1

THERMODYNAMICS OF SURFACES AND

INTERFACES

1. Introduction

Everything has to end somewhere. For solids, or liquids that

"somewhere" is a surface, or an interface between phases. For liquids, the

interface is between the liquid and a vapor phase or between two

immiscible liquids. For solids, the interface can be between the solid and

a vapor or liquid phase, between two chemically different solids, or

between crystals (grains) of the same chemical composition but differing

in crystallographic orientation. The properties of these interfaces differ

from bulk properties and they influence the properties of materials in

many ways.

An atom at a free surface of a solid has greater energy than an atom in

the interior (less tightly bonded). Surface atoms are at higher energy

levels. The sum of all excess energies (excess with respect to atoms

inside) of the surface atoms is the surface energy.

Surface energy () can be defined in terms of U, H, A or G depending on

the physical conditions:

U

A

H

A

A

A

G

AS V S P T V T P, , , ,

The most commonly used one is based on G. Then

dG = V dP - S dT + dA

There are several commonly observed manifestations of this surface

energy. As a liquid droplet tries to minimize its energy, it assumes a

spherical shape- the shape that minimizes the surface area. Smaller

droplets tends to agglomerate into larger droplets. In sintering small

particles reduce surface to volume ratios.

2. Surface Tension: The Mechanical Analog of Surface Energy

A film of liquid pulled by a force F, thereby creating new surface (two

surfaces top and bottom), the balance of forces and the work done can be

expressed as

2

l

dx

F

2 l dx = F dx

F

l2

The units of surface tension, , are force/length. By multiplying both the

numerator and denominator of by a length term, the dimensions become

energy/area. Thus , can be thought of as surface tension or surface

energy.

3. Approximate Calculation of Solid Surface Energies Assuming binding energy of an atom to a solid is the result of discrete bonds to its

nearest neighbors, then energy of one bond, ,

H

z N

s

A0 5.

Hs: molar enthalpy of sublimation (breaking all the bonds)

z: coordination number

NA: Avagadro's number

There are 0.5 z NA bonds per mole.

If we cleave an FCC crystal along (111) plane, 3 bonds per atom will be broken.

Because there are two surfaces formed, the work required to form the surfaces will be

3/2 per surface atom. Then the work per surface atom is

3

WH

Nfor FCC zS

A

3

2 412

( )

The surface energy is then

H

N

N

A

S

A4

where (N/A) is the number of atoms per unit area. For FCC, along (111) plane

N/A=4/( 3 02a ). Where ao is the lattice parameter.

For a (100) plane there are 4 broken bonds per atom, and the number of atoms per unit

area is 2/ a02

(a) (b) Atomic packing (a) on the (111) plane, and (b) on the (100) plane of an FCC crystal.

4

For copper Hs = 170000 J/mole and ao = 3.615 A.

Using (100) plane, = 1440 ergs/cm2

Experimental results indicate that this value is about 1600 ergs/cm2.

This shows that computation gives an inaccurate but reasonable values. As expected a

simple model of this kind is useful in determining approximate values and visulizing

the physical phonemanon of the of the system. However, for accurate determinations

other changes that takes place in the vicinity of the surface has to be taken into

account.

4. Effect of Surface Curvature In physical systems surface between two phases is not infinitely thin, geometrical

plane. They extend to sevral atomic layers. In a system with only one component, C, a

geometrical surface can be located by assigning no mass to the surface. That is the

shaded area on both sides of the surface are set equal.

[C]

Phase I Phase II

Distance

Surface

A B

Surface between two phases: I and II, S. If this surface moves between A and B, the

volumes of the phases will change: phase I increases, phase II decreases.

Schematically

phase I

A

B

S

phase II The total change of Gibbs energy

dG = GI dnI + GII dnII + dA = Wrev

5

dnI and dnII are changes in the masses as infinitesimal movement of surface takes

place. At equilibrium Wrev is zero.

For one component system, G = = Go

I dnI + II dnII + dA = 0

dnI = -dnII = dn

Then, II - I = dA

dn

If the surface is flat (infinite radius of curvature), then dA/dn is zero, because the area

of the interface does not change as mass moves from phase I to phase II. Thus, I = II,

this is sometimes written as

I = II =

chemical potential of a material with an infinite radius of curvature.

For a curved surface, the area changes as the boundary moves. Introducing the molar

volume

VdV

dn

II - I = dA

dn = V

dA

dV

The ratio dA/dV is not zero for a curved surface.

For a spherical surface

phase I

phase II

V = r3/3 and A = r2

where is the solid angle subtended by the portion of the surface being considered:

dV = r2 dr and dA = 2 r dr

dA/dV = 2/r

Therefore, II - I = V 2

r

where V molar volume and is the interfacial (surface) energy between phases I and

II.

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Compare two conditions for the same material;

I-flat surface,

II- surface with finite radius of curvature.

