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Angular velocity of an oxygen molecule.

  1. Jun 9, 2014 #1
    1. The problem statement, all variables and given/known data

    The atoms in the oxygen molecule [O][/2] may be considered to be point masses separated by a distance of 1.2 x 10^-10 m. The molecular speed of an oxygen molecule at s.t.p. is 460 m/s. Given that the rotational kinetic energy of the molecule is two-thirds the of its translational kinetic energy, calculate its angular velocity at s.t.p. assuming that molecular rotation takes place about an axis through the centre of, and perpendicular to, the line joining the atoms.



    2. Relevant equations

    K.E. = 1/2 mv^2
    K.E. = 1/2 Iω^2
    I = ∑mr^2
    n = N/Na
    m = nμ



    3. The attempt at a solution

    no. moles of oxygen = 1/6.02 x 10^23 = 1.66 x 10^-24 mol

    mass of oxygen molecule = (1.66 x 10^-24) x 32 = 5.32 x 10^-23 g = 5.32 x 10^-26 kg

    translational K.E. of oxygen molecule = 0.5 x (5.32 x 10^-26) x 460 = 1.22 x 10^-23 J

    rotational K.E. of oxygen molecule = 2/3 x 1.22 x 10^-23 J = 8.15 x 10^-24 J

    I = 2(mr^2) = 2((5.32 x 10^-32) x (6 x 10^-11)^2) = 3.83 x 10^-46 kg m^2

    angular velocity of oxygen molecule, ω = sqrt(((8.15 x 10^-24) x 2)/3.83 x 10^-46) = 2.06 x 10^11 rad/s


    I think the correct answer is 6.3 x 10^12 rad/s. I'm not sure where I have gone wrong here. I'm not sure if I used the correct method for working out the moment of inertia. Any guidance would be much appreciated.

    Thanks.
     
  2. jcsd
  3. Jun 9, 2014 #2

    CAF123

    User Avatar
    Gold Member

    You do not have to compute the mass of the oxygen atoms to obtain your answer. You can treat the two oxygen atoms as point masses rotating about an axis through the centre of rotation. So, if you carry the calculation through symbolically, the mass term should cancel.

    Should you not have ##E_{\text{rot}} = \frac{2}{3} E_{\text{trans}}##?
     
  4. Jun 9, 2014 #3
    OK, I've got it now, thanks!
     
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