Section 15.6: Exploring the Equipartition of Energy

    # of monatomic particles =      
# of diatomic particles =

Please wait for the animation to completely load.

The equipartition of energy theorem says that the energy of an atom or particle is, on average, equally distributed between the different modes (different degrees of freedom) available.  The way to count the modes, or degrees of freedom, is to count the number of quadratic terms in the energy expression.  For example, for a monatomic gas (without any external forces), the energy of a particle is given as 1/2 mvx2 + 1/2 mvy2 + 1/2 mvz2.  There are three terms that are quadratic.  However, for a particle in a three-dimensional simple harmonic potential well, there are six different modes (degrees of freedom).    The energy per particle has an average value of (f/2)kBT, where f is the number of degrees of freedom, kB is the Boltzmann constant and T is the temperature. 

You can verify this result by taking the distribution function for a monatomic ideal gas and finding the total energy:5 

n(ε)dε  = 2πN/(π kBT)3/2 ε1/2 exp (−ε/kBT) dε , 

where ∫ n(ε)dε = E  and E = (3/2) NkBT.  For a harmonic oscillator, the distribution is different, but in the end it gives you E = 3kBT.  Restart.

  1. In this animation of a monatomic gas in a box, why do the particles only have 2 degrees of freedom?  The table shows the total kinetic energy of all particles in the box, as well as the average kinetic energies of particles in the box (the animation averages over a 10-s period, so you need to wait 10 s to read the averages).
  2. Record the total energy.  What is the energy per particle? If the energy is given in joules/kB, what is the temperature inside the box?

Try this animation of a diatomic gas with 20 particles.  Notice that the graph shows the total kinetic energy of the diatomic particles and the kinetic energies of translation (motion in x and y directions) and rotation. 

  1. Why is the translational kinetic energy, on average, about two times the rotational kinetic energy?
  2. From the total energy, what is the energy per particle? If the energy is given in joules/kB, what is the temperature in the box? Hint: Remember that <energy>/particle = (f/2)kBT and in this case, f = 3 (Why?).

Now, try a mixture of 20 monatomic particles and 20 diatomic particles. 

  1. Why is the temperature of the gas in the box a single value (not one value for atoms and another for molecules)? 
  2. After waiting at least 10 s, compare the average values of the kinetic energies.   What value is the average monatomic kinetic energy close to?  Why should those two values, averaged over a long period of time, be equal to each other and greater than the rotational kinetic energy of the diatomic particles? 
  3. From the total energy (given in joules/kB), what is the temperature?  Hint: Explain why the total energy should be equal to (2/2)20kBT + (3/2)20kBT
  4. In this animation, if a mixture has 15 atoms, how many diatomic particles should it have so that the average kinetic energies of both particles are the same?  Try setting the number of monatomic particles and diatomic particles to check your answer.

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5This is a classical result because we are using the classical distribution (Section 15.5).  If the temperature is very low and the density high, we will need to use a quantum mechanical distribution and these results do not hold (see Section 15.7 for an example of low temperature calculations in a solid).

 

Applet authored by Ernesto Martin and modified by Wolfgang Christian.