Question

When components A and B are mixed, this system has a maximum temperature azerotrope at 1-bar pressure. Suppose 2 moles of pure A is in vapor-liquid equilibrium and 0.1 moles of liquid B is added at constant pressure. What happens? Explain why.

Answer 1

When mixture solution shows negative deviation from Roult's law and forms maximum boiling azetrope (boiling point higher than either of the components)

For this case when 2 mol of A in vapor liquid equilibrium mixed with 0.1 mol of B the interaction between molecule of A and B which mean A-B bond will be stronger than A-A bond or B-B bond. As a result of strong A-B interaction total vapor pressure of resultant mixture will be less than what we calculate from Roult's law hence shows negative deviation and forms maximum boiling azeotrope. Also total volume of mixture less than sum of individual volume of A and B.