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Degree of freedom

  1. Sep 18, 2005 #1
    Please I need help.

    Can any body can give me the formula for determine the number of degree of freedom of molecules undergoing translation,rotation and vibration of n number of atom.
  2. jcsd
  3. Sep 18, 2005 #2


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    it depends on what the molecules are. monatomic? (it's 3n.) diatomic? (it's 5n). more than that, i dunno.
  4. Sep 18, 2005 #3
    rbj can u clariyfy more if that 3n for monotomic and 5n for diatomic is for translation,rotation or vibration
  5. Sep 20, 2005 #4


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    sorry, kidia. they moved the thread but they didn't leave a little "moved" arrow behind.

    for monatomic gases (He Ne Ar, the inert gasses), each molecule is a single atom and virtually all of the mass of the molecule is concentrated at the nucleus. they're a simple ball with the mass all concentrated at the center. even if you were to spin the ball, there would be very little rotational kinetic energy in that spin because the mass is all at the center. no moment of inertia. so these molecules have 3 degrees of freedom of translation, and only those three. up-down (z-axis), left-right (x-axis), and forward-backward (y-axis). no rotation or vibration.

    diatomic gasses (O2 N2) have two identical atoms bonded together. for each atom, the mass is concetrated at the nucleus. so this structure is like a dumbell structure. besides the 3 translational motions (x, y, z-axis) that the monatomic gasses have, there are 2 more rotational degrees of freedom. imagine the dumbell lined up on the z-axis. there would be a non-zero moment of inertia along the x-axis and along the y-axis, but not along the z-axis.

    it's obvious (due to symmetry) why the 3 translational degrees of freedom should be the same (contain the same average kinetic energy) and why the 2 identical rotational degrees of freedom contain the same average kinetic energy, but someone else will have to explain why the 2 rotational degrees of freedom contain the same amount of average kinetic energy per degree of freedom as the 3 translational degrees of freedom. can a real physicist explain that?
  6. Sep 23, 2005 #5
    thanx very much rbj I catch u.
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