Thermodynamics of surface and interfaces

Define :

Consider to be a force / unit length of surface perimeter.

(fluid systems)

If a portion of the perimeter moves an infinitesimal of distance in the plane

of the surface of area A, the area change dA is a product of that portion of

perimeter and the length moved.

dAdNpdVTdSdU ii

i

Work term - dA; force x distance, and appears in the combined 1st

and 2nd laws of thermodynamics as

For a system containing a plane surface this equation can be readily integrated

:

i ii

U TS PV N A

Strictly speaking , is defined as the change in internal energy when

the area is reversibly increased at constant S, V and Ni (i.e., closed sys-

tem).

i

ii NPVTSU

A 1

where U – TS + PV is the Gibbs free energy of the system, i.e., the actual

energy of the system

and rearranging for yields.

And ii

i N is the Gibbs free energy of the materials comprising the

system, i.e., the energy of the system as if it were uni-

form ignoring any variations associated with the surface

def Surface Excess Quantities

Macroscopic extensive properties of an interface separating bulk phases

are defined as a surface excess.

Thus is an excess free energy due to the presence of the surface.

There is a hypothetical 2D “dividing surface” defined for which the pa-

rameters of the bulk phases change discontinuously at the dividing sur-

face.

def The excess is defined as the difference between the actual value of the

extensive quantity in the system and that which would have been

present in the same volume if the phases were homogeneous right up

to the “ Dividing Surface ” i.e.,

xxxx totals

The real value of x in

the system

The values of x in the homogeneous

and phases

Concept of the Gibbs Dividing Surface

For a 1 component system the position of the dividing surface is chosen such

that the two shared areas in the figure are equal. This yields a consistent value

(equal to zero ) for the surface excess.

Extensive property

Density

Distance perpendicular to the surface

For a multicomponent system the position of the dividing surface that makes

some Ni equal to zero will be unlikely to make all the other Nj ≠i = 0.

By convention, N1, the surface excess of the component present in the largest

amount (i.e., the solvent) is made zero by appropriate choice of dividing sur-

face.

Alternatively if we consider a large homogeneous crystalline body containing

N atoms surrounded by plane surfaces then if U0 and S0 are the energy and en-

tropy / per atom, the surface energy per unit area Us is defined by

0 0 :s sU NU AU NU A U

where U is the total energy of the system.

Similarly

sASNSS 0

Consider once again the combined form of 1st and 2nd laws including the sur-

face work term.

dAdNpdVTdSdU ii

i

Substitution of the definition of G leads to

dAdNVdpSdTdG ii

i

If the surface is reversibly created in a closed system (Ni fixed) at constant T

and P.

iNPTA

G

,,

is always the free energy change appropriate to the constraints imposed

on the system.

Since for the bulk phases a and b the surface terms vanish, the combined

1st and 2nd law take the form

i ii

dU TdS pdV dN and

i ii

dU TdS pdV dN and for the total system

i ii

dU TdS pdV dN dA From the definition of surface excess:

sX X X X

0

ss s si i

i

dU TdS dNpdV dA

By Def.

Integration yields,

s si i

i

U TS N A Forming the Gibbs-Duhem relation :

AddNdTS ii

is 0

so

i ii

d sdT d Gibbs-Adsorption Equation

wherei; i

ss NSs

A A

Solid and liquid Surfaces

In a nn pair potential model of a solid, the surface free energy can be thought

of as the energy/ unit -area associated with bond breaking. :

work/ unit area to create new surface = 2A

n

Then letting A = a2 where a lattice spacing 22a

where n/A is the # of broken bonds / unit-area and the is the energy per

bond i.e., the well depth in the pair-potential.

pair potential

r

U(r)

AddAdU

and dU df A

dA dA

If the solid is sketched such that

the surface area is altered

the energy d

dAAA

daaa

The total energy of the surface is changed by an amount..surfS AU

Surface Stress and Surface Energy

Shuttleworth cycle relating surface stress, f and surface energy, g.

fxx

W1=2g

Split

StretchW2w1

The difference in the work per unit area required for the constrained stretching (fix dimension in the y direction while stretching along the x-direction) is defined as the surface stress, fxx. This is the excess work owing to the presence of the surfaces.

w2=2(g+dg)(1+dx)1+dx

fxx

Unit Cube

1

Surface Stress and Surface Energy

Relation between fij and g

Consider 2 paths to get to the same final state of the deformed halves.

