Vapour pressures of pure liquids A and B are | vedclass

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(B) Calculate the theoretical vapour pressure using Raoult's law: $P_{	ext{calc}} = X_A P^{\circ}_A + X_B P^{\circ}_B = 0.1(300) + 0.9(800) = 30 + 72

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$\Delta H_{	ext{mix}} < 0$

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(B) Calculate the theoretical vapour pressure using Raoult's law: $P_{	ext{calc}} = X_A P^{\circ}_A + X_B P^{\circ}_B = 0.1(300) + 0.9(800) = 30 + 720 = 750 \ torr$. Since the observed pressure $P_{	ext{obs}} = 720 \ torr$ is less than $P_{	ext{calc}} = 750 \ torr$,the solution shows negative deviation from Raoult's law. For solutions showing negative deviation,the intermolecular forces between $A-B$ are stronger than those between $A-A$ and $B-B$. Consequently,$\Delta H_{	ext{mix}} < 0$ and $\Delta V_{	ext{mix}} < 0$.

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