CL461Lecture Slides Van Der Waals Forces

8/4/2019 CL461Lecture Slides Van Der Waals Forces

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CL461: Introduction to Colloidal and InterfacialEngineering

Colloidal Forces:The van der Waals Forces

Electric double layer forces

Short range forces

Electrokinetic Phenomena:

ElectrophoresisElectroosmosis

Streaming Potential

Stability of colloidal dispersions and interfaces:

DLVO theory

Slow flocculationStability of interfaces

Reference:

Chapters 10, 11, 12 & 13, Principles of Colloid and

Surface Chemistry, P.C. Hiemenz and R. Rajagopalan,

3rd

ed., Marcel Dekker (N.Y.), 1997.J.N. Israelachvili, Intermolecular and Surface Forces,

Academic Press, San Diego, 1985.


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Colloidal Forces

Govern the microscopic and macroscopicbehaviors of colloidal dispersions

(flocculation, clouding, ordering and related

phenomena) Important in interfacial systems and devices

Magnitude in the range nN to mNTwo categories:

Short-range forces, acting over distances in the

range 0.1 to 1.0 nm

(steric repulsion, solvation and depletion forces)

Long-range forces acting over large distances -

up to 1000 nm

(van der Waals and electric double layer forces)

van der Waals Forces

One of the most important forces in surface and

colloid chemistry Basically electrostatic in nature


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Involve interaction between electric dipoles at

the atomic level: due to the induced or

permanent polarities created in molecules by theelectric fields of neighboring molecules or due to

instantaneous dipoles caused by the positions

of the electrons around the nuclei Always present (between atoms, molecules and

macromolecules)

Electric Dipole: Two equal charges of different

signs separated by a finite distance.

Dipole moment r

is a vector

quantity Magnitude=charge separation

distance, i.e., qR (in C.m)

direction is conventionally taken from negative to

positive charge

A bond between two atoms with differentelectronegativities always results in a dipole

moment. E.g. O-H bond.


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In H 2O, the

bond angle

has to be 109o

for tetrahedral

symmetry. But the lone-pairs occupy more space

and reduce the bond angle to 104.5 o .

The dipole moment of O-H bond is 1.6 D

1 D = 1 Debye = 3.336 10-30

C.mThe dipole moment of H 2O is 1.85 D.

A polar molecule is one with a permanent

electric dipole moment. Eg., H 2O, HCl, HI etc. The

existence of a dipole moment requires an

asymmetric molecule.

Dipoles carry no net charge. A dielectric is a

polarizable nonconducting medium.

The permanent dipole-permanent dipole

interactions known as

Keesom interactions

occur when molecules

with permanent dipoles


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interact. These interactions are thermally

sensitive.

Dipole moment of CO 2 is zero in spite of thedifference in electronegativity between carbon

and oxygen because the molecule is symmetric

(O=C=O).

Even in molecules such as CO 2 which have no

net permanent dipole moment, temporary dipolemoment may be induced by the presence of

external electric field.

The magnitude of induced dipole moment

depends on the polarizability of the neutral

molecule. The induced dipole moment,

E ir

r

= wherev

is the electric field.

The SI unit for is C 2.m 2.J -1.

Permanent dipoles can also induce a dipole

moment in a neutral molecule. The permanentdipole-induced dipole

interactions are called

Debye interactions. The


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interaction energy is independent of the

temperature because the induced dipole follows

immediately the motion of the permanent dipoleand is thus not affected by thermal motion.

London proposed that even in non-polar

symmetric molecules an instantaneous dipole

moment can be induced. Due to constant motion

of electrons, the electronic distribution at anyinstant of time may not be perfectly symmetrical.

Lack of symmetry gives rise to instantaneous

(temporary) dipole moment in a neutral molecule

which can induce a dipole moment in a nearby

molecule leading to induced dipole-induceddipole interactions, called London interactions.


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There are three components of the van der Waals

interactions permanent dipole - permanent dipoleinteractions Keesom interaction permanent dipole-induced dipole interactions

Debye interaction induced dipole-induced dipole interactions

London interaction

The relative contribution of each to the total

force varies and depends on the type of

molecules. Of the three, the London force is

always present (like gravitational force) becauseit does not require permanent or charge-induced

polarity in the molecules.

