GATE CH 2023 | Question: 57

Question

Distillation of a non-reactive binary mixture with components $A$ and $B$ is carried out in a batch still as shown in the figure below. The initial charge of the mixture in the still is $1 ~ \mathrm{kmol}$. The initial and final amounts of $A$ in the still are $0.1 ~ \mathrm{kmol}$ and $0.01 ~ \mathrm{kmol}$, respectively. Use a constant relative volatility of $4.5.$ The mole fraction of $B$ remaining in the vessel is____________(rounded off to three decimal places).

 

Answer 1

Mole of A = 0.01 Kmol      Total Mol Assumed = Nt     So Xa = 0.01/Nt             =    a (assumed)

Mole of B = Nt-0.01                                                     So Xb = (Nt-0.01)/Nt      =     (1-a) (Accordingly)

We Know that Y= 4.5X/(1+3.5X)  (relative volatilty given 4.5)

Put the assumed value of Xa & Xb in above Eq. & get Ya & Yb

Then Ya+Yb=1 So find the value of a 

Now a = .25 (calculated) so 1-a= 0.75 ( that is Xb Ans.)