Dear colleague,

1/ The reciprocal solubilities are linked to the affinity of each element (Al, Ti) in the solid solutions. It is somehow linked to the mixing enthalpies inside these solid solutions but it is also linked to the tielines and relations with the other phases of the diagram.

2/ There are 2 subscribes for TiAl3 because this phase may undergo a transition : 1st order (allotropic), or sometimes 2nd order (ordering). For further information, please read :

Phase equilibria in the Ti–Al binary system by Ohnuma et al., Acta Materialia, Volume 48, Issue 12, 17 July 2000, Pages 3113-3123.

3/ In a phase diagram, the boundary drawn with dotted lines represent transitions that are not proved experimentally. Nevertheless, these lines are accepted because they follow the logical thermodynamics behavior of phase diagrams.

Regards

I agree with prof. Stéphane, the mutual solubility between two elements, A and B, it is not necesarily the same, if that was case all the phase diagrams would be symmetrical, which generally is it not the case.

In attach another phase diagram where some boundaries are not in dotted lines.

Al has FCC crystal structure and +3 common oxidation state. While, Ti has HCP crystal structure and +4 common oxidation state. These differences make the enthalpy of mixing of Al-rich solid solution different from that of Ti-rich solid solution in Al-Ti system.

The question why can be answered on different levels. The phase boundaries are of course determined by the thermodynamics, i.e. the Gibbs-energy functions of the various phases. Al substitution by Ti is unfavourable as compared to formation of intermediate phases (here TiAl3, specifically ordered fcc superstructure), which cannot be compensated at elevated temperatures sufficiently by the configurational entropy in the terminal solid solution Al1-xTix. This trend is present for all 3d transition metals. Only when the 3d shell gets full like in Cu and Zn, you get much more solubility in Al-base solid solution, where, however, the s and p states bond to Al. If you ask why Al easy dissolves in Ti (and in other 3d metals!): This is (it is only a very qualitative answer) because the state of bonding of Al in Ti solid solution is different than in pure Al. This is not so much a question of the crystal structure. The main difference is (so much I remember) the partial molar volumes. Moreover, dissolution is very exothermic. There is a chapter about this issue in Westbrook and Fleischer: INtermetallic Compounds from the late 1990s. However, I cannot find my copy at the moment.

Just to comment the dashed lines: these are simple less-well established boundaries. If the boundaries are non-dashed in another publications, it does not mean these lines are better there. The present diagram is one of the newest!