If vapour pressures of pure liquids A and B a | vedclass

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(B) Given: $P_A^{\circ} = 300 \, torr$,$P_B^{\circ} = 800 \, torr$. Mole fraction of $B$ $(x_B)$ = $0.92$,so $x_A = 1 - 0.92 = 0.08$. According to Rao

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$\Delta H_{mix} < 0$

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(B) Given: $P_A^{\circ} = 300 \, torr$,$P_B^{\circ} = 800 \, torr$. Mole fraction of $B$ $(x_B)$ = $0.92$,so $x_A = 1 - 0.92 = 0.08$. According to Raoult's Law,the expected total pressure $(P_{ideal})$ is: $P_{ideal} = P_A^{\circ} x_A + P_B^{\circ} x_B = (300 	imes 0.08) + (800 	imes 0.92) = 24 + 736 = 760 \, torr$. Since $1 \, atm = 760 \, torr$,the observed pressure is $0.95 \, atm = 0.95 	imes 760 = 722 \, torr$. Since the observed pressure $(722 \, torr)$ is less than the ideal pressure $(760 \, torr)$,the solution shows a negative deviation from Raoult's Law. For solutions showing negative deviation,$\Delta H_{mix} < 0$ and $\Delta V_{mix} < 0$.

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