important points from last lecture: the thermodynamics of polymer phase separation is similar to...
TRANSCRIPT
Important Points from 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 N=2, with the result that most polymers are immiscible because N >2.
• As N decreases toward 2, the interfacial width of polymers becomes broader.
• The Stokes’ drag force on a colloidal particle is Fs=6av.
• 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 diffusion coefficient for a particle moving in three directions is defined as D = <x>2/6t, relating the mean-squared displacement to the time of the travel.
• The Stokes-Einstein diffusion coefficient of a colloidal particle is given by DSE= kT (6a)-1.
PH3-SM (PHY3032)
Soft Matter
Lecture 7
Colloids under Shear and van der Waals’ Forces
22 November, 2011
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 ).
In a dispersion with a volume fraction of particles of in a continuous liquid with viscosity o, the dispersion’s is given by a series expression proposed by Einstein:
A typical value for the constant b is 2.5; the series can usually be truncated after the first two or three terms, since must be << 1 for the equation to hold.
...)1( 22 bbo
Einstein equation:
...)5.25.21( 22 polpolo
pol
22
polm
mo
-
m=0.64
m=0.60
Viscosity of Colloidal Dispersions at Higher Concentrations: Particle Packing
Particle packing at m
At higher volume fractions, the particle pack together and the viscosity diverges towards , at the point of solidification.
Viscosity of Soft Matter Often Depends on the Shear Rate
Newtonian:
(simple liquids like water)
s
Shear thinning or thickening:
s
s s
s s
Flow of Concentrated Colloidal Dispersions
In some systems at higher , decreases with increasing shear strain rate, , and the flow is non-Newtonian and shear-thinning. 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-strain rates, Brownian motion is able to randomise the particle arrangement and to destroy any ordering imposed by the shear stress. Viscosity is higher.
Effects of Shear Stress on Colloidal Dispersions
With no shear Under a shear stressConfocal Microscope Images
MRS Bulletin, Feb 2004, p. 88
Steady shear flow: http://youtu.be/n4KjvbCYAWk
Oscillatory shear flow: http://youtu.be/ERlpTZ080wU
The Characteristic Time for Shear Ordering, S
Both the shear strain rate and the Brownian diffusion are associated with a particular characteristic time, .
Slower shear strain rates thus have longer characteristic shear times. One can think of s as the time available over which the particles can be re-distributed under the shear stress.
AF
s =
A
A
y
Fxv
dtd
=
The characteristic time for the shear strain, s, is simply: 1
Characteristic Time for Brownian Diffusion, D
A characteristic time for Brownian diffusion, D, can be defined as the time required for a particle to diffuse the distance of its radius, a.
aa
Substituting in an expression for the Stokes-Einstein diffusion coefficient, DSE:
kT
aD
36
So,SE
D D
a2
DSE
a
t
RD
22)(
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:
S
DPe
=
Substituting in values for each characteristic time:
kT
a
kT
aPe
33 6
)1
(
6
Thus, when Pe >1, diffusion is slow (D is long) relative to the time of shear strain (S). Hence, the shear stress can order the particles and lower the . Shear thinning is observed!
(a unitless parameter)
A “Universal” Dependence of on Pe
When Pe <1, D is short in comparison to S, and the particles are not ordered because Brownian diffusion randomises them.
Shear thinning region
Large a; HighSmall a; Low
Data for different colloids of differing size and type
van der Waals’ Energies between ParticlesThe 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 at short distances.
Recall the London result for the interaction energy between pairs of non-polar molecules:
662
443
r
C
r
hrw
o
o =)(=)(
The total interaction energy between colloidal particles is found by summing up w(r) for the number of pairs at each distance r.
Remember that all types of van der Waals interaction energies will follow this general form (r -6) but the value of C will vary depending on the particular type of molecules.
Interaction Energy between a Molecule and a Ring of the Same Substance
x
Israelachvili, p. 156
is the molecular density in the condensed state: number/volume
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 = 2xdxdz.
If the substance contains molecules per unit volume in the condensed phase, then the number of molecules in the ring is N = V = 2xdxdz.
The distance, r, from the molecule to the ring is:2122 /)+(= zxr
The total interaction energy between the molecule and N molecules in the ring can be written as:
621226
2
))+((=)(=)(= /zx
xdxdzC
r
CNrNwW
Interaction Energy between a Molecule and a Semi- Slab of the Same Substance
x
Semi-
slab
Interaction Energy between a Molecule and a Slab of the Same Substance
Let the molecule be a distance z = D from a semi- slab.
For a ring of radius, x: 62/122 ))((
2
zx
CxdxdzNwW
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.
x
x
z
Dz zx
xdxdzCDW
0322 )(
2)(
33
z
Dz4 6
1)
42
(31
42
)(DC
zC
zdzC
DWz
Dz
Attractive Force between a Molecule and a Slab of the Same Substance
Force is obtained from the derivative of energy with respect to distance:
42D
CdD
DdWF
=
)(=
• D
36D
CDW
=)(
Interaction Energy between a Colloidal Particle and a Slab of the Same Substance
x
zD
R
dz
Slice Thickness = dz
z =0
z
z =2R
2R-z
D+z R = particle radius
x
Interaction Energy between a Colloidal Particle and a Slab of the Same Substance
For a slice of thickness dz and radius x, the volume is x2dz.
