By measuring the equilibrium between liquid and vapor phases of a solution at 30 °C and1.00 atm, it was found that xA 0.220 when y 0.3 14. Calculate the activity andactivities and activity coefficients of both components in this solution on the Raoult's lawbasis. The vapor pressures of the pure components at this temperatures arep 73.0 kPa and p 92.1 kPa. (x4 is the mole fraction in the liquid and ya the molefraction in the vapor.)

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Asked Nov 30, 2019
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By measuring the equilibrium between liquid and vapor phases of a solution at 30 °C and
1.00 atm, it was found that xA 0.220 when y 0.3 14. Calculate the activity and
activities and activity coefficients of both components in this solution on the Raoult's law
basis. The vapor pressures of the pure components at this temperatures are
p 73.0 kPa and p 92.1 kPa. (x4 is the mole fraction in the liquid and ya the mole
fraction in the vapor.)
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By measuring the equilibrium between liquid and vapor phases of a solution at 30 °C and 1.00 atm, it was found that xA 0.220 when y 0.3 14. Calculate the activity and activities and activity coefficients of both components in this solution on the Raoult's law basis. The vapor pressures of the pure components at this temperatures are p 73.0 kPa and p 92.1 kPa. (x4 is the mole fraction in the liquid and ya the mole fraction in the vapor.)

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Expert Answer

Step 1

Given,

Total pressure, PT = 1.00 atm = 101.325 kPa

Mole fraction in liquid phase, xA = 0.220

Mole fraction in vapour phase, yA = 0.314

Vapour pressure of pure component A, PA* = 73.0 kPa

Vapour pressure of pure component B, PB* = 92.1 kPa

Step 2

Firstly, we need to find the partial pressures (PA and PB) of components A and B as:

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Step 3

The activities (a) of the compone...

For component A:
РА
ад
(РА)"
31.81605 kPа
ад
73.0 kPa
ад — 0.436
For component B:
Рв
ав
(Рв)"
69.50895 kPa
ав
92.1 kPa
ав — 0.755
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For component A: РА ад (РА)" 31.81605 kPа ад 73.0 kPa ад — 0.436 For component B: Рв ав (Рв)" 69.50895 kPa ав 92.1 kPa ав — 0.755

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