1. Thermodynamic Potentials
Combined 1st and 2nd Law
dU = TdS - pdV
Enthalpy is a function of S and p
Potential Function in terms of S and p, Enthalpy
Lengendre Transform  subtract a -d(pV) term
from dU
dU + d(pV) = TdS - pdV + d(pV)
d(U + pV) = TdS + Vdp
where H = (U + pV) is the Enthalpy
and H = H (S, p)
Potential Function in terms of T and p,
Gibbs Free Energy
Lengendre Transform  add a -d(TS) term
to dH
d ( H-TS ) = Vdp - SdT
where G = (H-TS) is the Gibbs Free Energy
and G = G (p, T)

2. Four Fundamental Thermodynamic Potentials
Potential Function in terms of T and V,
Helmholtz Free Energy
Four Fundamental Thermodynamic Potentials
dU = TdS - pdV
dH = TdS + Vdp
dG = Vdp - SdT
dA = -pdV - SdT
The appropriate thermodynamic potential
to use is determined by the constraints
imposed on the system. For example,
since entropy is hard to control (adiabatic
conditions are difficult to impose) G and A
are more useful. Also in the case of solids
p is a lot easier to control than V so G is
the most useful of all potentials for solids.
Lengendre Transform 
subtract a -d(TS) term from dU
d(U-TS) = -pdV - SdT = dA
where A = A(V, T) is the
Helmholtz Free Energy
The Maxwell relations are useful in that
the relate quantities that are difficult or
impossible to measure to quantities that
can be measured.

3. The Thermodynamic Potentials
Four Fundamental Thermodynamic Potentials
Equilibrium
dU = TdS - pdV
dH = TdS + Vdp
dG = Vdp - SdT
dA = -pdV - SdT
fixed V,S
fixed S,P
fixed P,T
fixed T,V
The appropriate thermodynamic potential to use is determined by the constraints imposed on the system. For example, since entropy is hard to control (adiabatic conditions are difficult to impose) G and A are more useful. Also in the case of solids p is a lot easier to control than V so G is the most useful of all potentials for solids.

4. The Gibbs Phase Rule P  C + 2
For a system of C independent components not more
than C + 2 phases can co-exist in equilibrium.
Trivial Example: C = 1, P = 3 solid, liquid vapor
If P is less than C + 2 then C + 2 – P variables can take on arbitrary values
(degrees of freedom) without disturbing equilibrium.
def.
Thermodynamic degrees of freedom, f
Gibbs Phase Rule

5. 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.
Work term dA; force x distance, and appears in the combined 1st and 2nd laws of thermodynamics as

6. 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 system).
For a system containing a plane surface this equation can be readily integrated :
and rearranging for yields.
where U – TS + PV is the Gibbs free energy of the system, i.e., the actual
energy of the system

7. And
is the Gibbs free energy of the materials comprising the system, i.e., the energy of the system as if it were uniform ignoring any variations associated with the surface
Thus is an excess free energy due to the presence of the surface.
def
Surface Excess Quantities
Macroscopic extensive properties of an interface separating bulk phases are defined as a surface excess.

8. There is a hypothetical 2D “dividing surface” defined for which the parameters of the bulk phases change discontinuously at the dividing surface.
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.,
The real value of x in the system
The values of x in the homogeneous
and phases

9. Concept of the Gibbs Dividing Surface
Extensive property
Density
Distance perpendicular to the 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.

10. 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 surface.
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 entropy / per atom, the surface energy per unit area Us is defined by
where U is the total energy of the system.

11. Similarly
Consider once again the combined form of 1st and 2nd laws including the surface work term.
Substitution of the definition of G leads to

12. If the surface is reversibly created in a closed system (Ni fixed) at constant T and P.
is always the free energy change appropriate to the constraints imposed on the system.

13. Since for the bulk phases a and b the surface terms vanish, the combined 1st and 2nd law take the form
and
and for the total system
From the definition of surface excess:
By Def.

14. Integration yields,
Forming the Gibbs-Duhem relation : so
Gibbs-Adsorption Equation
where

15. 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 =
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.
Then letting A = a2 where a lattice spacing

16. pair potential
If the solid is sketched such that the surface area is altered
the energy r
U(r)
The total energy of the surface is changed by an amount.
and

17. Surface Stress and Surface Energy
Unit Cube 1
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.
fxx
W1=2γ
Split
Stretch W2 w1
1+dx
fxx
w2=2(γ+dγ)(1+dx)
Shuttleworth cycle relating surface stress, f and surface energy, γ.

18. 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) (γ + Δγ)
= w1 + 2 γ + 2 Δγ + 2 γ dx
where xx (= dx/1) has caused a change Δγ in γ.
Path II - The cube is first separated
and then stretched.
WII = W1 + W2
= 2 γ + W2
Since WI = WII,
w1 + 2 g + 2 Dg + 2 g xx = 2 g + W2
work/unit area = (W2 - w1)/2xx = fxx =  + D /Dxx

19. 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)
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}.

20. The pressure difference between the finite-size solid in equil
The pressure difference between the finite-size solid in equil. with the liquid is
Consider the equilibrium between a solid sphere and a fluid containing the dissolved solid. r

21. 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.

22. Consider the variation dU = 0 under the indicated constraints,
Making the substitutions
and for a sphere, gdA = (2g/r)dVs

24. Now consider an N component solid of which components 1, …
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
and

25. ⓐ ⓑ For interstitial exchange : fluid --interstitial--- solid
For substitutional exchange : fluid -- substitutional --- solid
and defining
and as
the molar volume. ⓑ

26. Examples of how finite – size effects alter equilibria
(1) Vapor pressure of a single – component solid
using ⓑ
same result as earlier

27. (2) Solubility of a sparingly soluble single component solid :
using ⓑ
(3) Melting point of a single component solid :
see
Clausius – Clapyron
Equation
where Sl and Ss are molar entropies.

28. using ⓑ
(4) Vapor pressure of a dilute interstitial component in a non-volatile matrix
( H in Fe….)
If the interstitial vaporizes as a molecule:
or if it reacts with a vapor species, A, forming a compound AmXn

29. The chemical potential of X in the vapor is related to the partial pressure P of Xn or AmXn by
and for the solid
when Vx is the molar volume of X in the solid.
Using ⓐ
indicating that f determines the change in vapor pressure