Please show all work. Thanks! Consider a solution with xA = 0.220 at 30?C in equ

Question

Please show all work. Thanks!

Consider a solution with xA = 0.220 at 30?C in equilibrium with the vapor. The vapor pressures of the pure components at this temperature are: p? A = 73.0kPa and p? B = 92.1kPa.

(a) Predict the vapor pressure and its composition at these conditions assuming ideal solution.

(b) Experimentally it was found that the actual vapor pressure was 1.00atm and yA = 0.314. Calculate the activities and activity coe?cients of both components in this solution on the Raoult’s law basis.

Explanation / Answer

a)

composition :

xA = 0.220 ,

so xB = 1-0.220 = 0.780

Total vapor pressure = xA * vp of A in standard state + xB * vp of B in standard state

=   0.220* 73.0 + 0.780* 92.1

= 87.898 kPa

b)

po (A) = 73.0 kPa

po(B) = 92.1 kPa

xA = 0.220 so xB = 1- xA = 1- 0,220 = 0.78

yA = 0.314 ., so yB = 1- .314 = 0.686

actual vapor pressure = 1 atm = 101325pa

so activity coefficient of A = actual VP/roult's law VP = 101325 / 73.0*10^3

= 1.3380

activity coeff of B = actual VP/roult's law VP = 101325 / 92.1*1000   

= 1.100

activity of A = xA * activity coeff of A = 0.220* 1.3380

= 0.29436

activity of B = xB * activity coeff of B  

= 0.78 * 1.100

= 0.858

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