1

2

THE THERMODYNAMICS OF SURFACES

PVnonrev WSdTVdPdG −δ−−=

dAAGdG

PT ,

∂∂

=

PTAG

,

∂∂

=γ

Non pressure-volume work done by the system

Surface energy

3

THE SURFACE ENERGY OF ICE

• Calculate the number of bonds broken per unit area of new surface• Multiply by the energy per bond

( )( ) 322323

cmmolecules/ 100.3g/mol 18

g/cm 9.0molmolecules/ 1002.6ˆ ×=×

=ρ

( ) 2153/2 cmmolecules/ 1097.0ˆ ×=ρ=Γ

[ ]b

sub EH 4ˆ

senergy/mas =

ρρ∆

( )( )( ) ( )ergs/cal 102.4

cmmolecules/ 103.04g/cm 9.0cal/g 677 7

322

3

×⋅×

=bE

ergs 101.2 13−×=bE

Related to the number of bonds per unit areaGood number to remember

Energy per bond

4

( )surfacenewof cm 2

area bonds/unit bonds of cm 12

2bE⋅Γ

≅γ

( )( )( )surfacenewofcm2

ergs 101.2cmmolecules/ 1097.0bonds of cm 12

132152 −××≅γ

2ergs/cm 103≅γ

2exp ergs/cm 109≅γ

THE SURFACE ENERGY OF ICE (cont.) Assume breaking a 1 cm2 cube in half

SURFACE TENSION AND SURFACE ENERGY

LF τ= 2( ) xFxxF ∆=−= 12work

xL∆τ= 2work

1212 22 LxLxAA −=−

xLAA ∆=− 212

( )12work AA −τ=

( ) ( )12work AA γ−γ=

( ) ( ) ( )1212 AAAA γ−γ=−τ

( )PTA

A

,

∂γ∂

=τ

PTAA

,

∂γ∂

+γ=τ

From the definition of surface energy

5

PTAA

,

∂γ∂

+γ=τ

• Surface tension and surface energy have the same units

• They are not numerically equal in general

• The surface energy will change with area for an elastic solid

• Surface energy will not be a function of area for a perfect fluid

( )dndAVdPSdTdG Pµ+γ++−=

APT

P

nG

,,

)(

∂∂

≡µ

pTAPT nG

nG

,,,

∂∂

≠

∂∂

SURFACES AND CHEMICAL POTENTIAL

change in Gibbs Energy when we add dn moles of material to the system

Note that

PTnG

,

∂∂

≡µ

reflects the fact that one cannot add matter to a system without, in general, changing its surface area

6

Vapor pressure over a sphere

dnVdV =

3

34 rV π=

drrdV 24π=

24 rA π=

rdrdA π= 8

rdVdA 2

=

dVr

dA 2=

dnrVdA 2

=

In general, the surface area and the amount of material in an object are related.The relationship for a sphere of radius r, and n is easy to derive

( )dndAVdPSdTdG Pµ+γ++−= dnrVdA 2

sphere =

( )dndnrVVdPSdTdG Pµγ +++−=

2

( ) dnrVVdPSdTdG P

µ+

γ++−=

2

( )P

PTsphere r

VnG

µ+γ

=

∂∂

=µ2

,

Note that the µ(P) is the chemical potential when r is very large

µ(P) is identified with the chemical potential of a material when it occupies a semi infinite medium and when it has a flat (planar) surface.

r

7

rVP

sphereγ

=µ−µ2)(

o

vapO P

PRT ln=µ−µ

( )( ) r

VPP

RT PsphereP

sphereγ

==µ−µ2ln

( ) RTrV

PP

Psphere γ

=2ln

moleccV

P/P(P)

