1. Phase
Transformation by
Dr.Srimala

2. Phase Transformation Content 1.0 Introduction
2.0 Homogeneous Nucleation
2.1 Gibbs Free Energy
2.2 Energies involved in homogeneous nucleation
2.3 Critical radius & Critical free energy
3.0 Nucleation rate
4.0 Heterogeneous Nucleation
4.1 Energies Involved in heterogeneous nucleation

3. 1.0 Introduction
Phase transformation proceed by nucleation followed by its growth.
Nucleation can be divided into 2 types:
Homogeneous nucleation
Heterogeneous nucleation

4. 2.0 Homogeneous Nucleation
Homogeneous nucleation occurs when there are no special objects inside a phase which can cause nucleation.
For instance when a pure liquid metal is slowly cooled below its equilibrium freezing temperature (undercooling) to a sufficient degree numerous homogeneous nuclei are created by slow-moving atoms bonding together in a crystalline form.

5. 2.1 Gibbs Free Energy G = U - TS Where U = Internal Energy,
T = Absolute Temperature, and
S = Entropy
• Materials Scientists refer to the difference in G between the old and new phases as the “ driving force” for the phase transformation.
• The release of heat when a metal solidifies indicates that the crystalline phase has a lower Gibbs Free Energy, G, than the liquid.
• At the equilibrium freezing temperature the Gibbs Free Energy of the liquid and the crystalline phase are equal.

6. 2.2 Energies Involved in Homogeneous Nucleation
Homogeneous nucleation needs undercooling to get G for formation of new phase
Assume that the new, product phase appears as spherical particles upon solidification
(energy)
Volume free energy
surface energy 2r
Free energy released by transformation is proportional to the volume of the particle multiplied by the volumetric free energy,

7. volumetric free energy is actually the free energy change per unit volume
Note:
approximately
If a cube, volume free energy =
Surface energy =
Note: a

8. Therefore, net change in free energy per particle,
Free energy consumed by creation of interface is proportional to the surface area of particle and the interfacial energy, .
Therefore, net change in free energy per particle,
Negative, and represent
the decrease in free energy
upon solidification
involves the increase in
energy required to form
a new surface

9. Schematic diagram of the energies involved in homogeneous nucleation
Volume free-energy change
surface free-energy change
Retarding energy
Driving energy
Total free-energy change
particle radius (r)
free-energy change (G)

10. 2.3 Critical radius & Critical free energy Free Energy Change Gr
The radius at which the free energy curve is at maximum is called critical radius r* and the critical Gibbs energy corresponding to the r* is G*.
Free Energy Change Gr
Differentiate to find the stationary point (at which the rate of change of free energy turns negative). r

11. From this we find the critical radius and critical free energy.

12. Replace r* in the following equation to obtain critical free energy.

13. 3.0 Nucleation rate
Although we now know the critical values for a nuclei to become a nucleus, we do not know the rate at which nuclei will appear in a real system.
To estimate the nucleation rate we need to know the concentration of nuclei of the critical size and the rate at which such nuclei are formed.
Concentration of critical nuclei, Cr is given by a Boltzmann factor, where C0 is the number of atoms per unit volume (NA/V):
The rate of nucleation is determined by the concentration of nuclei of the critical size and the rate at which they are activated through the addition of atoms or molecules to their surfaces.
ns is number of atoms or molecules to their surfaces.
v is collision frequency of molecules with nuclei

14. In the case of transformation in condensed phase, the collision frequency at the nucleus interface expressed as follow :
vo is the jump frequency of the atoms or molecules to the surface of nucleus and ∆GM is the activation energy for the movement to the nucleus

15. Summary of nucleation rate:
Free energy of critical nucleus
Activation energy (generally taken as small)
Nucleation rate
Vibration frequency
Density (no of atom/vol)
No of atoms at nucleus
Probability factor
k =1.38 x J/K

16. 4.0 Heterogeneous Nucleation
Heterogeneous nucleation occurs when there are special objects inside a phase which can cause nucleation.
example : grain boundaries triple junctions dislocations (existing) second phase particles
 = contact angle
Liquid
Solid
Nucleating agent

17. 4.1 Energies Involved in Heterogeneous Nucleation
In the case of a heterogeneous nucleus in the form of spherical cap, the surface energy term involves the surface energy between the solid nucleus and the liquid, and the change in surface energy of the catalyst surface as it is coated by the nucleus.
 s - l
liquid
Solid
Catalyst surface r 
 c - s 
 c - l 2R

18. The surface energy term is derived as follows:
Solid-liquid surface
Catalyst-solid surface
Where is the radius of curvature of the nucleus. Then we write
Solid-liquid interfacial energy
Solid-catalyst interfacial energy
Liquid-catalyst interfacial energy

19. Liquid-catalyst interfacial energy term can also be expressed as
The volumetric Gibbs free energy change is the product of the volume of the cap and , the specific Gibbs free energy change.
Volume in terms of its radius of curvature and contact angle is :

20. Liquid-catalyst interfacial energy term can also be expressed as
Therefore,

21. Volume in terms of its radius of curvature and contact angle is :

22. G het=G hom

23. As for solidification, the radius of the spherical caps depends only on the
interfacial energy, so:
But the shape factor modifies the critical free energy:
90o
0.5
60o
0.16
30o
1.3 x 10-2
10o
7.0 x 10-4

24. Exercise 1.1
Take 108 J/m3 as a typical value of chemical (volume) free energy chage for formation of second phase particle and 1 J/m2 as surface energy change. Calculate the radius of the spherical particle for which surface energy is 1% of the volume free energy.

25. Solution 1.1

26. Show that for T~Tm, where H is the enthalpy change on solidification per unit vol.
b) For the solidification of a metal (Tm =1000 K) with undercooling of 200K, calculate the rate of homogeneous nucleation in Nuclei/m3/s. Neglect activation energy. Assume v = 1012/s and s*pd estimated as 1028/m3 , H = -1.26x109 J/m3, LS=0.16J/m2
Exercise 1.2

27. Solution 1.2
proven
(a)
(b) 8 9 3

29. Quiz 1
At one atmosphere pressure pure germanium melts at 1232 K and boils at 2980 K. The pressure at the triple point (S,L,G) is 8.4 x 10-8 atm. Estimate the heat of vaporization of germanium.