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Archived Vapor composition from bubble-point temp.

  1. Oct 31, 2013 #1
    Suppose you have a mixture of 30 mol% pentane and 70 mol% hexane at 1 atm and 60oC.

    If the Tbubble-point = 55.1oC, calculate the composition of vapor of the first bubble as the mixture is heated.

    I found p*pentane = 1392.95 and p*hexane = 484.96 using Antoine's equation with T = 55.1.

    Now I suppose I must use Raoult's law as:

    yaP = xap*a(@55.1oC)

    Using this I get xpentane = (0.7*760)/1392.96 = 0.382

    However, the answer sheet says xpentane = 0.55.

    Can anyone figure where I'm mistaken?
     
  2. jcsd
  3. Feb 4, 2016 #2
    You applied the equation incorrectly. y is the mole fraction in the vapor, and x is the mole fraction in the liquid.

    y = (0.3)(1392.96)/760=0.550
     
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