Even neutral atoms or molecules such as helium

or hydrocarbons give rise to the Londoninteraction. As a consequence, the London

interaction plays a special role in colloid and

surface chemistry.


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It is important in adhesion, physical adsorption, surfacetension, wetting phenomena structure of macromolecules such as proteins

and other biological and nonbiological

polymer molecules

stability of foams and thin films determining the strength of solids, properties

of gases and liquids, heat of melting and

vaporization of solids etc.

Some important points:

1. Between two identical molecules across water or

any other solvent, the van der Waals force is

always attractive.

2. It is a relatively long ranged force compared to

other atomic or molecular level forces. Theeffective range of this force is from about 0.2 nm

to over 10 nm.


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3. The London force is also called the dispersion

force due to the role it plays in the dispersion of

light in visible and UV wavelengths.4. For interaction between two microscopic bodies,

pairwise interactions between constituent

molecules are added up neglecting the influence

of other neighboring molecules, which is an

approximation.5. The functional form that relates the potential

energy to the distance of separation x of a pair

of molecules for all three types of interactions,

namely Debye, Keesom and London interactions

is 6 )1( x indicating that attraction results from

these interactions always and an inverse sixth

power law is obeyed.

6. An inverse seventh power law is a special case

of the induced dipole induced dipoleinteraction that applies to the case of large

separations.


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At contact, i.e., x 0, the van der Waals interaction

energy . But before this happens, a strong

repulsive force arises, called steric repulsion or

Born repulsion, which varies as 1/ x 12 .

Net interaction energy:

612 x x

= (1)


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is a measure of the strength of the attractive van

der Waals force.

The total potential energy function will display aminimum which corresponds to an equilibrium

situation.

Eq. (1), which is a 12-6 power law, is called the

Lennard-Jones potential.

The interaction force F ( x ) is given by,

x x

x F = )()( (2)

Exercise 1:

Through what distance is the tip of an AFMscantilever maximally deflected by a single

inter-atomic bond of the Lennard-Jones type,

given that the cantilevers spring constant is

10 -3 Nm -1, the Lennard-Jones energy and

distance parameters being = 0.2 10-18

J andr 0 = 0.15 nm respectively.


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Hints:

(i) The Lennard-Jones potential function:

=

60

120 2)(

r r

r r r

(ii) Calculate the force and the distance r at

which the force is maximum from the first and

second derivatives of (r) with respective to r .

Debye interaction:

A molecule with a dipole moment 1 produces an

electric field in the space around it.

This field can induce a dipole in a second

molecule.

The potential energy of the second dipole

620

212

21

)4( x

=

(3)

x is centre-to-centre distance between the

molecules.

2 is the polarizability of the second molecule.


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0 = 8.85 10-12 C2.J -1.m -1 = the dielectric constant

of vacuum which is the intervening medium

between the two molecules.The second dipole acts on the first one in a similar

fashion giving a second contribution to the

interaction energy that is identical to Eq. (3) except

that the subscripts are interchanged.

The total potential energy of attraction is the sumof the two contributions:

620

221

212 1

)4(

)(

x D

+=

(4)

This is the Debye equation for the attraction

between a permanent dipole and an induced

dipole.

Keesom interaction:

Interactions involving permanent dipoles are angle

dependent and are often less than the thermal

energy, k BT.

kB is the Boltzmann constant (1.38 10-23 J.K -1)

T is the absolute temperature.


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When this is true, the molecular rotation becomes

more or less free.

The orientation-dependent dipole-dipole interactionenergy is averaged over all angles with a

Boltzmann weighting factor [exp(- /k BT)].

The orientations that have a lower energy have a

larger weighting factor; the net interaction energy

due to dipole-dipole interaction is non-zero and isgiven by,

620

22

21 1

)4(32

xT k B K

=

(5)

This is Keesom equation for the interaction

between two permanent dipoles.

