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Rms speed

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the rms speed of one nitrogen molecule at 27 Celcius

    2. Relevant equations
    sqrt(v^2) = sqrt((3kT)/(m))

    3. The attempt at a solution
    sqrt(v^2) = sqrt((3kT)/(m))
    k = 8.62*10^-5 eV/K
    T = 300 K
    3kT = 0.078 eV
    m = 14 g/mol
    m = 0.014 kg/mol

    3kT/m = 0.078 eV/0.014 kg/mol = 5.571
    if 5.571 was 557.1 m/s I would assume it was correct but I doing something completely wrong here, any help will be appreciated. edit nevermind that statement since I have to take the sqaure root also

    sqrt(5.571)
     
    Last edited: Feb 17, 2016
  2. jcsd
  3. Feb 17, 2016 #2

    Bystander

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  4. Feb 17, 2016 #3
    Yea sorry it is 27 Celcius which is 300 K :)
     
  5. Feb 17, 2016 #4

    Bystander

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    Diatomic molecule.
     
  6. Feb 17, 2016 #5
    increasing the weight of the molecule does not solve the problem when you would expect it going about 500 m/s on average.
     
  7. Feb 17, 2016 #6

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    300 m/s: eV conversion, or something equally fatal.
     
  8. Feb 17, 2016 #7

    DrClaude

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    Units?
     
  9. Feb 17, 2016 #8
    3kT/m = 0.078 eV/0.014 kg/mol = 5.571 eV*kg/mol
     
  10. Feb 17, 2016 #9

    DrClaude

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    And how do you go from there to (m/s)2?
     
  11. Feb 17, 2016 #10
    1 eV = 1.6 * 10^-19 J, so adding this to the 3kT/m = (0.078 eV *(1.6 *10*-19 J))/0.014 kg/mol = 8.9264^-19 kg*J/mol
    J = N*m = (kg*m^2)/s^2
    8.9264^-19 kg^2*m^2/s^2*mol
    hmm something fishy is going on here, no idea how kg^2/mol relate if they do at all

    Oh wait.
    it is 8.9264^-19 m^2/s^2*mol
    Still have the 1/mol problem though
     
  12. Feb 17, 2016 #11

    DrClaude

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    Yes, because you are taking m as the mass of one mole of atomic N, not as the mass of one molecule of N2.
     
  13. Feb 17, 2016 #12
    According to my book Boltzmann's constant is 8.62*10^-5 eV/K
    but every place online I look at problems similar they use 8.3145 J/mol*K
    and if I do this problem method I have seen online, I get 3*(8.3145 J/mol*K)(300 K) / 0.014 kg/mol = 5.345^5 J/kg which makes the units a lot easier to process for me, kg *m^2/ kg*s^2
    which gives sqrt(5.345^5 m^2/s^2) = 66.049 m/s , which is still quite slow but better.
     
  14. Feb 17, 2016 #13

    DrClaude

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    Your value for the Boltzmann constant is correct. The other you cite is the gas constant, defined as ##R = N_A k_B##, where ##N_A## is the Avogadro constant. It is used mostly by chemists.

    You are still not taking into account that you have diatomic molecules of N2.

    That should be 5.345×105, not 5.3455.
     
  15. Feb 17, 2016 #14
    Oh yea im missing Avogadro's constant and I also forgot 1 atom is not 1 molecule.

    (28.0134 g/mol) / (6.02*10^23 mol^-1) = 4.65*10^-23 g = 4.65*10^-26 kg

    3*(8.62*10^-5 eV/K)*(300 K) / (4.65*10^-26 kg) = 1.6092^29 m^2/s^2

    v = sqrt(1.6092^29 m^2/s^2) = 990.479 m/s
    that is more like it, now it is super fast. thank you very much for all your help :)
     
  16. Feb 17, 2016 #15
    took me some time to write the above, but thank you for clarifying :)
     
  17. Feb 17, 2016 #16

    DrClaude

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    The number is not correct for units of m2/s2

    That speed is not the correct one, it is too big. Are you converting from eV to J correctly?
     
  18. Feb 17, 2016 #17
    No it looks like I made a mistake with the power.
    28.0134/6.02*10^23 = 4.6533^-23
    (3(8.62*10^-5)(300))/(4.65*10^-26) = 1.66838*10^24
    sqrt(1.66838^24) = 465.09 m/s
     
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