At 90 degree C, the vapour pressure of 1, 2-dimethylbenzene is 20 kPa and that o

Question

At 90 degree C, the vapour pressure of 1, 2-dimethylbenzene is 20 kPa and that of 1, 3-dimethylbenzene is 18 kPa. What is the composition of a liquid mixture that boils at 90 C when the pressure is 19 kPa? What is the composition of the vapour produced? It is found that the boiling point of a binary solution of A and B with xA-0.4217 is 96 degree C. At this temperature the vapour pressures of pure A and Bare 110.1 kPa and 76.5 kPa, respectively. (a) Is this solution ideal? (b) What is the initial composition of the vapour above the solution?

Explanation / Answer

(1)

Lets denote 1,2-dimethylbenzene by A and the other one by B.

Given:

P0A = 20 kPa

P0B = 18 kPa

Using Raoult's law of partial pressures:

P = PA + PB = P0A*xA + P0B *xB

19 = 20*xA + 18*xB

Thus, required composition is: xA = xB = 0.5

Vapor phase composition: yA = PA/P = 10/19 = 0.53 ; yB = PB/P = 9/19 = 0.47

(2)

For calculating this part's answer, I need to know the pressure exerted by the binary solution at the given temp.

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