I'm fully aware of the various questions asked about Raoult's Law, and I'm asking this after having gone through all of them and not finding a suitable answer to my question.
I have studied, that at a certain temperature, the equilibrium vapour pressure of a liquid remains constant, irrespective of the amount of liquid taken. From this, I conclude that if I take a mixture of two liquids, the equilibrium vapour pressure should just be the sum of their individual equilibrium vapour pressures, since both liquids attain equilibrium when the partial pressures of their vapours are equal to their equilibrium vapour pressures. However, Raoult's Law states that the vapour pressure depends on the amount of each liquid in the solution.
This question almost answered my question, but I do not understand why the concentration of the liquid matters when it's in a solution. I understand that this is not due to any intermolecular forces, since the mixture in consideration is assumed to be ideal.
We had a long discussion in class about this, and in the end we concluded that the equilibrium vapour pressure would depend on the surface area of the liquid, and in a mixture the surface is not completely available to a single component of the mixture. I'm not sure if this explanation is correct, and I would like your opinions about this.
This question talks about a different explanation, but I believe it talks about solutions of solids in liquids. Is a similar explanation valid for mixtures of liquids? I can't seem to get my head around the explanation that the liquid molecules occupy a fraction of the surface which is equal to their mole fraction, which reduces the equilibrium vapour pressure of each component of the mixture.