At 300 K, the vapour pressure of toluene and | vedclass

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(D) Given: $P_{T}^0 = 3.63 \ kPa$, $P_{B}^0 = 9.7 \ kPa$, $X_{T} = 0.4$. Since $X_{T} + X_{B} = 1$, the mole fraction of benzene is $X_{B} = 1 - 0.4 =

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$0.20$

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(D) Given: $P_{T}^0 = 3.63 \ kPa$, $P_{B}^0 = 9.7 \ kPa$, $X_{T} = 0.4$. Since $X_{T} + X_{B} = 1$, the mole fraction of benzene is $X_{B} = 1 - 0.4 = 0.6$. The total vapour pressure of the solution is given by $P_{total} = P_{T}^0 X_{T} + P_{B}^0 X_{B}$. $P_{total} = (3.63 	imes 0.4) + (9.7 	imes 0.6) = 1.452 + 5.82 = 7.272 \ kPa$. The composition of toluene in the vapour phase $(y_{T})$ is calculated using Dalton's Law: $y_{T} = \frac{P_{T}}{P_{total}} = \frac{P_{T}^0 X_{T}}{P_{total}}$. $y_{T} = \frac{1.452}{7.272} \approx 0.1996 \approx 0.20$.

At $300 \ K$, the vapour pressure of toluene and benzene are $3.63 \ kPa$ and $9.7 \ kPa$ respectively. What is the composition of vapour in equilibrium with the solution containing $0.4$ mole fraction of toluene? (Assume the solution is ideal)

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