Then, the diffrence in chemical potentials of flat surface, , and the surface with

finite radius of curvature, r is given by

r - = V 2

r

Thus, chemical potential of a material with a finite radius of curvature is greater than

the chemical potential for that same material with an infinite radius of curvature. This

difference in manifests itself in several ways.

For nonspherical, curved surface with principal radii of curvatures r1 and r2,

r r1 2, = V (1 1

1 2r r )

Equation that shows the difference between the chemical potentials of flat surface, and

the surface with finite radius of curvature can also be derived using mechanical

argument combined with thermodynamics. If we think the surface as a membrane

surrounding the condensed phase, and the surface tension as the stress in the

membrane, we can establish the pressure difference accross the curved surface

through a force balance.

Pi

Po

The force tending to push the two halves of the sphere apart is the product of the

cross-sectional area at the mid point ( r2) and the difference in pressure between

inside and outside of the sphere (Pi - Po = P). This force ( r2 P) is balanced by the

surface tension force (2 r ). Thus

r2 P = 2 r

Then, P = 2

r

This increased pressure on the condensed phase translates into a difference in

chemical potential.

d = V dP (at constant T)

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Integration between two limits yields above relationship

r - = V 2

r

Compute the pressure in a one micron diameter water droplet formed in supercooled

water vapor at 80oC. The vapor pressure of water at 80oC is 0.52 atm. The surface

energy of water may be taken as 80 ergs/cm2.

P = 2 2 80 10 10

0 5 10320000

27 4 2

2

6 2

r mPa

ergs

cm

Jerg

cm

m N

m

.( )

P = 320000

1013333 16

Paatm

atmPa

. .

Pi = 3.16 + Po = 3.16 + 0.52 = 3.68 atm.

4.1. Effect of Surface Curvature on Vapor Pressure

A condensed material (solid or liquid) is in equilibrium with its vapor. At equilibrium,

chemical potential of the condensed phase, c, is equal to that of its vapor, v. This is

valid for both flat (a) and curved (b) surfaces.

L

L

L

L

LL

L

V

L

V

(a) (b)

c, = v,

c,r = v,r

c,r - c, = V 2

r

From chemical potentials

c,r - c, = v,r - v, = R T ln P

P

ro

o

8

where Pro is the vapor pressure of a particle of radius r. Po

is the equilibrium vapor

pressure over a flat surface. Therefore,

ln P

P

ro

o

= V

RT r

2

Calculate and plot as function of radius the vapor pressure in a system with liquid zinc

droplets suspended in zinc vapor at 900 K.The vapor pressure of zinc over bulk liquid

at this temperature is 2.5x10-2 atm; the molar volume of the liquid is 9.5 cc/mol and

= 380 ergs/cm2.

ln P

P

ro

o

= V

RT r

2 =

9 5

8 314 900

2 38010

9 6 103

2 78.

. ( )

.

( )

cmmol

Jmol K

ergs

cm Jerg

K r cm r cm

P atm ero r cm( ) .

.

( )

2 5 10 2

9 6 10 8

r = 10-3 cm; Pro = 2.5002x10

-2 atm

r = 10-4 cm; Pro = 2.5024x10

-2 atm

r = 10-5 cm; Pro = 2.524x10

-2 atm

r = 10-6 cm; Pro = 2.75x10

-2 atm

Pro x102 (atm)

2.75

2.7

2.65

2.6

2.55

2.5

2.45

0.01 0.1 1 10 rx104 (cm)

4.2. Phase Boundary Shift in Unary Systems

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Consider a unary two phase (+) system in which the curvature of the - boundary

plays a role in the conditions for equilibrium.

T = T

P = 2

r; P = P +

2

r

=

d = V dP - S dT

d = V dP - S dT = V d(P + 2

r) - S dT

V dP - S dT d = V d(P + 2

r) - S dT

rearrangement yields,

(S - S) dT - (V - V) dP - 2 V dr

r 2 = 0

By seting S - S = Str(/) and V - V = Vtr(/), above equation can be

Str(/) dT - Vtr(/) dP - 2 V dr

r 2 = 0

Integrating between two limits (radius of curvature) gives the result of curvature on

phase boundary shift due to curvature. The first two terms are identical with the

Clausius-Clapeyron equation in which variations due to geometric effects are not

considered; the third term contains the effect of curvature on the equilibrium

conditions.

4.3. Solubility of Small Particles

To simplify the problem, consider an A-B system with a solubility line as shown

(assuming vanishingly small solubility for A in B).

T

XB

XB

aB

1

0

T1

at T1 the composition of the phase in equilibrium with the phase is XB,. Hence

aB = 1 at XB = XB,. phase is taken as pure B, for simplification.

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If material B exists as small spherical particles, the chemical potential of B as a

function of its radius of curvature is

B,r - B, = V 2

r

The surface energy, - is the interfacial energy between pure B and .

