Answer to Question #97034 in Physical Chemistry for Abhishek

Question #97034
The vapour pressures of pure liquids . A and B at 300 K are 4.5×10⁴ Pa and 6.5× 10⁴ Pa respectively. calculate the mole fractions of of A and B in vapour and liquid phases of a solution when the liquid equilibrium total vapour pressure of the binary liquid solution is 5.5× 10⁴ Pa at 300 K . Assume that the liquid and the vapour are ideal .
1
Expert's answer
2019-10-31T07:29:12-0400

Let mole fraction of A in liquid phase is xax_a and B in liquid phase is xb=1xax_b=1-x_a

Then according to question,


0.55=0.45xa+0.65(1xa)0.55=0.45x_a+0.65(1-x_a)

0.2xa=0.1orxa=0.50.2x_a=0.1\\or\\x_a=0.5

xb=1xa=0.5x_b=1-x_a=0.5

Composition in vapour phase will be-

pa=0.45xa=0.225 barpb=0.65xb=0.325 barp_a=0.45x_a=0.225\ bar\\p_b=0.65x_b=0.325\ bar

Mole fraction of A in vapour phase , ya=0.2250.55=0.4090y_a=\frac{0.225}{0.55}=0.4090

Mole fraction of B in vapour phase,yb=0.3250.550=0.5909y_b=\frac{0.325}{0.550}=0.5909


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