson-Boltzmann equation is obtained:
d 2y
dx
2
=
zeno
 o
zey
+ zey
[exp(
) exp(
)]
kT
kT
Solutions of the Poisson-Boltzmann Equation
ex
e x
We recall that sinh( x ) =
2
zey
+ zey
+ zey
So, [exp(
) exp(
)] = 2 sinh(
)
kT
kT
kT
The P-B Equation then becomes:
d 2y
dx
2
=
2zeno
 o
zey
sinh(
)
kT
But when x is small, sinh(x)  x, and so for small y:
2z 2e 2no
d 2y 2zeno zey
(
)=
y = ky
2 = 
kT
 o kT
dx
o
In this limit, a solution of the P-B equation is y = y o exp( kx )
where k-1 is called the Debye screening length.