# Why in positive deviation from Raoult's Law partial vapour pressure of a liquid component is greater than what is predicted by Raoult's Law?

When my teacher was teaching me positive deviation from Raoult's Law, he told me that in this case $$P_{solution}>P_A^°χ_{A} + P_B^°χ_{B}$$ He told that here the solute-solvent interactions are weaker than the solute-solute interactions and solvent-solvent interactions due to which more vapours can be formed from the solution and hence the observed vapour pressure is greater than the theoretical vapour pressure. I was pretty satisfied with the answer and was happy. But after that he told me that in this case, $$P_{A} > P_A^°χ_{A}$$ and $$P_{B} > P_B^°χ_{B}$$. He did not give any explanation about it. Okay, the previous explanation makes sense, but how are the above two equations possible? I mean in positive deviation more vapours are formed because solute-solvent interactions are weaker. But there is no change in the interactions between solute-solute and solvent-solvent. So, why are $$P_{A} > P_A^°χ_{A}$$ and $$P_{B} > P_B^°χ_{B}$$ true? Shouldn't these equations actually follow Raoult's Law and follow $$P_{A} =P_A^°χ_{A}$$ and $$P_{B} = P_B^°χ_{B}$$? Please someone explain as to why $$P_{A} > P_A^°χ_{A}$$ and $$P_{B} > P_B^°χ_{B}$$? I am so confused. Please help.

• Baruah I don’t think (Maybe) you actually understood what a solution is in first place, take a solution of A and B, here molecules of B are surrounded by not only B but A as well. Now consider an ideal solution of A and B components$\ce{Interaction_{B-B}=Interaction_{A-B}}$. So here vapour pressure of B is as expected according to its mole fraction in the solution. But for a positive deviation we know that interaction of A-B will be less, thus Since B is also surrounded by A molecules as well. Thus B will have more vapour pressure than expected. – Rishi May 9 at 17:49
• Also the inequality you seem to accept implies the second one, at least for one term. – Alchimista May 10 at 8:13