Review of Thermodynamics in Phase Transformations in metals and Alloys

In this first chapter, there are many lectures covered it
we start from a brief review of thermodynamics necessary for understanding of phase diagrams. We will then apply the thermodynamic concepts to the analysis of phase equilibria and phase transformations in one-component and multi-component systems. We will learn how to read and analyze phase diagrams of real materials and how to construct phase diagrams from thermodynamic data.
Effects of pressure:
the effect of pressure on equilibrium temperature for pure iron shown in fig(1) bellow:



Fig (1): The effect of pressure on equilibrium temperature for pure iron.
From fig (1):
with increasing pressure has effect of decreasing the ?/? equilibrium temperature and raising Tm
At very high pressure hcp ?-Fe become stable
The reason for these changes are:
(?G/?P )T = V
The eq. which give the change in temperature required to maintain equilibrium between two phases if pressure increased by dP called Clausius-Clapeyron equation:
(?P/?T)eq. = ?H/T?V
Since close - packet ? -Fe has smaller volume than ? -Fe
?V= V_m^? - V_m^? < 0
? H= H^? -H^? >0
So that (?P/?T)eq. is negative P lower the equilibrium transition temperature
?/L equilibrium transition temperature raised with increasing P due to larger molar volume of liquid phase.
It can be seen that the effect of increasing pressure is to increase the area of the phase diagram over which the phase with the smallest molar volume is stable (?- Fe) as shown in fig (1)

The driving force for solidification
If a liquid metal is under cooled by ?T below Tm before it solidifies, solidification will be accompanied by adecrease in free energy (?G)
This ?G is driving force for solidification.
The G for solid and liquid at T given :
G^L = H^L - T S^L
G^S = H^S - T S^S
Therefore at temperature T
?G^ = ??H?^ - T ??S?^
At Tm ? G?^L = G^S i.e. ?G^=0
So:
?G^ = ??H?^ - Tm ??S?^= 0 therfore :
??S?^ =?H/Tm = L/Tm
??G?^ ?L- T L/Tm
i.e. for small ?T
??G?^ ?L?T/Tm
H.W: The molar volumes of solid and liquid lead under the atmospheric pressure are: 18.92 cm3 and 19.47 cm3 respectively ,Tm=600 k , molar heat of melting ?Hm =4810 J. calculate the pressure which must be applied to lead in order to increase its melting temperature by 20 °C.

Binary solutions
In alloys : composition is also a variable with P&T
Gibbs free energy of binary solutions
imagine that 1mol of homogeneous solid solution is made by mixing XA mol of A and XB mol of B respectively in the alloy.
Because there are a total of 1 mole of solution so:
XA + XB =1
In order to calculate the free energy of the alloy, the mixing can be made in two steps:
1- bring together XA mol of pure A and XB mol of pure B.
2. allow the A and B atoms to mix together to make a homogenous solid solution.
The change in Gibbs free energy caused by the mixing (?Gmix) is giving by following eq:
?Gmix = ?Hmix – T?Smix
?Hmix : heat absorbed or evolved during step 2
?Smix : is the differences in entropy between the
mixed and unmixed state





Ideal solution
When ?Hmix = 0 the resultant solution is said to be ideal solution
In ideal solution the free energy change on mixing is only due to the change in entropy:
?Gmix = – T?Smix
?Smix = - R (XA ln XA +XB ln XB)
The actual free energy of the solution G will also depend on GA & GB:
G= G2 = XAGA + XBGB +RT (XA ln XA +XB ln XB)








Fig(5):free energy for mixing for ideal solution Fig (6): molar free energy for mixing for ideal solution combination of fig (4) & fig(5)







Regular solution
In practice ?Hmix ? 0 and usually mixing is endothermic (heat absorbed) or exothermic (heat evolved)
Quasi-chemical approach: in this approach ?Hmix is only due to bond energies between adjacent atoms.
The simplest way to consider the energy (heat) produced by mixing (regular solution model) as follow :
??H?_mix^AB = ?XA XB
? : Regular solution interaction parameter
?Gmix = ? XA XB + RT (XA ln XA +XB ln XB)
?Hmix – T ?Smix
The actual free energy of alloy depend on values chosen for GA & GB so the total free energy is:
G= G2 = XAGA + XBGB + ? XA XB +RT (XA ln XA +XB ln XB)


Intermediate faces
Intermediate faces: it is faces which does not have the same crystal structure as either of the pure components.
Intermediate faces are often based on ideal atoms ratio that results in a minimum Gibbs free energy.
When small composition deviation cause a rapid rise in G the phase referred to intermetallic compound as show in fig (a)
in other structures the fluctuations in composition can be tolerated by some atoms occupying “wrong” position or by atom sites being left vacant, and in these cases the curvature of G curve is much less fig (b).

Binary solutions with unlimited solubility
the simplest case when binary phase diagram in which A and B component are mutually soluble in any amount in both S and L phases and both are ideal solution


















Fig (8): the derivation of asimple phase diagram from the free energy curves for the L&S

T1 is above equilibrium melting temperature of both pure components fig(8-a):
T1> Tm(A) > Tm(B)
Decreasing temperature below T1 will have two effects:
G_A^L and G_B^L more rapidly than G_A^S and G_B^S
The curvature of G(XB) will decrease
At Tm(A) the G_A^L=G_A^S fig (8-b)
At T2 the Gibbs free energy curves for S and L phases will cross fig(8-c)
The common tangent construction can be used to show that for composition near cross-over of GS and GL the total Gibbs free energy can be minimized by separation into two phases.
Between T2 and Tm(B) GL continues to increase faster than GS so that the points b and c fig(8-c) will both move to right
Tm(B ) G_B^L=G_B^S so b and c will meet at single point d fig(8-d)
Below Tm(B ) the GS < GL and all the alloy are stable in single phase (solid)


Binary solutions with ?Hmix > 0 ( Intermediate faces)
If attraction between unlike atoms is very strong the ordered phases may be extend to liquid

Fig (9):
Phases that are separated from composition extremes (0% and 100%) are called intermediate phases .they can have crystal structure different from structure of component A & B
Examples of intermediate S.S phases : in Cu – Zn , ?,? are terminal S.S , ?,??`?^ ? ? ? are intermediate solid solution.