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Pressure Drop In Bulb : Application of Graham's Law

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  1. Sep 5, 2016 #1
    1. The problem statement, all variables and given/known data
    The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 50 minutes when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight 72 in the molar ratio of 1 : 1 at a total pressure of 6000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 70 minutes.


    2. Relevant equations
    04a232a4-d3ac-4133-b605-28de576a003d.gif

    3. The attempt at a solution
    Since the evacuated bulb contains a mixture of oxygen and another gas in the molar ratio of 1 : 1 at a total pressure of 6000 mm, the partial pressure of each gas is 3000 mm.


    The drop in the pressure of oxygen after 70 minutes


    = 500/50*70 = 700 mm of Hg


    ∴ After 70 minutes, the pressure of oxygen


    = 3000-700=2300 mm of Hg


    Let the rate of diffusion of other gas be rn, then


    Rate of O2/Rate of gas =1.5 (graham's formula)


    ∴ Drop in pressure for the other gas = 700/1.5=1400/3 mm Hg


    ∴ pressure of the other gas after 70 minutes


    = 3000 – 1400/3 mm = 7600/3 mm of Hg


    Molar ratio = Moles of unknown gas/Moles of O2


    = (7600/3)/(2300) = 76/69


    [Partial pressure ∝ mole fraction]

    But the answer is 39/46.

    Where am I going wrong?
     
  2. jcsd
  3. Sep 5, 2016 #2

    TSny

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    When the initial pressure of the oxygen is 2000 mm Hg, the pressure drops by 500 mm Hg after 50 minutes.

    If the initial pressure of the oxygen is changed to 3000 mm Hg, would the pressure still drop by 500 mm after 50 minutes?
     
  4. Sep 5, 2016 #3
    So should I use that rate of diffusion is directly proportional to pressure?
     
  5. Sep 5, 2016 #4

    TSny

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    Yes. Here you are dealing with "effusion". Kinetic theory shows the rate is proportional to the pressure.
    https://en.wikipedia.org/wiki/Effusion
     
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