# Thermodynamics and potential energy

## 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?

Related Introductory Physics Homework Help News on Phys.org
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).

Thanks! If I would have looked ahead in the textbook a few more pages, I could have saved myself a lot of frustration...