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Phase transformation Question

  1. May 4, 2010 #1
    My question is related to free energy of nucleation in heterogeneous systems and homogeneous systems.

    I've been given to prove delta_G* (Hetero.)= 27/64 * (delta_G*(Homo))

    The stuff I know is :
    Total (delta G ) = [-4/3 *Pi*r^3* delta Gv] + [4*Pi*r^2*sigma] - [Pi*r^2 * sigma]

    1st term is related to energy of volume destroyed in liquid during formation of spherical nucleus
    2nd term is surface area created by the sphere which is a positive interfacial energy
    3rd term is grain boundary area destroyed as the nucleus forms. (im not quite sure of this though)

    so total delta G = delta hetero.

    when i substitute r* =2*sigma/delta Gv

    I get delta G* het= 1/4 *delta G* homo.....

    it'd be great if anyone can let me know where I'm going wrong!!
    Thanx!!
     
  2. jcsd
  3. Jun 4, 2010 #2
    My immediate observation is that your formula for heterogeneous nucleation is partly for a spherical nucleus and partly for a hemispherical nucleus.

    You can assume that the nucleus is fully coherent with the surface on which is forming. That is a fancy way of saying the lattices in the heterogeneous nucleus and surface match so there is no surface energy associated with that boundary.

    If you rewrite your equation for a hemispherical heterogeneous nucleus, the volume is
    2/3(pi)(r)^3

    The surface area is
    2(pi)(r)^2

    The area (pi)(r)^2 is the circular area of the join between the hemispherical nucleus and the surface on which it forms. When there was previously liquid in contact with it, there was an associated energy of

    (pi)(r)^2(sigma)
    that has been replaced by a boundary with no surface energy associated with it.

    Bear in mind that your radius of maximum energy will be different for homogeneous and heterogeneous nucleation.
     
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