On Nov 14, 2012, at 6:25 PM, Adam Stone wrote:

> I'm going through the quaternary phase field example and having some 
> difficulty understanding the calculation of standard potentials.
:
:
> But I don't understand how you "cook" these standard potentials:
> 
> for Cj in interstitials + substitutionals + [solvent]:
> ...     Cj.standardPotential = R * T * (numerix.log(Cj.L/rhoL) 
> ...                                     - numerix.log(Cj.S/rhoS))
> 
> Since this is the difference between the solid and liquid potentials, any 
> terms common to both would have disappeared and we should be left with the 
> remaining terms unique to each phase. But if these are indeed "standard" 
> potentials for the solid and liquid, it seems like none of them should 
> include R*T*ln(C/p) terms in the first place? So I don't see how this 
> expression is obtained (or why the solid term is being subtracted from the 
> liquid term rather than vice versa). 

This is just a statement of equality of chemical potentials in equilibrium. The 
difference between the standard potentials must be equal to the log of the 
ratio of the concentrations in equilibrium (for the ideal solution considered 
in that example, anyway). We start this example with the ansatz that an 
0.3-0.4-0.2-0.1 solid is in equilibrium with a 0.4-0.3-0.1-0.2 liquid. 

The standard states of different phases are not in equilibrium; if they were, 
then vapor pressure would be a useless concept (pure bcc iron would be in 
equilibrium with 1 atm of Fe vapor under the same conditions that pure liquid 
helium was in equilibrium with 1 atm of He gas). We pick standard states 
because we need something to compare to, but they can't all have the same 
values or there would be no chemical specificity.

It is also not true that the R*T*ln(C/p) terms don't appear in the standard 
state. It's just that, as a matter of convention, we usually take the standard 
state of condensed phases to have a unit activity and of gaseous phases to be 
at 1 atm and so the logs are zero. That's only because we chose to compare 
pressures in units of atmospheres, though; you don't have to.


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