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Finding velocity of gas molecule

  1. Sep 10, 2013 #1
    Equations:
    1. vrms = √(3RT/M)
    2. PV = nRT
    3.M = mNA

    I want to answer part a.
    I am given T,P, and the density of gas molecules (d).


    From equation 3. I want to solve for m.
    m = dV = d(nRT/P)
    m[NA] = M = d(nRT/P)[NA] = dNRT/P; N = nNA.

    From equation 1.
    vrms = √(3RTP/(dNRT)) = √(3P/dN).
    I get stuck at this part because I dont know the value of N.
     

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  3. Sep 10, 2013 #2

    DrClaude

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    What is that equation?
     
  4. Sep 10, 2013 #3
    Is M molar mass or mass?
    I thought it was molar mass...
     
  5. Sep 10, 2013 #4

    DrClaude

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    I'm asking you! The equation doesn't make sense to me. I think it is wrong and that this is where your problem lies.

    Please detail what are M, m and NA.
     
  6. Sep 10, 2013 #5
    I added the equations in the attachment from the book
     

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  7. Sep 10, 2013 #6
    I think I figured it out. M=mNa and m = M/Na = d*v, so M = d*v*Na
     
  8. Sep 10, 2013 #7

    DrClaude

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    If ##m## is the mass of one molecule, then how can ##m = d \times V##?
     
  9. Sep 10, 2013 #8
    Well it could be the integral of a single mass through volume V. But I am not sure. Do you have a suggestion?
     
  10. Sep 10, 2013 #9
    And it didn't work because the variable V from equation 2 contains the value n which is unkown
     
  11. Sep 10, 2013 #10

    DrClaude

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    What about ##M_\mathrm{sam}##?
     
  12. Sep 10, 2013 #11
    Msam = M*n
     
  13. Sep 10, 2013 #12
    And M = dRT/P
     
  14. Sep 10, 2013 #13
    Thank you!
     
  15. Sep 10, 2013 #14
    I am now solving for M for part b.

    M = 3RT/(Vrms)2

    The units of this RHS is Joul/(mol*Kelvin)*Kelvin/(meter/second)2 = Joul*Second2/(mol*meter2)

    The LHS has units of

    kilograms/mole.

    How does the RHS turn into kilograms/mole?
     
  16. Sep 10, 2013 #15

    DrClaude

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    I didn't understand ##M = m N_\mathrm{A}## because you had written ##m = d \times V##, so I had taken ##m## to be the total mass of the gas. Using the notation of the book, ##m## is the mass of one molecule and ##M_\mathrm{sam}## the total mass of the gas, therefore the correct relation is ##M_\mathrm{sam} = d \times V##, which I guess is what you now have figured out, which you used along with ##n = M_\mathrm{sam} / M##.

    Go back to the definition of the joule in terms of base units:

    J = kg m2 s-2
     
  17. Sep 10, 2013 #16
    Thank you for all your help.
     
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