# Degrees of freedom of diatomic molecule and Fermi energy

Sample of question (Please refer to attached PDF document for full details):

We have at our disposal an amount of energy equaling 3.5kJ and we wish to allocate 1 mole of gas which occupies volume V.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

In A) you made mistake in the first part. The internal energy for a diatomic molecule near or below room temperature is

5/2 N k T = 5/2 n R T.

This is because the molecule has besides kinetic energy also internal energy in the form of rotational energy. In a purely classical approach, it should also have vibrational energy, but that contribution is frozen at and below room temperatures. You can see this as follows. You can count the degrees of freedom for the molecule in two ways. Any possible motion of the molecule, can always be described as the motion of the two atoms separately. Then because each atom has 3 degrees of freedom (each atom can move in 3 independent directions), the total number of degrees of freedom is 6. Now these 6 degrees of freedom can also be accounted of by consideing that the two atoms form a molecule and then considering the possible degrees of freedom for the molecule.

The center of mass of the molecule has 3 degrees of freedom.

There are 2 independent ways to choose a rotation axis for the molecule (it has to be orthogonal to the length of the molecule), so 2 degrees of freedom for rotation.

Then, since there must be 6 degrees of freedom in total, that leaves us with one degree of freedom for vibration.

The vibrational degree of freedom doesn't contribute, so the average energy per molecule is 5/2 k T. If the temperature is so high that the vibrational degree of freedom can be treated classically, then it makes a contribution of k T and not 102 k T, as there is a potential energy and a kinetic energy component here, both of which make a contribution of 1/2 k T. You would then get 7/2 k T. That seems at odd with the extreme high temperature limit where the molecules should completely disassociate and then each oxygen molecule becomes 2 oxygen atoms, each of which has an average energy of 3/2 k T, yielding a total energy of 3 k T per disassociated molecule on average. The difference is explained by the binding energy of the molecule. The 7/2 k T is relative to the ground state at 0 K according to the harmonic approximation (which isn't exact). So, when the temperature is raised, part of the energy goes into the potential energy, which was negative to begin with. However, when you consider a gas of oxygen atoms, the 3 k T is relative to a hypothetical state at 0 K where you would still have free atoms, so a total energy of zero.

Anyway, you should change 3/2 to 5/2 in Eq. (1).

The average kinetic energy of the molecule is given by 3/2 k T, so you did the next problem correct, except that the temperature should be corrected. The last problem of part A is also correct, but there too the temperature should be replaced by the correct one.

Then part B:

What you write there is correct, let's derive that from first principles.

The number of available states for electrons in a volume V with magnitude of their momenta less then pf is:

N = 2 V times [volume in momentum space]/h^3

The factor of 2 comes from the two possible spin states. The volume in momentum space is 4/3 pi Pf^3, so we have:

N = 8 V/(3h^3) pi Pf^3

Then at zero temperature all the lowest energy states are filled, and then N will be equal to the number of molecules. The Fermi energy is the energy of the elctrons with the highest eenrgies in this case, so,

Ef = Pf^2/(2m) ---->

Pf = (2 m E)^(1/2)

Inserting in the equation for N gives:

N = 8 V/(3h^3) pi (2 m Ef)^(3/2) ----------->

Ef = h^2/(2 m) [3 N/(8 pi V)]^(2/3)

As you write, The average energy per electron is 3/2 Ef. To see this, you can just compute the total energy and divide that by N. The number of states with momenta of magnitude between p and p + dp is 2 V volume of momentum space/h^3 = 2 V 4 pi p^2 dp/h^3. We can then multiply that by p^2/(2m) and integrate from zero to Pf. Now, N can be written as 2 V total volume of momentum space/h^3 = 2 V 4/3 pi pf^3/h^3, so we can cancel the factor of 2 V 4 pi /h^3, leaving us with:

1/(pf^3/3) Integral from 0 to Pf of p^2/(2m) p^2 dp = 3 pf^2/(5*2 m) = 3/5 Ef

The total energy is thus 3/5 N Ef. Equating that to the 3.5 Kj and taking N to be the 1 mole allows you to solve for V.

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