Material γ(erg/cm2 10µm 1µm 0.1µm

Water (298K) 72.8 18 1.00 1.01 1.11

Hg(298K) 480 14.7 1.01 1.06 1.77

Cu(1000K)TMp=1083°C

1670 7.12 1.00 1.03 1.33

( ) RTrV

PP

Psphere γ

=2ln

8

Effect of Drop Size (r) on Vapor pressure

Material γ(erg/cm2) Vm(cc/mole) r(µm) p/poWater 72.8 18 0.1 1.11T = 298 K 1 1.01

Mercury 480 14.7 0.1 1.77T = 298 K 1 1.06

10 1.01

Cu (Solid) 1670 7.12 0.1 1.33T = 1000 K 1 1.03Tm = 1350 K

γ

=RTr

V2exppp m

o

rdVdA 2

=

dnVdV −=

( )

RTrVP

bubbleγ

−=µ−µ2

( ) RTrV

PP

Pbubble γ

−=2ln

r

Vapor pressure within a bubble

9

21

11rr

K +=

rKsphere

2=

rKbubble

2−=

( ) RTKV

PPPK =ln

Curvature and vapor pressure

Principle radii

r1

r2

Surface topology and chemical potential

r r

r > 0 (+)

µ > µo

P > Po

r =

µ = µo

P = Po

r < 0 (-)

µ < µo

P < Po

∞

convexconcave

Convex – surface normals diverge.

γ

=RTr

V2exppp m

o rV2 m

oγ

=µ∆=µ−µ

10

Surface topology and matter transport

flat surface (r = )

valley (r < 0)

hill (r > 0)

pore (r < 0)

particle (r > 0)

∞

Hydrostatic Pressure across an Interface

• Consider sphere (radius = r) of material suspended in media of pressure Po.

P

Po

+γ=∆

γ=∆

21 r1

r1P

r2P

Young-La Place Equation

11

liquid

solid

vapor

θ

∆A

γSV

γLV

γSL

Wetting and Contact Angle

• Liquid (or solid) drop on a solid surface.• Usually, does not spontaneously “wet” (spread across) the

surface.• Instead, remains a drop with definite contact angle (θ).• Contact angle dictated by balance of interfacial energies.

LV

SLSVcosγ

γ−γ=θ

Young-Dupre Eq.

Drop

Solid surface

θ

ΔA

Contact angle

12

θ ΔA

SVγSLγ

LVγ

θγ∆+γ∆−γ∆=∆ cosLVSVSL AAAG

θγ+γ−γ= cos0 LVSVSL

LV

SLSV

γγ−γ

=θcos

At equilibrium ∆G = 0

LV

SLSV

γγ−γ

=θcos

1≥γ

γ−γ

LV

SLSV

1−≤γ

γ−γ

LV

SLSV

Completely wetting (θ = 0)

Nonwetting (θ = 180º)

13

Pb on Cu{111}Pb(Ni) on C(graphite)

Au on SiC

Acta mater. 48 (2000) 4439–4447THE EFFECTS OF INTERFACIAL SEGREGATION ON WETTING

IN SOLID METAL-ON-METAL AND METAL-ON-CERAMICSYSTEMS

P. WYNBLATT*Department of Materials Science and Engineering, Carnegie Mellon University,

Pittsburgh, PA 15213,USA

Work of adhesion

cermetmetceradW −γ−γ+γ=

metal

ceramic

Definition of contact angle θγ+γ=γ − cosmetcermetcer

( )θ+γ= cos1metadW

14

Purified Ar

As-received Ar

Air

Molten Cu on BaTiO3 at 1100C

Wettability of electrode metals on barium titanate substrate, J. Mat. Sci. 36(2001) 825

Liquid metals on Barium Titanate

15

Capillarity (Capillary Rise)

• Capillary tube submerged into a liquid. If liquid, at least, partially wets the capillary (θ < 90), then liquid will rise up the capillary.

hPoθ

R

r

θ

0h0cos900h0cos90

<<θ>θ>>θ<θ

o

o Capillary riseCapillary depression

rcos2gh θγ

=ρ∆

Capillarity (Liquid Phase Sintering)

• Wetting liquid phase between two solid particles.• Pressure difference tends to “pull” the particles together.

δr

liquid R

Po

x

Po

o

o

PP

rPPP

r1

x1P

<

γ−≅−=∆

−γ=∆

Since pressure pushing spheres together (Po) is greater than that in the liquid (P), spheres will be pulled closer together.