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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