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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.

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    $\begingroup$ 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. $\endgroup$ – Rishi May 9 at 17:49
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    $\begingroup$ Also the inequality you seem to accept implies the second one, at least for one term. $\endgroup$ – Alchimista May 10 at 8:13

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