Molar free energy plot

At the initial compositions 1 and 2 of $α$ and $β$ respectively the chemical potentials of A and B atoms in each phase can be found by extrapolation of the tangents to the free energy curves at 1 and 2 to the corresponding sides of the free energy diagram, as shown above.

All atoms diffuse so as to reduce their chemical potential. Therefore, A atoms will have a tendency to diffuse from $α$ to $β$ $(μ_\ce{A}^α > μ_\ce{A}^β)$ and B atoms will have a tendency to diffuse from $β$ to $α$ $(μ_\ce{B}^β > μ_\ce{B}^α)$

The point marked 1 and 2 are basically the compositions of two metal alloys that have been welded. Now annealing is performed at a particular temperature $T_1$ to bring about the change now why will A atoms diffuse from $\beta$ phase to $\alpha$ phase and B atoms from $\alpha$ phase to $\beta$ phase?

How will drawing a common tangent be different in interpretation than drawing two individual tangents at compositions 1 and 2?

  • $\begingroup$ Maybe i’m missing something, but I don’t see the combined molar fee energy curve on your diagram. $\endgroup$ Jul 19 '19 at 15:16
  • $\begingroup$ @ChetMiller This is all that I know. $\endgroup$
    – user586228
    Jul 19 '19 at 15:18
  • $\begingroup$ @ChetMiller - the dashed line is the common tangent for equilibrium between the phases. OP - the common tangent line, and the tie to chemical potential, as well as the drive to have the same chemical potential for the components in the phases, is likely discussed in your text. If not, find a copy of Porter and Easterling in the library and read the first chapter or two. $\endgroup$
    – Jon Custer
    Jul 19 '19 at 17:20
  • $\begingroup$ @Jon Custer I'm not the OP. But, the part that threw me a little is that each phase has its own total molar free energy curve in the figure. The two curves have a common tangent at the equilibrium concentration combination. I'm more used to seeing a single total molar free energy curve for the two species, with 2 minima in the curve, and a common tangent to the two minimum portions of the curves. The rest of the explanation makes sense to me. $\endgroup$ Jul 19 '19 at 17:28
  • $\begingroup$ @ChetMiller Ok ignore what you do not know explain the rest to me. $\endgroup$
    – user586228
    Jul 19 '19 at 17:30

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