B,r - B, = R T (ln aB,r - ln aB,)

If B obeys Henry's law in phase, then aB, = Bo XB, and aB,r = B

o XB,r

Combination of relationships yields, B,r - B, = R T ln X

X

B r

B

,

,

Then, ln X

X

B r

B

,

,

= V

RT r

2

where XB,r is the solubility of B in A () when radius of curvature for B particles is r,

XB, is solubility when B has an infinite radius of curvature (flat surface). The

solubility of B increases as its particle size decreases. If B exists as small spheres and

large spheres, the solubility of B will vary with particle size. In the material

immediately surrounding the small particles, the concentration of B will be higher. If

sample is held at high temperature where B atoms are mobile, the small particles will

tend to dissolve and the larger particles will grow. This is coarsening.

4.4. Melting Temperature of Small Particles

To examine the effect of curvature upon solid-liquid equilibria, compare the

solid/liquid equlibria for (a) infinite and (b) finite radius of curvature.

S

S

S

S

SS

S

L

S

L

(a) (b)

When a particle of solid in equilibrium with its liquid.

Ts = Tl

s = l

Pressure difference in solid and liquid are related by; P = 2 l s

r

Ps = Pl + 2 l s

r

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dl = Vl dPl - Sl dT

ds = Vs dPs - Ss dT = Vs d(Pl + 2 l s

r

) - Ss dT

at equilibrium, s = l, hence ds = dl

Vl dPl - Sl dT = Vs d(Pl + 2 l s

r

) - Ss dT

rearrangement yields,

(Sl - Ss) dT - (Vl - Vs) dPl - 2 Vs l-s dr

r 2 = 0

Assuming that we are not concerned with overall pressure changes on the liquid (dPl)

and Sl - Ss = Sm

Sm dT = 2 Vs l-s dr

r 2

Integrating from the melting temperature of a large particle (r=, T, radius of

curvature) to melting point of a particle with radius r (r, Tr). Assuming Sm constant,

Sm dT = 2 Vs l-s dr

r 2

Sm T = -2 Vs l-s 1

r

Then, T = 2V

S

l s

m

1

r

When a liquid is cooled below its bulk melting point the solid phase forms and grows

as dendrites. The scale of the resulting microstructure, indicating whether it is fine

grained or a coarse structure and the rate at which it solidifies, is strongly influenced

by the size of tips as they grow. The tip radius is in turn determined by the

temperature difference between the liquid adjacent to tip and that of the supercooled

liquid surrounding it. Calculate this temperature difference for a silicon dendrite with

a tip radius of 0.1 microns growing into surrounding liquid that is undercooled 5oC

below the bulk melting point. The surface energy of the solid-liquid interface may be

taken as 150 ergs/cm2.

Assuming the liquid and solid at the tip of the dendrite are in local equilibrium. The

temperature of the liquid at the interface can be calculated from

T = ergJ

molJ

cm

ergsmol

cm

m

sl

cm

K

rS

V 7

410

101.0

1

1683

46500

1502.11212 2

3

T = -1.22 K

Ti = 1683 - 1.22 = 1681.8 K

Tl = 1683 - 5 = 1678 K

5. The Equilibrium Shape of Crystals

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If surface energies of a condensed phase is isotropic, as in liquid, then based on

common exprience the condensed phase will assume a spherical shape at equilibrium.

If surface energy varies with crystallographic orientation, as in a solid, it is expected

that the equilibrium shape to be determined by the relative surface energies of the

crystallographic planes. The presence of low surface energy planes would be prefered

and the equlibrium shape would not be a sphere.

The equlibrium shape of a crystal is determined by minimizing its surface free energy.

If a crystal may be found by planes 1, 2, 3, ... whose exposed area will be A1, A2, A3,

..., the equilibrium shape will be the one that minimizes i Ai. A graphical method

for doing this uses a Wulff plot.

In this method, the surface energy of a crystal is ploted in polar coordinates. Planes are

drawn at each point normal to the radius. The equilibrium shape will be defined by the

interior surface of the planes erected perpendicular to the polar plot.

The development of facetes in two dimensions can be illustrated by a simple example.

Consider the two-dimensional cubic crystal shown

(110)

(100) (100)

a

c

a - 2 /2 c

When looking in (100) direction, planes (110) and (100) constitutes the exterior of the

surface. a and c dimensions of the crystal will be calculating by minimizing the

13

surface energy Es at constant crystal area, A. Crystal is symmetrial in the vertical and

horizontal directions; hence only the upper right quadrant may be isolated for analysis.

Es = 2 a c2

2 o + c 1

A = a2 - c2

4

The surface energy is minimized by setting the differential of Es with respect to c

equal to zero at costant A. The result is

c

a 2 2 1

0

the ratio c/a is a measure of the relative amounts of exposed (110) and (100) planes. If

the surface energies of planes are equal, o = 1, then

c = 2 a c2

2

and the size of the two facets will be equal.