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28 June, 11:43

By measuring the equilibrium between liquid and vapour phases of a solution at 30.0 degree C at 1.00 atm, it was found that xA = 0.220 when yA = 0.314. Calculate the activities and activity coefficients of both components in this solution on the Raoult's law basis. The vapour pressures of the pure components at this temperature are: pA = 73.0 kPa and PB 92.1 kPa. (xA is the mole fraction in the liquid and yA the mole fraction in the vapour.)

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  1. 28 June, 14:44
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    Answer:The activities of A and B are and 0.44 and 0.76 the activity coefficients are 2 and 0.97 respectively.

    Explanation:

    Given that at equilibrium

    Xa = 0.220

    Ya = 0.314

    Vapour pressure of a = Pa * = 73kPa

    Vapour pressure of b = Pb * = 92.1kPa

    Total pressure Pt = 1atm = 101.325kPa

    From Partial Pressure Law

    Pt = Pa+Pb ... Eqn (1)

    Where Pa and Pb are partial pressures of a and b.

    Pa = Ya * Pt

    =0.314 * 101.328

    Pa = 31.82kPa

    From Eqn (1)

    Pb = Pt - Pa

    = 101.325 - 31.82

    =69.51kPa

    From Raoult's Law

    Aa = Pa/Pa*

    Where Aa is the activity of component a

    = 31.82/73

    Aa=0.44

    Also Ab is the activity of component b

    Ab=69.51/92.1

    =0.76

    The activity coefficients of each component is given by:

    Yi = Ai/Xi

    For component a

    Ya = Aa/Xa

    = 0.44/0.220

    =2

    For component b

    Yb = Ab/Xb

    Xb is unknown.

    But at equilibrium Xa+Xb = 1

    Therefore, Xb = 1-0.22 = 0.78

    Yb = 0.755/0.78

    = 0.97
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