asdf1 said:
Why for systems containing several components and phases does it require that the chemical potential of ecah component must be idential in every phase?
You can mathematically derive this result using the differential gibbs free energy of each phase. I'll give an outline and you can derive the rest.
If you have n moles of chemical species (ie, [itex]n_i[/itex] moles of the [itex]i^{th}[/itex] species), the differential of the gibbs free energy, for this
single phase, open system is
[tex]d(nG) = \left( \frac{\partial nG}{\partial P} \right)_{T,n} dP + \left( \frac{\partial nG}{\partial T} \right)_{P,n} dT + \sum_i \left( \frac{\partial nG}{\partial n_i} \right)_{P,T,n_j} dn_i[/tex]
Since we define chemical potential as
[tex]\mu_i = \left( \frac{\partial nG}{\partial n_i} \right)_{P,T,n_j}[/tex]
and using maxwell's relation, the equation becomes
[tex]d(nG) = nVdP - nSdT + \sum_i \mu_i dn_i[/tex]
If you now have k phases with each phase having n species, you can think of each individual phase as an open system (cause it's free to transfer mass). So, you can apply the above equation to each phase individually.
Now here's the important part. At equilibirium P & T are same throughout the whole system. Can you see that if you look at the entire system (with all the species and phases), it's closed? This means you can say that for the entire system,
[tex]d(nG) = nVdP - nSdT[/tex]
For the
whole mass of the entire closed system.
Now, the total gibbs free energy is given by the
sum over each phase. But, this value must be equal to the the gibbs free energy of the
entire closed system. So, compare the two equations and consider pairs of phases.
Can you take it from here and get the result?