Question

Liquids A and B form ideal solution for all compositions of A and B at 25°C Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 bar, respectively. What is the vapor pressure of pure liquid B in bar?

Answer 1

Answer:

0.50 bar/

Explanation:

Apply Roults Law

Pt = Xa*Pa + Xb*Pb

Case 1 --> Xa = 0.25 Xb=0.75

0.3 = 0.25* Pa + 0.75*Pb

Case 2

0.4 = 0.5*Pa + 0.5*Pb

Solve simultanoeous equation

Pb= 0.5 bar

Answer 2

ANSWER.

The vapour pressure of pure liquid B = 0.2 bar.

EXPLANATION.

$\sf : \implies \: liquid \: a \: and \: b \: from \: ideal \: solution \: for \: all \: compositions \: at \: 25 {}^{0} c \\ \\ \sf : \implies \: two \: such \: solution \: with \: 0.25 \: \: and \: \: 0.50 \: mole \: fraction \: of \: a \: \\ \\ \sf : \implies \: total \: vapour \: pressure \: of \: 0.3 \: \: and \: \: 0.4 \: \: bar$

$\sf : \implies \: p_{t} \: = X _{A} P {}^{0} _{A} + \: X _{B} P {}^{0} _{B} \\ \\ \sf : \implies \: X _{A} + X _{B} = 1 \\ \\ \sf : \implies \: 0.3 = 0.25 \: P {}^{0} _{A} \: + \: 0.75 \: P {}^{0} _{B} \\ \\ \sf : \implies \: \frac{3}{10} = \frac{25 \: P {}^{0} _{A}}{100} + \: \frac{75 \: P {}^{0} _{B} \: }{100} \\ \\ \sf : \implies \: 30 =25 \: P {}^{0} _{A} \: + \: 75 \: P {}^{0} _{B} \: \: \: \\ \\ \sf : \implies \: 6 = 5P {}^{0} _{A} \: + 15P {}^{0} _{B} \: \: ......(1)$

$\sf : \implies \: 0.4 = 0.50 P {}^{0} _{A} \: + \: 0.50 \: P {}^{0} _{B} \\ \\ \sf : \implies \: \frac{4}{10} = \frac{50P {}^{0} _{A} }{100} + \: \frac{50P {}^{0} _{B}}{100} \\ \\ \sf : \implies \: 40 ={ 50P {}^{0} _{A} } \: + { 50P {}^{0} _{B} } \\ \\ \sf : \implies \: 4 = 5P {}^{0} _{A} \: + \: 5P {}^{0} _{B} \: ......(2)$

$\sf : \implies \: from \: equation \: (1) \: \: and \: \: (2) \: \: we \: get \\ \\ \sf : \implies \: 2 = 10 P {}^{0} _{B} \\ \\ \sf : \implies \: P {}^{0} _{B} \: = \frac{2}{10} = 0.2 \: bar$