# 27.1: The Average Translational Kinetic Energy of a Gas

### Equipartitioning of Energy

When the mechanical energy is

$\mathcal{E}(x) = \mathcal{H}(x) = \sum_{i=1}^N \dfrac{\textbf{p}_i^2}{2m_i} + U \left( \textbf{r}_1, \ldots, \textbf{r}_N \right) \label{1}$

the canonical distribution function nicely separates into a product of momentum and coordinate distributions:

\begin{align} f(x) &= \dfrac{C_N}{Q(N, V, T)} \left[ e^{-\beta \sum_{i=1}^N \textbf{p}_i^2/2m_i} \right] \left[ e^{-\beta U \left( \textbf{r}_1, \ldots, \textbf{r}_N \right)} \right] \\ &= \dfrac{C_N}{Q(N, V, T)} \left[ \prod_{i=1}^N e^{-\beta \textbf{p}_i^2/2m_i} \right] \left[ e^{-\beta U \left( \textbf{r}_1, \ldots, \textbf{r}_N \right)} \right] \\ &= \left[ \prod_{i=1}^N P \left( \textbf{p}_i \right) \right] \left[ \dfrac{1}{Z} \: e^{-\beta U \left( \textbf{r}_1, \ldots, \textbf{r}_N \right)} \right] \\ &= \left[ \prod_{i=1}^N \mathcal{P} \left( p_{x, i} \right) \: \mathcal{P} \left( p_{y, i} \right) \: \mathcal{P} \left(p_{z, i} \right) \right] \left[ \dfrac{1}{Z} \: e^{-\beta U \left( \textbf{r}_1, \ldots, \textbf{r}_N \right)} \right] \end{align} \label{2}

where $$P \left( \textbf{p}_i \right)$$ is the Maxwell-Boltzmann distribution as a function of the momentum $$\textbf{p}_i$$.

Given this decomposition of the momentum distribution into products of distributions of individual components, we always have, for any system,

$\left< \sum_{i=1}^N \dfrac{\textbf{p}_i^2}{2m_i} \right> = \dfrac{3}{2} N k_B T \label{3}$

However, in addition to this, we also have the following averages:

$\left< \dfrac{\textbf{p}_i^2}{2m_i} \right> = \dfrac{3}{2} k_B T \label{4}$

and

$\left< \dfrac{p_{x, i}^2}{2m_i} \right> = \dfrac{1}{2} k_B T, \: \: \: \left< \dfrac{p_{y, i}^2}{2m_i} \right> = \dfrac{1}{2} k_B T, \: \: \: \left< \dfrac{p_{z, i}^2}{2m_i} \right> = \dfrac{1}{2} k_B T \label{5}$

The fact that these averages can be written for every individual component tells us that each degree of freedom has an average of $$k_B T/2$$ energy associated with it. This is called equipartitioning of energy.

How does equipartitioning happen? It occurs through the interactions in the system, which dynamically, produce collisions, leading to covalent bond breaking and forming events, hydrogen bond breaking and forming events, halogen bond breaking and forming events, diffusion, and so forth, all depending on the chemical composition of the system. Each individual collision event leads to a transfer of energy from one particle to another, so that, on average, all particles have the same energy $$k_B T/2$$. This does not mean that, at any instant, this is the energy of a randomly chosen particle. Obviously, there are considerable fluctuations in the energy of any one particle, however, the average over these fluctuations must produce the value $$k_B T/2$$.

A system obeys a well defined set of classical equations of motion, so that we can, in principle, determine exactly when the next collision will occur and exactly how much energy will be transferred in the collision. However, since we do not follow the detailed motion of all of the particles, our description of collisions and their consequences must be statistical in nature. We begin by defining a few simple terms that are commonly used for this subject.