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RMS Speed

  1. Sep 10, 2005 #1
    I have been having difficulties with this problem and I was wondering if I could get some help with it.

    Q. What is the rms speed of nitrogen molecules contained in a 7.0 m3 volume at 4.20 atm if the total amount of nitrogen is 1600 mol?

    I figured I would have to use this equation to solve it:

    V rms = (3KT/m)^(1/2)

    But first I would need to find T, so I found it through this equation:

    PV=nRT
    (4.20 atm)(1.013e5 N/m^2/atm)(7.0m3) = (1600 mol)(8.315 J/mol K)(T)
    and T = 223.9 K

    I also needed to find m

    m (N2) = (28)(1.66 x 10e-27) = 4.648e-26 kg

    Then I subsitituted them all back into the original equation

    V rms = [(3*1.38e-23 J/K*223.9 K)/(4.648e-26 kg)]^(1/2)
    V rms = 446.5 m/s

    But that isn’t the right answer. I don’t know what I am doing wrong, but I was wondering if anyone could help point me in the right direction, it would be greatly appreciated.

    Thank you!!
     
  2. jcsd
  3. Sep 10, 2005 #2
    for the PV = nRT part, see if all your units cancel out. Keep in mind that the R value you have used will only cancel out pascals.

    also how do you fin d out mass from the number of moles
    number of moles = mass / molar mass
    whati s the molar mass of N2??
     
  4. Sep 10, 2005 #3
    that's why I used the conversion factor from atms to pascals (1.013e5). Is that the wrong conversion?
     
  5. Sep 10, 2005 #4
    i didnt see you had already done that. OK now what should the units of volume be? Volume is always given in litres. not cubic metres. However 1L = 1 dm^3 (decimeter)
     
  6. Sep 10, 2005 #5
    Ok I took all your suggestions and I found the new Volume to be 7000 L, which would make the new T about 223859 Kelvin.

    Then the molar mass I found by dividing (4.648e-26 kg)/1600 mol and got 2.905e-29

    I subsitituted all those in and got 564826 m/s and that still seems far off.

    I think I messed up the molar mass, any ideas on what I am doing wrong?

    Thanks so much for all your help!! I really appreciate it! :smile:
     
  7. Sep 11, 2005 #6
    I'm still having troubles with this problem, any help would be appreciated, since some of the last suggestions didn't work out too well.


    Q. What is the rms speed of nitrogen molecules contained in a 7.0 m[tex]^{3}[/tex] volume at 4.20 atm if the total amount of nitrogen is 1600 mol?


    I figured I would have to use this equation to solve it:

    V rms = [tex]\sqrt{\frac{3KT}{m}}[/tex]



    But first I would need to find T, so I found it through this equation:

    PV=nRT
    (4.20 atm)(1.013e5 [tex] \frac{\frac{N}{m^{2}}}{atm}[/tex])([tex]7 m^{3}[/tex]) = (1600 mol)(8.315 [tex]\frac{J}{mol K}[/tex])(T)
    and T = 223.9 K

    (noting that the conversion:1.013e5 [tex] \frac{\frac{N}{m^{2}}}{atm}[/tex] to change atms to Pascals was used and the volume should NOT be converted to L because it makes the temperature to large and it doesn't cancel with my conversion factor )



    I also needed to find m

    m (N[tex]^{2}[/tex]) = [(28)/(6.02e23)] = 4.65e-23 g or 4.65e-26 kg



    Then I subsitituted them all back into the original equation

    V rms = [tex]\sqrt{\frac{(3)(1.38e-23 J/K)(223.9 K)}{(4.65e-26 kg)}}[/tex]
    V rms = 446.4 m/s



    Am I going in the right direction?
    Thank you!!!
     
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