The inverse dependence on temperature reflects

that the greater thermal motion overcomes the

mutual orientating effects of the dipoles at higher

temperatures.

London (dispersion) interaction :

London interaction between a pair of induced

dipoles is a quantum mechanical result.


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All molecules possess transient dipoles as a result

of fluctuations in the instantaneous positions of

electrons.The instantaneous dipole produced in one

molecule can induce an instantaneous dipole in

another molecule.

The dipoles are formed by symmetrical vibration of

electrons in the two molecules. A vibrating dipolemay be regarded as a harmonic oscillator and the

interaction energy may be found by solving

Schrodinger equation.

The London interaction energy between two

molecules is given by:

620

2,01,0

21

21 1)4(2

3 x

h L

+

=(6)

h is the Plancks constant (6.63 10 -34 J.s)

is the characteristic vibrational frequency

0 is the polarizability

eme

0

2

21

= (7)


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e = 1.6 10 -19 C = electronic charge

m e = 9.11 10-31 kg = electronic mass

h is a characteristic energy of the system theionization energy I

Eq. (6) can be rewritten as:

620

2,01,0

21

21 1)4(2

3 x I I

I I L

+

=(8)

van der Waals interaction:

In general, we may think of any molecule as

possessing a dipole moment and polarizability.

So, all three types of interaction may operate

between any pair of molecules.

(See Table 10.2 of Hiemenz & Rajagopalan)

For nonpolar molecules with zero dipole moment,

two of the interactions make no contribution. E.g.,

benzene.

Since all molecules are polarizable, London

interaction is the most prevalent of the three.


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Only in highly polar molecules such as water is the

dipole-dipole interaction greater than the

dispersion component.The mixed interaction described by Debye

equation is generally the smallest of the three.

For a pair of identical molecules, the net van der

Waals interaction A is the sum of the three

contributions (Eqs. 4, 5 and 6) and is given by

611

62

1,01

412

11,020

143

32

2)4(

1 x x

hT k B

A

=

++= (9)

The interaction parameter 11 is defined as,

++= 2

1,01

412

11,020

11

4

3

3

22

)4(

1

h

T k B (10)

The subscript 11 indicates that a pair of identical

molecules is being considered.

11 is measured in J.m6

The molar polarization P of a substance can be

related to its relative dielectric constant r by


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+=

+=

T k N M

P B

A

r

r

3321 2

00

M = Molecular weight; = density; N A = Avogadronumber

From studies of r as a function of T, 0 and can

be evaluated. For substances with no dipole

moment, = 0 and r = n2 where n is the refractive

index at long wavelengths.

Therefore, 00

2

2

321

A N

nnM

P =+=

To use this equation, it is necessary to extrapolate

to infinite wavelength (zero frequency) as electric

field of light also alters the molecule.

Example 10.1

Relative magnitudes of van der Waals forces and

relation to heat of vaporization

Taking = 1.0 D and = 10 -39 C 2.m 2.J -1, calculate

the amount of energy needed to separate a pair of

molecules from 0.3 nm to . Scaled up by


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Avogadro number, how does this energy compare

with typical enthalpies of vaporization.

Contribution to 11 by Debye term = 20 )4(

1

211,02

=23039

212 )10336.30.1(102)1085.84(1

= 1.81 10 -78 J.m 6

Contribution to 11 by Keesom term = 20 )4(

1 T k B3

24

1

= 2931038.13)10336.30.1(2

)1085.84(1

23

430

212

= 1.66 10 -78 J.m 6

The characteristic frequency in the London term

eme

0

2

21

=

= 3139219

1011.910)106.1(

21

= 8.44 10 14 HzContribution to 11 by London term

= 20 )4(

1

21,014

3 h


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=2391434

212 )10(1044.81063.643

)1085.84(1

= 3.41 10-77

J.m6

Contribution from London term is more than one

order of magnitude compared to other terms.

