Hi,
I know the attached Txy diagram has an error but can't figure it out. Could you please look at the diagrams and explain it to me?
Hi Arvin,
Well, there are several mistakes in these phase diagrams.
You show the simplest class of binary phase diagrams, which is called a Class or Type I phase diagram according to the Scott and van Konnenymburg classification:
R.L. Scott. P.H. van Konynenburg. Discuss. Faraday Soc.. 49 (1970) 87 -97.
Diagram 1.
This is wrong because, even though it is correctly attached to both T-axes, there cannot be phase equilibria between a pure compound and its mixture. This is due to the entropic mixture term, which makes it mathematically impossible to establish a tangent plane between a point of composition greater than zero, with an ending point of zero composition and infinite slope. Therefore, from either axis, both branches of the phase diagram must go either up or down, but it is not allowed that one goes up and the other down. So the attachment on the left of diagram 1 is erroneous.
Diagrams 2 & 3
The CP must be approached from both sides of the phase diagram with a zero slope (unless you have a critical azeotrope, but this is not the case for this system).
Additionally, in diagram 3, in a Type I phase diagram you cannot have a critical point in coexistence with one of the phases. However, this could be possible for systems involving three coexisting phases, such as Type III. For Type I, both coexisting branches must separate from the CP with a zero slope and move in the same direction, up or down.
Please have a look at figure 16 in the following publication by Privat and Jaubert:
Classification of global fluid-phase equilibrium behaviors in binary systems
Romain Privat, Jean-Noël Jaubert
Chemical Engineering Research and Design
Volume 91, Issue 10, October 2013, Pages 1807-1839
Article Classification of global fluid-phase equilibrium behaviors i...
In Figure 16 of this publication, you can find the equivalent to the three cases you are asking for:
DIagram 1 is equivalent to the P1 case.
Diagram 2 is equivalent to the P3 case.
Diagram 3 is equivalent to the P5 or P6 cases.
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