opening valve separating two volatile solutions Let there be two containers A and B containing two volatile solutions (ideal) of  pure liquids P and Q  respectively. The containers are connected by a tube such that only vapours can travel through it but it is initially closed by a valve. Now the valve is opened at $t=0$.
What will be the final equilibrium pressure over the containers?

My  Approach:
I think that final pressure will be somewhere between the pressure of pure liquid P and pure liquid Q, might their average but I can't say for sure.
vapour pressure is a function of temperature so might be there is no change ,if the liquids would have been mixed inn a single container , forming ideal solution with specified mole fraction of each component then I know how to use Raoult' s law but here I don't see the situation being an equivalent one.
There is not enough information to decide, and the question is not clearly worded.
As to the wording, &quot;two volatile solutions (ideal) of pure liquids P and Q respectively,&quot; implies there is no vapor present, yet &quot;vapours can travel through&quot;. What is the volume of each volatile liquid, and the total volume of the containers, including the connecting tubes? If the whole apparatus has no place for vapor, then the pressure is determined by uncompressible liquids.
As to insufficient information, do the two chemicals react? For example, if the liquids are $\ce{HCl(aq)}$ and $\ce{NH3(aq)}$, the resulting $\ce{NH4Cl(aq)}$ would have a far lower vapor pressure than either of the initial reactants.