Solution to <PERSON>'s Law for a volatile binary solution (no pun intended)
<PERSON>'s Law
In a closed container, $$P_A=P_A^\circ x_A$$ where $P_A$ is the pressure of compound in gaseous phase, and $x_A$ is the mole fraction of the liquid component A in the liquid mixture. If the liquid is alone in the container, $x_A=1$.
Problem I'm trying to solve
$A$ moles of compound $A$ and $B$ moles of compound $B$ are stored in container closed with a piston. External pressure on the piston is maintained as $P_e$. Find moles of $A$ and $B$ that will be found in the gaseous phase at equilibrium. 474
($P_A^\circ$ and $P_B^\circ$ are known values)
Attempt at solving
Let $a$ moles of A and $b$ moles of B exist in gaseous phase. This means $A-a$ and $B-b$ moles of respective compounds are leftover in the liquid phase.
Now Raoult's Law gives us
$$a \frac{RT}V = P_A^\circ \left(\frac{A-a}{A+B-a-b}\right)$$ $$b \frac{RT}V = P_B^\circ \left(\frac{B-a}{A+B-a-b}\right)$$
(Ideal gas equation used implicitly)
Also total pressure of gases inside should equal $P_e$
$$(a+b) \frac{RT}V = P_e$$
These are not ordinary linear equations, so I'm not totally sure my calculations are correct, but I'm getting the solution as
$$P_A^\circ \left(\frac Aa -1\right) = P_B^\circ \left(\frac Bb -1\right) = \frac{AP_B^\circ (P_e-P_A^\circ)+BP_A^\circ(P_e-P_B^\circ)}{(A+B)(P_A^\circ+P_B^\circ-P_e)}$$
So far so good, but it turns out that on plugging in specific values for $A,B,P_A^\circ,P_B^\circ,P_e$, $a$ and $b$ attain negative values.
Mathematically it can be proved that this happens when the rightmost expression is negative. The rightmost expression can be rewritten as
$$P_e \left[\frac {\frac{\frac{A}{A+B}}{P_A^\circ}+\frac{\frac{B}{A+B}}{P_B^\circ}-\frac{1}{P_e}} {\frac{1}{P_A^\circ}+\frac{1}{P_B^\circ}-\frac{1}{P_e}} \right]$$
Let $N$ mean numerator and $D$ mean denominator of this expression. We can prove that $N0$ if and only if $N>0$ or $D<0$ holds.
Question 1
Assuming all the calculations are correct, this seems to be a rather arbitrary mathematical condition. Can someone explain the physical significance of this? Like a more intuitive way to understand why we must check such an arbitrary condition rather than something as simple as $P_e>P_A^\circ+P_B^\circ$?
Question 2
If $\frac ND <0$, is the solution always $a=A$, $b=B$ (which means no liquid evaporates)? 565