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Thermodynamics and potential energy

  1. Apr 30, 2008 #1
    1. The problem statement, all variables and given/known data

    From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of an oxygen molecule at 300 K?

    2. Relevant equations

    E_avg=3/2*kb*T (kb is the boltzmann's constant)
    PE = mgh = KE_final = 1/2*m*v_rms^2

    3. The attempt at a solution

    I know that E_avg is equal to 6.21*10^-21. This is then equal to mgh.

    I must be having difficulties calculating the mass of O2. There are 32g/mol = .032kg/mol. Dividing by Avogadro's number, this gives 5.3135*10^-26 kg per O2 molecule. Then plugging this into the equation h = E_avg/(mg) returns the height as 1.19*10^4 m... This is wrong. Can anyone help me figure out where I made an error?
     
  2. jcsd
  3. Apr 30, 2008 #2
    Your problem is not in calculating the mass of O2, rather your problem is that it's O2 and not O.

    The equipartition theorem states that a system has an average energy of kT/2 in each quadratic degree of freedom (mode in which the gas can store the energy). A monatomic gas can store energy as kinetic energy (Vx^2, Vy^2, Vz^2), so the average energy of an atom in this gas is 3*(kT/2).

    A diatomic gas can store the energy in the three translational degrees of freedom and two additional modes of rotation, so they have 5 degrees of freedom. Hence the average energy of a diatomic gas is 5*(kT/2).
     
  4. Apr 30, 2008 #3
    Thanks! If I would have looked ahead in the textbook a few more pages, I could have saved myself a lot of frustration...
     
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