A single atom is free to move in space along the \(\mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) axis. However, each of these movements requires energy. This is derived from the energy held by the atom. The Law of Equipartition of Energy defines the allocation of energy to each motion of the atom (translational, rotational and vibrational). Before we understand this law, let’s understand a concept called ‘Degrees of Freedom’.

Degrees of Freedom

Imagine a single atom. In a three-dimensional space, it can move freely along the \(\mathrm{X}\), \(\mathrm{Y}\), and \(\mathrm{Z}\) axis. Motion from one point to another is also known as translation. Hence, this movement along the three axes is called translational movement. If you have to specify the location of this atom, then you need three coordinates ( \(\mathrm{x}, \mathrm{y}\), and \(\mathrm{z})\).
We can also say that a single atom has 3 Degrees of Freedom. Most monoatomic molecules (i.e. molecules having a single atom-like Argon) have 3 translational degrees of freedom, provided their movement is unrestricted.

Let’s now imagine a diatomic molecule (a molecule having two atoms like \(\mathrm{O}_2\) or \(\mathrm{N}_2\) ). Apart from the three translational degrees of freedom, these molecules can also rotate around the centre of mass. Two such rotations are possible along the axis normal to the axis that joins the two atoms.
This adds two additional degrees of freedom (rotational) to the molecule. In simpler words, to specify the location of the molecule, you would need the X,Y, and Z coordinates along with the rotational coordinates of the individual atoms.
It is important to note here that these diatomic molecules are not rigid rotators (where molecules do not vibrate) at all temperatures. Along with the translational and rotational movements, diatomic molecules also oscillate along the interatomic axis like a single-dimensional oscillator. This adds a vibrational degree of freedom to such molecules.

Hence, to specify the location of a diatomic molecule, you would finally need the \(\mathrm{X}\), \(\mathrm{Y}\) and \(\mathrm{Z}\) coordinates along with the rotational and vibrational coordinates. So, in a nutshell, Degrees of Freedom is nothing but the number of ways in which a molecule can move. This forms the basis of the Law of Equipartition of Energy.

Law of Equipartition of Energy

The law states that: “In thermal equilibrium, the total energy of the molecule is divided equally among all Degrees of Freedom of motion”. Before delving into the calculations, let’s understand the law better. If a molecule has 1000 units of energy and 5 degrees of freedom (which includes translational, rotational, and vibrational movements), then the molecule allocates 200 units of energy to each motion.

The kinetic energy of a single molecule is
\(
\varepsilon_t=\frac{1}{2} m v_x^2+\frac{1}{2} m v_y^2+\frac{1}{2} m v_z^2 \dots(13.22)
\)
For a gas in thermal equilibrium at temperature \(T\) the average value of energy denoted by \(<\varepsilon_t>\) is
\(
\left\langle\varepsilon_t\right\rangle=\left\langle\frac{1}{2} m v_x^2\right\rangle+\left\langle\frac{1}{2} m v_y^2\right\rangle+\left\langle\frac{1}{2} m v_z^2\right\rangle=\frac{3}{2} k_B T \dots(13.23)
\)
Since there is no preferred direction, Eq. (13.23) implies
\(
\begin{aligned}
\left\langle\frac{1}{2} m v_x^2\right\rangle & =\frac{1}{2} k_B T,\left\langle\frac{1}{2} m v_y^2\right\rangle=\frac{1}{2} k_B T \\
\left\langle\frac{1}{2} m v_z^2\right\rangle & =\frac{1}{2} k_B T \dots(13.24)
\end{aligned}
\)

In case of a diatomic molecule, translational, rotational, and vibrational movements are involved. Hence the Energy component of translational motion

\(
\varepsilon_t=\frac{1}{2} m v_x^2+\frac{1}{2} m v_y^2+\frac{1}{2} m v_z^2
\)

The energy component of rotational motion 

\(
\varepsilon_{\mathrm{r}}=\frac{1}{2} I_1 \omega_1^2+\frac{1}{2} I_2 \omega_2^2\)

where \(\omega_1\) and \(\omega_2\) are the angular speeds about the axes 1 and 2 and \(I_1, I_2\) are the corresponding moments of inertia.

Molecules of a monatomic gas like argon have only translational degrees of freedom. But what about a diatomic gas such as \(\mathrm{O}_2\) or \(\mathrm{N}_2\)? \(\mathrm{A}\) molecule of \(\mathrm{O}_2\) has three translational degrees of freedom. But in addition, it can also rotate about its centre of mass. Figure 13.6 shows the two independent axes of rotation 1 and 2, normal to the axis joining the two oxygen atoms about which the molecule can rotate. The molecule thus has two rotational degrees of freedom, each of which contributes a term to the total energy consisting of translational energy \(\varepsilon_t\) and rotational energy \(\varepsilon_r\)
\(
\varepsilon_{\mathrm{t}}+\varepsilon_{\mathrm{r}}=\frac{1}{2} m w_x^2+\frac{1}{2} m w_y^2+\frac{1}{2} m v_z^2+\frac{1}{2} I_1 \omega_1^2+\frac{1}{2} I_2 \omega_2^2 \dots(13.25)
\)

We have assumed above that the \(\mathrm{O}_2\) molecule is a ‘rigid rotator’, 1.e., the molecule does not vibrate. This assumption, though found to be true (at moderate temperatures) for \(\mathrm{O}_2\), is not always valid. Molecules, l1ke \(\mathrm{CO}\), even at moderate temperatures have a mode of vibration, 1.e., its atoms oscillate along the interatomic axis like a one-dimensional oscillator, and contribute a vibrational energy term \(\varepsilon_v\) to the total energy:
\(
\varepsilon_v=\frac{1}{2} m\left(\frac{\mathrm{d} y}{\mathrm{~d} t}\right)^2+\frac{1}{2} k y^2
\)

\(
\varepsilon=\varepsilon_t+\varepsilon_r+\varepsilon_v \dots(13.26)
\)
where \(\mathrm{k}\) is the force constant of the oscillator and \(y\) is the vibrational coordinate.
Once again the vibrational energy terms in Eq. (13.26) contain squared terms of vibrational variables of motion \(y\) and \(\mathrm{d} y / \mathrm{d} t\).

At this point, notice an important feature in Eq.(13.26). While each translational and rotational degree of freedom has contributed only one ‘squared term’ in Eq. (13.26), one vibrational mode contributes two ‘squared terms’: kinetic and potential energies. According to the Law of Equipartition of Energy, in thermal equilibrium, the total energy is distributed equally among all energy modes. While the translational and rotational motion contributes \(1 / 2 \mathrm{K_BT}\) to the total energy, vibrational motion contributes \(2 \times 1 / 2 \mathrm{K_BT}=\mathrm{K_BT}\) since it has both kinetic and potential energy modes.