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Thermodynamics and potential energy

  • Thread starter kkaplanoz
  • Start date
  • #1
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Homework Statement



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?

Homework Equations



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

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?
 

Answers and Replies

  • #2
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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).
 
  • #3
4
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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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