For azeotropic mixtures, is the total vapour pressure at equilibrium equal to the partial pressure of one of the components?

I know, a ridiculous statement. But here's how I reached at it . Say we have a solution of two components liquids A and B.

By Raoults law,

$$p_A= p_A^o \times x_A \tag1$$

Also by Dalton's law,

$$p_A= P_T\times x^V_A \tag2$$

Now if the solution is azeotropic,

$$x_A= x_A^V$$

So that implies the total pressure equals the partial pressure of A.

Which is absurd. Please tell me where I am going wrong.

• What makes you assume that the total pressure equals the partial pressure of A just because the mole fractions in vapour and liquid are equal? I cannot seem to arrive at how you made that statement. – William R. Ebenezer Apr 6 at 17:36
• Both equations (1) and (2) calculate the partial pressure of component A. So they can be equated. Since the mole fractions in both phases are equal, the ther variables, total reassure and vapour pressure of A must be equal? – Fatimah Rashid Apr 6 at 17:51
• Yes, I see what you mean. But you are getting confused for actually nothing! The total pressure is equal to the vapour pressure of pure A, not the partial pressure of A over solution, so that is not a paradox at all. Nice observation though. – William R. Ebenezer Apr 6 at 18:06
• A mixture that exhibits an azeotrope does not follow Raolts law. Raolts law is only valid for ideal solutions. – Chet Miller Apr 6 at 18:45