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Molar fraction !

  1. Nov 6, 2005 #1
    Hi,

    here's the problem :

    A 60 gal tank contain pure Azote at 36,2 psig and 24,2 °C. We add 908,25 g of freon (CF2Cl2) that evaporate completely. If the temperature of the tank is 8,2 °C :

    a) What is the molar fraction of the freon.


    I've made some calculation, but it always result that I need the volume of the Azote to continue.
     
  2. jcsd
  3. Nov 6, 2005 #2

    Gokul43201

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    Show us your calculations, and we can tell you where the error is.
     
  4. Nov 8, 2005 #3
    The molar fraction is the (number of mol for the freon) / (total number of mol). I found the number of mol for the freon = 7,51161 mol.

    Now I need the total number of mol. So I need the
    (number of mol for the freon) + (number of mol for the azote) / (Molar weight of freon) + (Molar weight of azote).

    The only thing that is missing is the (number of mol for the azote). And I found 2 ways to find it

    1) volumic weight * Volume
    2) PV=nRT ===> n = PV/RT ==> M = m/n ==> m = n*M .....

    In each case, I don't have the volume of the azote so I cannot continue
     
  5. Nov 11, 2005 #4

    Gokul43201

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    Initially, you are told that the entire volume of the tank (ie : 60 gall) contains azote (or Nitrogen) at some P, T. So, the volume of nitrogen is 60 gall. From this you can calculate n as you've proposed above.
     
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