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Pressure ratio problem

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the pressure ratio of He to N2 at which helium would have the same density as nitrogen if their temperatures were the same.

    2. Relevant equations

    I used D = m/v

    3. The attempt at a solution

    DHe = mHe/v
    DN2 = mN2/v
    Both gases occupy the same volume, so just v for both.

    Since DHe = DN2,
    mHr/v = mN2/v and mHe = mN2

    For some x and y,
    x mol He(4.003 g/mol He) = mHe
    y mol N2(28.01 g/mol N2) = mN2

    4.003x g = 28.01y g
    x = 7y

    To me it looks like there are 7 times as many moles of He as N2 but I doubt that would directly apply to their pressure ratios. I think I'd have to use PV = nRT but I'm not sure how I'd put it in.
    I was actually helping some chemistry students earlier today with this and am hoping I can have the answer ready for them tomorrow morning. :smile:
     
    Last edited: May 1, 2009
  2. jcsd
  3. May 1, 2009 #2

    Borek

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    Staff: Mentor

    Honestly, I have no idea what the question asks. Oxygen? And why do you use neon in your calculations?

    Could be what you did is OK, but with all these typos/inconsistencies it is not.
     
  4. May 1, 2009 #3
    I fixed the Ne and oxygen; I have an older edition of the book than that which the students are using and I hadn't quite changed everything to match the problem in their book. Should be alright now.
     
  5. May 1, 2009 #4

    Borek

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    Staff: Mentor

    OK.

    Now, knowing ratio of numbers of moles try to calculate ratio of pressures using PV=nRT. Don't be surprised if everything cancels out :wink:
     
  6. May 4, 2009 #5
    I think I got it now (don't know why I didn't look at it like this before)

    I found that nHe/nN2 = 7/1, and using P = nRT/V,

    PHe/PN2 = (nHeRT/V)/(nN2RT/V)
    PHe/PN2 = nHe/nN2 = 7/1
     
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