11 = (0.181 + 0.166 + 3.41) 10-77 = 3.76 10 -77 J.m 6

VA at 0.3 nm = 6977

)103.0(

1076.3

= - 5.16 10 -20 J

VA at = 0

Energy required to separate the pair of molecules

to infinite distance = 5.16 10 -20 J

Energy required for 1 mole = 5.16 10 -20 NA

= 5.16 10 -20 6.02 10 23

= 31060 J.mol -1

When this energy is supplied the intermolecular

attraction becomes zero and the substance turns

into gaseous state.

The value of this energy is in the correct order of

magnitude for heats of vaporization.


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Example 10.3

The strength of van der Waals forces and the

structure of materialsNonpolar argon and methane are gaseous at room

temperature while substances made up of high

molecular weight hydrocarbons are liquids or

solids. Consider argon as example and show that

the typical attractive energy between two argonatoms separated by a distance of about 0.38 nm is

of the order of thermal energy at room

temperature. Discuss the implication of this result

to the structure of the substance at room

temperature. For argon, 0 /(4 0) is about 1.6 10-30

m 3 and the ionization energy is roughly 2.5 10 -18 J.

London dispersion energy for two identical

molecules (Eq. 8):

620

2 1)4(4

3 0 x

I L

=

= 6923018

)1038.0(1

)106.1(105.243

= -1.6 10 -21 J


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At room temperature, (~ 20 C),

Thermal energy = k BT = 1.38 10-23 293

= 4.04 10-21

JThus, L = -1.6 10

-21 J = - 0.4 k BT.

As L < k BT, the atoms remain well separated;

argon remains in gaseous state.

In larger molecules, it is possible that A >> k BT

and the material condenses into a liquid or solid.

van der Waals Forces between macroscopicbodies

Consider two blocks of identical material withplanar surfaces


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Fig (a)

Molecule O is at a normal distance z from thesurface of a bulk sample with a planar face and

infinite extension and of the same material.

x is the distance of O from all molecules in the

ring-shaped volume element, dV = 2 y dy d

The number of molecules within this volumeelement

dN = ( N A /M ) dV

M = molecular weight; N A = Avogadro number

= density of the material.

d A = Van der Waals interaction energy between

the molecules in dV and the molecule O

dN x

d A

=

6

=

ydyd

xM N A

6

2

(11)

But,222

)( y z x++=

(12)Combining (11) and (12),

[ ]322)(2

y z

ydyd M N

d A A++

=

(13)


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Total interaction energy for the entire block

[ ]

++

=

0 0322)(

2 y z

ydyd M N A

A

(14)

Integration over y yields,

40

2220

322 )(1

41

])[(1

21

21

])[( +=

++=

++

z y z y z

ydy

Integration over yields,

30

30

4

1121

)(1

31

41

)(41

z z z d =

+=

+

Thus Eq.(14),

33 61212

z M N

z M N A A

A

=

=

(15)

Fig. (b) Suppose point O is located inside a

second block of material.

A will be the interaction energy for all molecules

within a slice at a distance z from the first block.

The number molecules in a volume element of

thickness dz per unit area = ( N A /M ) dz .


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The interaction energy per unit area of this slice

and block 1 is given by

36 z M N

dz M N

dz M N d

A A

A A AA

=

=

(16)

Integrating between the z = d and z = , the

potential energy of attraction per unit area between

two identical blocks of infinite extension.

2

2

2

2

3

2

121

21

6

6

d M N

d M N

z dz

M N

A A

d

A AA

=

=

=

211

12 d

A AA

=

(17)

The Hamaker constant M N

A A11

2

=

It has energy units.The subscript 11 implies two identical bodies.

A 11 is in the range 10-21 to 10 -19 J for most materials.


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r

Eq.(17) shows that the volume of the block does

not influence the interaction energy per unit

surface area between the two blocks. In suchcases, the van der Waals force can be considered

as a surface force.

Hamaker has derived the potential energy of

interaction for several geometries (Table 10.4,Hiemenz and Rajagopalan).

Spheres of equal radius R ;

distance of separation of surfaces along the line of

centers = d

222

21

2

1

2

2

1

2

44),(4),(),(

),(ln

),(

2

),(

2

6 R Rd d d R f Rd d d R f

d R f

d R f

d R f

R

d R f

R A A

++=+=

++=

(18)

The distance between the centers of the spheres,

r = d + 2 R .


