# opening valve separating two volatile solutions [closed]

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.

• @Poutnik 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. – Ginger bread Sep 26 '20 at 6:44
• Remember you consider equilibrium, not the initial state. There are onviously 2 cases, for miscible and non miscible liquids. – Poutnik Sep 26 '20 at 6:46
• @Poutnik yes, I am thinking about equilibrium only – Ginger bread Sep 26 '20 at 6:47
• So there is nothing to solve, you already know all, what is needed, just write the expression for the final pressure. – Poutnik Sep 26 '20 at 7:01

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.