Is this a violation of Equipartition theorem?

Question

My book has a question asking to calculate average kinetic energy of the molecules in 8g of methane.

By the Equipartition theorem, I calculated that energy:
1. per molecule per degree of freedom = $(1/2)kT$ (by definition)
2. per molecule = $3kT$ (since methane is polyatomic => 6 degrees of freedom*)
3. per mole = $3RT$
4. for 8g methane ($=0.5$ moles) = $1.5\cdot RT$

But the answer stated by my book is $0.75\cdot RT$ - just half of what mine is.

This is not the only question where values have mismatched. I have started losing faith from Equipartition theorem.

Question:

What did I get wrong?

SUMMARY+UPDATE: Final conclusion reached is that while total kinetic energy is indeed the one Equipartition theorem gives (and so it's correct! :D) but the textbook assumes the molecule to be at at very low temperatures (although this is NOT mentioned) implying the rotational and vibrational energies are NOT counted.

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^{*=>my textbook says so. I understand that there's a lot of complex things in this topic, but let's please assume it is 6 as my book says.}

Answer 1

For a non-linear polyatomic molecule like methane, there are a total of $3N$ degrees of freedom, where $N = 4$ is the number of atoms in the molecule. This is broken down as follows:

$$\begin{array}{c|c} \hline \text{Component} & \text{# Degrees of freedom} \\ \hline \text{Translational} & 3 \\ \text{Rotational} & 3 \\ \text{Vibrational} & 3N - 6 \\ \hline \end{array}$$

(Note that for a linear molecule, or for an atom, this breakdown is different.)

Under "normal" temperatures (close to room temperature) the vibrational contribution to the internal energy can be neglected. (The vibrational gap $\Delta E = \hbar\omega \gg k_\mathrm{B}T$ for "normal" $T$, so higher vibrational states are not populated and the equipartition theorem does not hold.) This is probably the reason why your book states that there are six total degrees of freedom.

The key is to realise that the kinetic energy of a molecule arises only from the translational degrees of freedom. From my limited knowledge of physics, it seems to be a common thing to separate translational degrees from the "internal" degrees of freedom, i.e. rotational, vibrational, and electronic. This allows the translational motion of a body to be modelled as the movement of a point mass, located at the centre of mass of the body.

In this paradigm, rotations and vibrations preserve the centre of mass of a molecule, and therefore don't contribute to the kinetic energy of the molecule. Therefore, the correct answer is $(1.5~\mathrm{mol})\cdot RT$.

Another way of looking at it is to consider the definition of the kinetic energy

$$E_k = \frac{1}{2}m\langle v^2\rangle = \frac{1}{2}m(\langle v_x^2\rangle + \langle v_y^2\rangle + \langle v_z^2\rangle)$$

where $v$ is the velocity of the centre of mass, $v_\mathrm{cm}$. For each quadratic term the contribution to the energy is $kT/2$, for a total of $3kT/2$ per molecule. These three terms correspond to precisely the three translational degrees of freedom.

Answer 2

The equipartition theorem is a well established result of classical statistical mechanics. It states that the mean value of each independent quadratic term in the energy (per molecule) is equal to $\frac{1}{2}k_BT$ where $k_B$ is Boltzmann's constant and T the temperature. Being 'classical' means that if $\Delta E$ is the energy of a quantum then the thermal energy is far greater than this and effectively all levels are populated at the given temperature, i.e. $k_BT >> \Delta E$. (A derivation of equipartition can be found in this answer Derivation of mean kinetic energy )

The kinetic energy is split into three parts, translational , rotational and vibrational. As motion is possible in three directions (cartesian axes x, y, z thus three degrees of freedom) the average translational kinetic energy is $<E_{trans}> = \frac{3}{2}k_BT$.

The rotational kinetic energy also exists and occurs because a molecule can rotate as a whole body about each of the three axes (x, y, z, and assuming a non-linear molecule such as methane). ( The angular kinetic energy is $\frac{1}{2}I\omega^2$ where I is moment of inertia and $\omega$ the angular velocity). Thus the average rotational kinetic energy is $<E_{rot}> = \frac{3}{2}k_BT$.

If there are N atoms in a molecule then there are in total $3N$ degrees of freedom. We have accounted for $6$ of them in whole body translation and rotation, which leaves $3N-6$ internal degrees which are attributable to vibrations. As methane has 5 atoms the average vibrational kinetic energy is then $<E_{vib}> = \frac{9}{2}k_BT$. This value will not be reached at room temperature as typical vibrational quanta ($500 \cdots 3000 \pu{ cm^{-1}}$) are far greater than $k_BT \approx 208 \pu{ cm^{-1}}$ at room temperature. Thus the vibrational kinetic energy can often be set to zero with little error. However, the temperature is not specified in your question so in any answer it is necessary to justify removing the vibrational energy.

The rotational quantum for a polyatomic molecule is by contrast often only a few wavenumbers, or less, and $ \approx 10 \pu{cm^{-1}}$ for methane and so the average rotational energy will satisfy the energy criterion $k_BT >> \Delta E$ as will translational kinetic energy.

The total average kinetic energy (in Joules) is $<E_{tot}> = \frac{15}{2}k_BT$ per molecule. Removing the vibrational energy gives $<E_{tot}> = \frac{6}{2}k_BT$, which is $ \frac{6}{2}RT ~ \pu{J mol^{-1}}$ or $ \frac{3}{2}RT ~ \pu{J }$ for $0.5$ mol. It seems that to obtain the answer you quote, the rotational kinetic energy would also have to be removed, which then supposes that the temperature is very low.