What is the kinetic energy of one molecule of a polyatomic gas?

Question

According to Kinetic theory, the kinetic energy of one molecule of gas is 3/2kT. Is this true for diatomic and polyatomic gases as well?

I've read that a diatomic gas has 5 degrees of freedom (3 translational +2 rotational), and according to the Law of Equipartition of Energy, each degree of freedom contributes an energy of 1/2kT. So, shouldn't the kinetic energy of a diatomic gas like Chlorine actually be 5/2kT?

Or maybe the equation K.E. = 3/2kT only takes traslation motion into account and the K.E. they are talking about is actually translation one. Please tell me.

Answer 1

Kinetic theory assigns a degree of freedom to every quadratic term involving momentum (linear, rotational/angular, vibrational) and every quadratic term involving the cartesian co-ordinates appearing in the expression for the total energy for a molecule.

The Equipartition of Energy theorem then says that for a system in thermal equilibrium, each degree of freedom has an average energy of $k_BT/2$, where $T$ is the absolute temperature and $k_B$ is Boltzmann's constant.

If a molecule has $f$ degrees of freedom then the total energy of a molecule is

$$E_{molecule} = \frac{f}{2}k_BT\,. $$

However, there is a problem since each degree of freedom a molecule can possibly possess does not always contribute to its energy. This is because the contribution by the degrees of freedom to the energy of a molecule depends on the temperature of the gas.

In the case of H$_2$, for example, at low temperatures (30 K) only translational degrees of freedom contribute to a molecule's energy but at 300 K both translational and rotational degrees of freedom contribute. Looking at hydrogen, it has 3 translational, 3 rotational (rotation about x, y and z axes) and 2 vibrational degrees of freedom. The three translational degrees of freedom contribute at 300 K but only 2 rotational degrees of freedom contribute to the energy of a molecule and to the specific heat at constant volume of the gas. The third rotational degree of freedom does not contribute because the energy $k_BT/2$ is small compared to the energy levels that quantum mechanics says a molecule can have for rotation about the axis along the bond that joins the atoms making up the molecule. (See also this link https://physics.stackexchange.com/a/168945/168935 .) The vibrational degrees of freedom are activated when the temperature reaches 5000 K, and are activated for the same reason.

The same general argument applies to a chlorine gas molecule

Answer 2

Yes, you are correct. The total kinetic energy (including translational, rotational and vibrational) of a given molecule of a polyatomic gas is:

$$K =\frac f2 k_bT$$

Where $f$ is the number of degree of freedom of the molecule.

$$K =\frac 32 k_bT$$

Only takes the translational kinetic energy into account.

Answer 3

Equipartition is a classical approximation that holds for translational modes because they have very dense quantum energy levels. Rotational modes have wider energy spacing but still can typically be treated as classical at room temperature. So in this sense, counting 5 degrees of freedom for a diatomic gas is correct.

But the only reason not to include vibrational, electronic, etc. (ultimately degrees of freedom equal to 3 times the number of elementary particles in the molecule) is that the additional modes are "frozen" in their ground state, as the excited states are too high for thermal fluctuations to reach. At sufficiently high temperatures, these modes "unlock" and thermal energy goes into them as well.