Consider a binary mixture containing A (more volatile) and B. Solve for followin

Question

Consider a binary mixture containing A (more volatile) and B. Solve for following: 1. By considering 45.68 moles of feed containing 37% A is vaporized to produce 11.42 moles of vapor, what will be the vapor composition? 2. Instead of 11.42 moles, 22.84 moles are vaporized in the first step and the vapor is condensed, what will be the vapor composition? 3. In the second stage, 11.42 moles of the condensed part (coming from the top of stage 1) is vaporized, what will be the liquid composition? 4. In the third stage, 3.426 moles of the condensed part (coming from the bottom of the 2^nd stage) is vaporized, what will be the vapor composition? The x-y data needed is as follows: y = - (1 - f)/f x + (x_F/f) Sol:

Explanation / Answer

Question says A is more volitile but it is not mentioned how much so it's not possible to find correct composition.
If we assume that both A and B evaporate at the same rate then we can find the composition.
Ans(1):
37% of A is contained in feed then vapour will also contain same %
hence amount of A in vapour = 37% of 11.42 mole= 0.37* 11.42=4.2254 mole
amount of B in vapour = 11.42-4.2254 mole= 7.1946 mole
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Ans(2):
this time process is same but with 22.84 moles
37% of A is contained in feed then vapour will also contain same %
hence amount of A in vapour = 37% of 22.84 mole= 0.37* 22.84=8.4508 mole
amount of B in vapour = 22.84-8.4508 mole= 7.1946 mole
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you can proceed in same manner to find composition of each part

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