What is the degree of freedom for two mis-cible (non-reacting) substances in vapor-liquid equilibrium forming an azeotrope?

Solution (By Examveda Team)

The given number of component $$= 2$$
Given number of phases $$= 2$$
We know, $${\text{degree of freedom (number of independent variables) }} = C - \phi + 2$$
$$\left( {{\text{assuming no reaction taking place}}} \right)$$
$$f = 2 - 2 + 2 = 2$$
And since under azeotropic conditions the composition of a component in both the phases remains same one of the independent variable becomes dependent and hence the degree of freedom reduces to 1.