To calculate the total interaction energy between a colloidal particle and the slab, we need to add up the interactions between every slice (with N molecules) and the slab.
Each slice contains N = V = x2dz molecules.
For a single molecule in the slice at a distance of z+D, the interaction energy with the slab is:
36 )+(=
zD
CW
R
zD
dzxCNwDW
2
3
2
)(6)(
z
0z
-
Interaction Energy between a Colloidal Particle and a Slab of the Same Substance
For a sphere with a radius of R, the chord theorem tells us that x2 = (2R - z)z. Substituting in for x2:
2Rz
0z3
22
)(
)-2(
6)(
zD
zdzzRCDW
R
zD
dzxCNwDW
2z
Dz3
2
)(6-)(
0
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 the z in the brackets in the numerator because z <<R when energies are large.
D
CR
zD
RzdzCDW
6)(
2
6)(
22z
Dz3
22
0
Attractive Force between a Colloidal Particle and a Slab of the Same Substance
Note that although van der Waals interactions vary with molecular separation as r -6, the particle/surface interaction energy varies as D -1.
It is conventional to define a Hamaker constant as A = 22C. Then,
DAR
DCR
DW66
22
==)(
The force between the particle and a slab is found from the derivative of W(D):
26D
ARdD
DdWF =
)(=
Units of A: JJmm
=)( 623
1
Integrating:D
CR
zD
RzdzCDW
6)(
2
6)(
22z
Dz3
22
0
Hamaker Constants for Identical Substances Acting Across a Vacuum
Substance C (10-79 Jm6) (1028 m-3) A (10-19 J)
Hydrocarbon 50 3.3 0.5
CCl4 1500 0.6 0.5
H2O 140 3.3 1.5
A = 2C2
“A” tends to be about 10-19 J for many substances. Why?
If v = molecular volume, we know that 1/v and r3 v
So, roughly we see: A C2 22 v2/v2 = a constant!
For non-polar, uncharged molecules, recall the definition of the London constant: C o
2
Surface-Surface Interaction EnergiesThe attractive energy between two semi- planar slabs is !
Can consider the energy between a unit area (A) of surface and a semi- slab.
zD
z=0
dz
Unit area
In a slice of thickness dz, there are N =Adz molecules. In a unit area, A = 1, and N = dz.
We recall that for a single molecule: 36D
CDW
=)(
Surface-Surface Interaction Energies
zD
z=0
dz
Unit area
36D
CDW
=)(z
z
Dz z
dzCNwDW
36)(
To find the total interaction energy per unit area, we integrate over all distances for all molecules:
)(=)( 22
21
6z
CDW
z=D
z =
2
2
22
2
122
116 D
C
D
C =][=
212 D
A
For each molecule:
Summary of Molecular and Macroscopic Interaction Energies
Israelachvili, p. 177
D
AR
R
RR
D
AW
662
1
21
If R1 > R2:
Colloidal particles
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: 212 D
ADW
=)(
Typically for hydrocarbons, A = 10-19 J. Typical intermolecular distances at “contact” are D = 0.2 nm = 0.2 x 10-9 m.
23
29
19
1066102012
10
mJx
mx
JDW =
).(=)(
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 mJ m-2.
Adhesion Force for Planar SurfacesAs we’ve seen before, the force between two objects is F = dW/dD, so for two planar surfaces we find:
36 D
AdD
DdWF
=
)(=
As W is per unit area, the force is likewise per unit area. Thus, it has the dimensions of pressure, (P = F/A)
This pressure corresponds to nearly 7000 atmospheres! But it requires very close contact.
Using typical values for A and assuming “molecular contact”:
28
39
19
106610206
10
mNx
mx
JF .=
).(=
Geckos stick to nearly any surface – even under water – because of van der Waals’ attractive forces
Spatulae
SetaeA Gecko Pads on feet
Images from http://news.bbc.co.uk/1/hi/sci/tech/781611.stm
Synthetic “Gecko” Tape
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 ParticlesNumerous 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 ForcesColloidal 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” by the counter-ions.
-ve particle surface Ions are in a
solvent, such as water.
The Debye Screening Length,
kT
zne
o
o
222
It can be shown that depends on the ionic (salt) concentration, no (number/volume), and on the valency, z, as well as the dielectric constant, , for the liquid ( = 85 for water).
For one mole/L of salt in water, -1 = 0.3 nm. As the salt concentration increases,
the distance over which the particle charge acts decreases.
+ +
+
+
+
++
+
+
+
++
++
++
+
++
+
-
-
-
--
-
-
-
-
-
--
-
-
-
-
x
-1
)exp(= xo ---
--
X
Electric potential on surface
Colloidal particles with the same charge will repel each other.
But the repulsion is not significant at a separation distance of D > ~ 4-1
++
++
++ +
+
++
+ +D
Colloidal Interaction Potential
Also, there is usually an attractive van der Waals’ force between particles (with a positive A).