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Eq. (18) can be rewritten as,

++

=2

22

2

2

22

2 4ln

2

4

2

6 r

Rr

r

R

Rr

R A A (19)

For R >> d , i.e., r 2R ,

d R A

Rr R A

A

=

=

12212 (20)

For d >> R , or r >> R ,66

22

362

36

+

=

=

Rd R A

r R A

A (21)

van der Waals interaction energy between two

bodies depends on

(1) Hamaker constant which is a function of

molecular parameters

(2) a geometric function which depends on the size

of the bodies and the distance of separation.

The energy decreases with the increase in distance

of separation by an inverse power law and hence is

fairly long-ranged.


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Effect of the medium on the van der Waals

attraction

For dispersions of one phase in another, the effect

of medium must be taken into account.

The medium will influence the Hamaker constant

while the geometry related function remains

unaltered.Let A ijk be the Hamaker constant for interaction

between i and k in medium j .

Coagulation process represented in Fig (a):

Particles numbered 2 represent the dispersedphase and those numbered 1 are the solvent.

Assume a pseudochemical reaction - initially each

dispersed particle is accompanied by a solvent


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particle forming an independent kinetic unit; later

the dispersed particles form a doublet and the two

solvent particles form a kinetically independentdoublet.

The change in the potential energy is given by

122211 2 += (22)Since s are proportional to the Hamaker

constants, Eq.(22) can be written as,

122211212 2 A A A A += (23)

The subscript 212 indicates two particles of type 2

separated by medium of type 1.

A 11 , A 22 , and A 12 are the values of Hamaker

constants in vacuum.

Using a mixing rule type of approximation,

221112 A A A = (24)

Combining Eqs. (23) and (24),

( )22211212 A A A = (25)

From Eq. (25), the following important points may

be noted:


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1. A 212 = A 121

2. A 212 is always positive, i.e., identical particles

attract each other in a medium as well asunder vacuum.

3. A 212 is less than A 11 and A 22 . Thus the medium

diminishes the van der Waals attraction

between the particles.

4. A 212 is zero for A 11 = A 22 . Hence, by varying themedium, van der Waals interaction can be

reduced to zero.

Fig (b): Dissimilar particles (numbered 2 and 3) are

dispersed in the solvent. By analogy,The change in the potential energy is given by

13122311 += (26)For the contribution of molecular properties,

13122311213 A A A A A += (27)

Using a mixing rule type of approximation,

3311221133221111213 A A A A A A A A A += (28)

which factors to


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11221133213 A A A A A = (29)

From Eq. (29), the following important points may

be noted:

1. A 213 = A 312

2. When A 33 and A 22 are both either greater or

less than A 11 , A 213 will remain positive and the

particles will attract each other.

3. A 213 is zero for A 11 = A 33 or A 11 = A 22 in which

case there is no van der Waals interaction.

4. A 213 is negative when,

A 33 > A 11 > A 22 or A 22 > A 11 > A 33 .

In this case the interaction energy will be

positive; i.e., there will be net repulsion between

the particles.

Repulsive van der Waals occurs in a number of

practically important cases, such as different types

of polymers in organic solvents and for

hydrocarbon films on water.


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Hamaker constant from DLP theory:

Adding together molecular interactions to account

for macroscopic attractions is anoversimplification and overlooks several things:

Molecules nearer the surface screen the

interactions of those buried well inside the

material.

Because of the inverse power law, the moleculesnearer the faces make the predominant

contribution to the interaction.

If these molecules have permanent dipoles, they

may experience orientation effects under the

influence of the surface that are not described bythe theory.

The possibility of surface heterogeneity confuses

the choice of polarization parameters

The presence of intervening medium can further

aggravate the above complications.

A macroscopic theory based entirely on

measurable bulk property (dielectric constant and

refractive index) rather than molecular parameters


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is developed by Dzyaloshinskii, Lifshitz and

Pitaevskii from quantum field theory.

