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Equipartition theorem and chlorine

  1. Dec 20, 2007 #1
    [SOLVED] Equipartition theorem

    1. The problem statement, all variables and given/known data
    If the chlorine molecule at 290 K were to rotate at the angular frequency predicted by the equipartition theorem what would be the average centripetal force?
    (The atmos of Cl are 2*10^-10 m apart and the mass of the chlorine atom 35.45 a.m.u)

    (Correct answer is 1.6*10^-10 N)

    2. Relevant equations
    Kinetic energy
    [tex] E_{kin} = \frac{mv^2}{2} [/tex]

    Centripetal force
    [tex] F_c = \frac{mv^2}{r} [/tex]

    Boltzmann's constant
    [tex] k_B = 1.3807*10^{-23} JK^{-1} [/tex]

    3. The attempt at a solution
    Hi!

    I tried to solve it this way.
    I think of my Chlorine molecule as two points. One stationary and the other one circulating around it.
    From the equipartition theorem i get that the average energy of the molecule is

    [tex] E = \frac{7k_BT}{2} [/tex]

    I think that this is equal to the kinetic energy

    [tex] E = \frac{7k_BT}{2} = \frac{mv^2}{2} [/tex]

    so

    [tex] mv^2 = 7k_BT [/tex]

    Using this in the equation for the centriopetal force I get

    [tex] F_c = \frac{7k_BT}{r} = 1.4014^{-10}N [/tex]

    which is not the right answer and I havn't used the mass at all in the problem.
    What's wrong??
     
  2. jcsd
  3. Dec 21, 2007 #2

    Shooting Star

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    Homework Helper

    For diatomic molecules, Eavg = 5/2 KbT.
     
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