The sum of these contributions makes the DLVO interaction potential:
D
ARe
kTRnDW D
6
642
2
)(Van der WaalsElectrostatic
R
( is a constant.)
0 1 2 3 4 5 6
Interparticle Distance (nm)
0
5
10
15
Inte
ract
ion
Ene
rgy,
w(k
T)
wmax
dmax
Decreasing barrier
Decreasing spacing
Effect of Ionic (Salt) Concentration
DR
Low salt
High salt
D
ARe
kTRnDW D
6
642
2
)(
D
Charged particles are less stable at higher ionic concentrations.
Packing of Colloidal ParticlesWhen mono-sized, spherical particles are packed into an FCC arrangement, they fill a volume fraction of 0.74 of all space.
Can you prove to yourself that this is true?
When randomly-packed, is typically 0.64 for spherical particles. (Interestingly, oblate spheroidal 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 the dense packing of colloidal particles dispersed in a liquid (e.g. water).
= =Sphere V
Occupied V
)3
4(4 3a
e3
e
4aa
Effect of Salt on Ordering of Charged Particles
salt concentration
Disordered
Ordered: FCC arrangement
0.7
Short screening length
Long screening length
Lower ionic concentrations encourage colloidal particle ordering.
Electrokinetic EffectsIf particles have a charge, q, they can be moved by an electric field.
FEFS E
At equilibrium, the force from the applied electric field, FE, will equal the Stokes’ drag force, FS.
FE = Eq = FS = 6av
aq
The mobility, , of a particle is then obtained as:
a
q
E
v
6
Mobility measurements can be used to determine colloidal particle charge. Particles can also be separated by size or charge.
Studying Soft Matter Physics Allows Discrimination between Nanotechnology Science Fact versus Fiction
A vision of “nanorobots” travelling through the a blood vessel to make repairs (cutting and hoovering!).
An engine created by down-scaling a normal engine to the atomic level
http://physicsworld.com/cws/article/print/19961K Eric Drexler/Institute for Molecular Manufacturing, www.imm.org.
http://news.softpedia.com/news/Day-old-nanotechnology-503.shtml
An artist’s conception of a “nano-robot” landing on a red blood cell in flowing blood and injecting a “medicine”.
What’s wrong with this picture?
(1) Low Reynolds number, Re: viscosity is dominant over inertia. Flow is streamlined – not turbulent. Motion stops immediately when propulsion stops.
(2) Brownian and thermal motion: there are no straight paths for travel and nothing is static! (Think of the AFM cantilever beam.)
(3) Attractive surface forces: everything is “sticky” at the nano-scale. It is often not easy to slide one surface over another. Particles might stick to the walls. When particles make contact (D ~ 0), they can be difficult to separate them.
Key Limitations for Nanorobots and Nanodevices
Why not make use of the length scales and self-assembly of soft matter?
Alternative Vision of a Nano-Device
A channel that allows potassium ions to pass through a cell membrane but excludes other ions. The nanomachine can be activated by a membrane voltage or a signalling molecule.
Flexible molecular structure is not disrupted by thermal motion.
Closed state: K+ cannot pass through Open state: K+
can pass through
http://physicsworld.com/cws/article/print/19961
The Boltzmann Equation
Ions (both + and -) have a concentration, n (number/vol.) at a distance x from a surface that is determined by the electrostatic potential (x) there, as given by the Boltzmann Equation:
))(
exp(=+ kTxze
nn o
))(+
exp(=kT
xzenn o
Here e is the charge on an electron, and z is an integer value.
o
+++++ x
-n
x+
“Bulk” concentration
no
Charged surface
In turn, the spatial distribution of the electrostatic potential is described by the Poisson equation:
o
z
dx
d )(
=2
2
The Poisson Equation
But n+ and n- can be given by the Boltzmann equation, and then the Poisson-Boltzmann equation is obtained:
)]+
exp()[exp(=kTze
kTzezen
dx
d
o
o
2
2
The net charge density, , (in the simple case in which there are only ions to counter-balance the surface charge) is
)(= + nnze
We recall that
Solutions of the Poisson-Boltzmann Equation
)+
sinh(=)]+
exp()[exp(kTze
kTze
kTze
2So,
2
-)sinh(
-xx eex
The P-B Equation then becomes: )sinh(=kT
zezen
dx
d
o
o
22
2
But when x is small, sinh(x) x, and so for small :
222
2
2 2)(
2 kTnez
kTzezen
dxd
o
o
o
o
In this limit, a solution of the P-B equation is )exp( xo
where -1 is called the Debye screening length.
Problem Set 41. 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 Pas. 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 m 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.)
4. For charged colloidal particles, with a radius of 0.1 m, dispersed in a solution of NaCl in water, calculate the Debye screening length when the salt concentration is (a) 1 mole per litre; (b) 0.01 moles per litre; and (c) 10-5 moles per litre. In the absence of salt, the particles pack into a random, close-packed arrangement at a volume fraction of 0.64. For each of the three salt concentrations, calculate the volume fraction of particles when the particles are randomly close-packed. You should assume that the radius of the particles is equal to the true particle radius plus the Debye screening length.