Path I - The cube is first stretched

and then separated.

WI = w1 + w2

= w1 + 2(1+dx) ( + g Dg) = w1 + 2 + 2 + g Dg 2 g dxwhere xx (= dx/1) has caused a change Dg in g.

Path II - The cube is first separated and then stretched.

WII = W1 + W2

= 2 + g W2

Since WI = WII, w1 + 2 + 2 + g Dg 2 g xx = 2 + g W2

work/unit area = (W2 - w1)/2xx = fxx = + D /Dxx

Surface stress, surface free energy and chemical equilibrium of small crystals

Recall that for finite-size liquid drops in equilibrium with the vapor.

(see condensation discussion)

0

2l lV

r

Equil. cond.

where Vl is the molar vol. of the liquid.

For a finite-size solid of radius r the internal pressure is a function of the

size owing to the surface stress {isotropic surface stress}.

ss Vr

f20

The pressure difference between the finite-size solid in equil. with the liquid is

2s l

fP P

r

Consider the equilibrium between a solid sphere and a fluid containing the dissolved solid.

r

( ) ( ) ( )

1 1 1 12 2

+

ii

s solid l liquid surface s s l l

N Ns s l l s s l l

i i i ii i

dU TdS pdV dN dA

dU TdS TdS TdS p dV p dV pdV

dN dN dN dN dA

The total energy of the system is given by

=0

Gibbs dividing surface set for component 1, other components are not allowed to cause area changes.

( ) ( ) ( )

1 1 1 1 +

s solid l liquid surface s s l l

s s l l

dU TdS TdS TdS p dV p dV

dN dN dA

Consider the variation dU = 0 under the indicated constraints,

( ) ( ) ( )

;

1 1 1 1 1

0;

0,

0

0 0;

s solid l liquid surface

S l S l

S l S l

dU

dS dS dS

dV dV dV dV dV

dN dN dN dN dN

Making the substitutions

1 1 1( ) ( ) 0s l ss l sdU p p dV dN dA

and for a sphere, gdA = (2g/r)dVs

1 1 1

20 ( ) ( )s l s

s l s sp p dV dN dVr

1 1 since / / the molar volumes ss s odV dN V N V

1 1

2( )s l

s l op p Vr

2Also since, S l

fp p

r

1 1

2( )s lo

fV

r

Now consider an N component solid of which components 1, ….. k are

substitutional and k +1, …. N are interstitial.

Note that the addition or removal of interstitial atoms leaves AL unchanged.

Then s

N

issisil VrPTSUN

1

/2

s

N

issisis VrfPTSUN

1

/2

and

rVfN s

N

iisilis /)(2)(

1

For interstitial exchange : fluid --interstitial--- solid

0LdA

Nkiilis ,.......1 , ⓐ

For substitutional exchange : fluid -- substitutional --- solid

kirVfN s

k

iisilis ....1 ,/)(2)(

1

and defining 01

NNk

iis

and 0/ NVs as 0V the molar volume.

kirVfilis ....1 ,/)(2 0 ⓑ

Examples of how finite – size effects alter equilibria

(1) Vapor pressure of a single – component solid

( ) ln /l e ep RT P P

0 0

0 in comparision to f term

( ) 2 / ( )s e lp V f r V P P

using ⓑ

rVfppRTrfV lls /)(2/ln/2 00

rVppRT l / 2/ln 0

same result as earlier

( ) ln /il i e eC RT C C

0( ) 2 /is i eC V f r

using ⓑrVCCRT l / 2/ln 0

(3) Melting point of a single component solid :

TTST mlml )(

2( )s m s m o

fT S T T V

r

see

Clausius – Clapy-

ron

Equation

where Sl and Ss are molar entropies.

(2) Solubility of a sparingly soluble single component solid :

(4) Vapor pressure of a dilute interstitial component in a non-volatile matrix

( H in Fe….)

If the interstitial vaporizes as a molecule:

nxnx

or if it reacts with a vapor species, A, forming a compound AmXn

nm XAnXmA

using ⓑ

0 02 21,m

ml s f

fl s

m

LS S

T

V V TT T

r S S r L

when Vx is the molar volume of X in the solid.

Using ⓐrfVnppRT xl /2/ln

indicating that f determines the change in vapor pressure

The chemical potential of X in the vapor is related to the partial pressure

P of Xn or AmXn by

l

lxl p

p

n

RTp ln

and for the solid

rfVp xlxs /2