Unfortunately, in its general form, the DLP theoryis complicated and difficult to apply. According to

this theory,

d

iiii

iiii

h A +

+=

)()()()(

)()()()(

8

3

13

13

0 12

122213 (30)

In this expression, (i) is the dielectric constant

and is a complex function of frequency .

By matching the dielectric constants (or refractive

indices), the Hamaker constant can be made equal

to zero.

Although very general, the difficulty in applying the

DLP theory has limited its use and several semi-

classical theories have been developed.

Semi-classical approach based on surface tension:

This is applicable to blocks of nonpolar materials

at small separations.

Consider two identical blocks of material separated

by their equilibrium vapor. The vapor can be


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assumed to behave essentially as vacuum. The

facing planes become equilibrium surfaces

characterized by an interfacial free energy .d 0 = the equilibrium spacing between the

molecules.

Viewing the cohesion process as the one in which

two blocks of material are separated from d = d 0 to

d = , the work of cohesion = 2 .The van der Waals attraction energy between two

identical blocks is given by,

211

12 d

A

= (17)

The work done to separate the blocks from d = d 0

to d = is given by,

212 20

110 ===

d

Ad (31)

or 2

011 24 d A =

(32)For e.g., consider an alkane of = 25 mJ.m -2. With

d 0 = 0.2 nm, A = 7.510-20 J, which is very close to

the value predicted from molecular parameters.


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Hamaker constant A 213 as given by Eq. (29) can be

written in terms of , assuming that the d 0 value is

the same for all three materials. For a variety of polymers a reasonable estimate for d 0 is 0.2 nm.

( ( 121320213 24 = d A (33) A 213 becomes negative and hence the coagulation

of dissimilar particles becomes energetically

unfavorable when the surface tension of the

medium is intermediate between the surface

tension of the two kinds of dispersed units.

Practical Applications:

1. Particle engulfment

Solid particles are dispersed in a melt and the melt

is allowed to solidify in a channel along which a

suitable temperature gradient is maintained. The

dispersed particles are either engulfed or rejected

by the advancing solidification front.

This can be explained as follows:


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Let melt be component 1, the dispersed particles 2,

and solid front 3. Engulfment is equivalent to

coagulation that occurs when the van der Waalsenergy is attractive between the particles and the

advancing solid front, i.e., A 312 is positive.

Rejection occurs when A 312 is negative. Though the

solid and the melt are chemically identical, since

they are in different physical states, their Hamaker constants A 33 and A 11 are different.

From Eq. (33), rejection is predicted so long as the

surface tension of the melt is intermediate between

the values of the solid and the dispersed particles.

(See e.g. 10.5 and Table 10.6 Hiemenz &Rajagopalan)

2. Stability of thin liquid films

A drop of pentane spreads into a thin film when

placed on water surface, whereas a larger

hydrocarbon such as dodecane breaks up intosmaller droplets.

The van der Waals interaction energy between

planar interfaces of air (material 2) and water


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(material 3) through the film of thickness d of

material 1 (pentane/dodecane) = 2312

12 d

A

. In the

case of pentane, A 312 is negative, i.e., there is

repulsive interaction between air and water. Hence

pentane spreads on the water surface thereby

separating air and water as much as possible. In

the case of air-dodecane-water, the Hamaker

constant is positive and the van der Waals force is

attractive. Hence the film continues to thin and

breaks up into small droplets.

3. Soil removal

In many practical cases, the adhesion contact

force can be approximately estimated by the van

der Waals force of attraction. The adhesion force

between a soil particle and a fiber goes down by a

factor of 4 when the soil-fiber system is immersed

in water. From Eq. (29), we find that if the Hamaker constant of the intervening medium ( A 11 ) is

increased, then A 213 gets reduced. Since Hamaker

constant for water is larger than for air, there is a


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decrease in the van der Waals attraction between

the fiber and soil particle when immersed in water.

Exercise 2

Calculate the van der Waals energy and force of

interaction between two polystyrene spherical

particles of diameter 2 m, when separated by a

distance of 10 nm in water. Hamaker constants of water and polystyrene are respectively, 4.310 -20 J

and 8.810 -20 J.

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CL461Lecture Slides Van Der Waals Forces