Different volatilities in ideal solutions?

[SilverSoldier]

Suppose $A$ and $B$ are two liquids, and intermolecular forces between molecules of $A$ are of magnitude $f_{AA}$ and between molecules of $B$, $f_{BB}$. If $f_{AA}>f_{BB}$, then the pure liquid $A$ is volatile than $BB$, i.e., the tendencies that the molecules have to leave liquid phase to the gas phase depends on the intermolecular forces between them.

Say we mix $A$ and $B$ to make a solution. From my understanding, such a solution is said to be ideal, if the tendencies that the molecules of $A$ and $B$ have to leave the solution to the gas phase remain unchanged upon mixing.

Now certainly, once the liquids are mixed, molecules of $A$ will interact with those of $B$, and new intermolecular interactions will form between them; let us say of magnitude $f_{AB}$.

If the tendency of molecules of $A$ to leave the liquid must remain unchanged, the new interactions it forms with molecules of $B$ must be of the same magnitude as $f_{AA}$, so $f_{AA}=f_{AB}$, or the tendency should change. Similarly, for the tendency of molecules of $B$ to leave the solution to not change, $f_{BB}=f_{AB}$.

But then $f_{AA}=f_{BB}$, so the liquids must have the same volatilities. Is this so? Have I misunderstood anything?

 

[snorkack]

If the tendency of molecules of $A$ to leave the liquid must remain unchanged, the new interactions it forms with molecules of $B$ must be of the same magnitude as $f_{AA}$, so $f_{AA}=f_{AB}$, or the tendency should change. Similarly, for the tendency of molecules of $B$ to leave the solution to not change, $f_{BB}=f_{AB}$.

But then $f_{AA}=f_{BB}$, so the liquids must have the same volatilities. Is this so? Have I misunderstood anything?

Yes, you misunderstood something
The situation $f_{AA}=f_{BB}=f_{AB}$ is one option to have ideal solution, but not the only one.
Consider four molecules - two A and two B molecules.
When the pairs change, they form two pairs of AB.
Therefore, it is sufficient that
$f_{AA}+f_{BB}=2*f_{AB}$