Condition for formation of azeotropes Consider a binary azeotropic solution of two volatile components $A, B$. I wanted to try and find out the required concentrations of $A, B$ in the solution in terms of their vapour pressures. I tried using Raoult's Law to do this, but the I got an unexpected answer:

Let the vapour pressure of $A, B$ at the current temperature be $P_A, P_B$, respectively. Let their concentrations be $\chi_A, \chi_B$ and their partial vapour pressures over the solution be $p_A, p_B$.

Then, by Raoult's Law, 

$$p_A = P_A\chi_A$$

$$p_B = P_B\chi_B$$

By Dalton's Law of partial pressures, the mole fractions of $A, B$ in the vapor phase are given by $\frac{p_A}{p_A + p_B}$ and $\frac{p_B}{p_A + p_B}$ respectively. Since the solution is azeotropic, these two are equal to their mole fractions in solution.

That is:

$$\frac{P_A\chi_A}{P_A\chi_A + P_B\chi_B} = \chi_A \implies \frac{P_A}{P_B}(1-\chi_A) = \chi_B\tag1$$ 

Similarly, 

$$\frac{P_B}{P_A}(1-\chi_B) = \chi_A \tag2$$

Substituting $(1)$ in $(2)$, yields

$$ \frac{P_B}{P_A}\left(1-\frac{P_A}{P_B}(1-\chi_A)\right) = \chi_A$$

$$ = \frac{P_B}{P_A}-1+\chi_A = \chi_A$$

$$\implies P_B = P_A$$

But there are numerous azeotropes where this condition is not satisfied. How come?
It is not that surprising that you got this answer: the existence of azeotropes is due to non-ideal behaviour of mixtures (see for instance the Wikipedia page and the associated diagram). It is not surprising that when you impose an ideal behaviour of the solvent that follows Raoults law, you get some unexpected results. Basically, what your calculation leads to it that for ideal solvent mixture, the only situation in which the gas phase molar ratio can be equal to the solution molar ratio is when both solvent have equal vapour pressures.