# How would the equipartition theorem be used to estimate the average kinetic energy of molecules?

May 17, 2017

At high enough temperatures,

$\left\langle\kappa\right\rangle \equiv {K}_{a v g} / \left(n {N}_{A}\right) \approx \frac{N}{2} {k}_{B} T$

where $\left\langle\kappa\right\rangle$ is the average molecular kinetic energy in $\text{J/molecule"cdot"K}$, $n$ is the mols, ${N}_{A}$ is Avogadro's number in ${\text{mol}}^{- 1}$, $N$ is the number of degrees of freedom, ${k}_{B}$ is Boltzmann's constant in $\text{J/K}$, and $T$ is temperature in $\text{K}$.

At room temperature, this holds for simple molecules such as ${\text{N}}_{2}$ and ${\text{O}}_{2}$, but overestimates for more complicated molecules like ${\text{CH}}_{4}$ and ${\text{NH}}_{3}$.

THE EQUIPARTITION THEOREM AT THE CLASSICAL LIMIT

Well, the equipartition theorem for the free particle at high enough temperatures is:

$\boldsymbol{{E}_{a v g} \approx {K}_{a v g} = \frac{N}{2} n R T}$,

or

${K}_{a v g} = \frac{N}{2} \cdot \stackrel{\text{Number of Molecules}}{\overbrace{n {N}_{A}}} \cdot {k}_{B} T$

where:

• $N$ is the number of degrees of freedom in the molecule.
• $n$ is the $\text{mols}$ of substance.
• ${N}_{A} = 6.0221413 \times {10}^{23} {\text{mol}}^{- 1}$ is Avogadro's number.
• ${k}_{B} = 1.38065 \times {10}^{- 23} \text{J/K}$ is the Boltzmann constant.
• $T$ is the temperature in $\text{K}$.

If we define $\left\langle\kappa\right\rangle = {K}_{a v g} / \left(n {N}_{A}\right)$ as the average molecular kinetic energy, then:

$\textcolor{b l u e}{\left\langle\kappa\right\rangle = \frac{N}{2} {k}_{B} T}$

EXAMPLE USING NITROGEN MOLECULE

For instance, let's say we were looking at ${\text{N}}_{2}$, a linear molecule. Here is how we could determine its degrees of freedom:

• $3$ translational dimensions ($x , y , z$)
$\to$ $3$ translational degrees of freedom
• $2$ rotational angles ($\theta , \phi$ in spherical coordinates) for linear molecules
$\to$ $2$ rotational degrees of freedom

(it would have been $3$ for nonlinear molecules)

• Hardly any vibrational degrees of freedom, because of a very stiff triple bond.

So, for ${\text{N}}_{2}$, we expect its average kinetic energy to be:

$\textcolor{b l u e}{\left\langle\kappa\right\rangle} \approx \frac{3 + 2 + 0}{2} {k}_{B} T$

$= \textcolor{b l u e}{\frac{5}{2} {k}_{B} T}$ $\textcolor{b l u e}{\text{J/molecule"cdot"K}}$

or,

$= \textcolor{b l u e}{\frac{5}{2} R T}$ $\textcolor{b l u e}{\text{J/mol"cdot"K}}$

CHECKING LITERATURE VALUES

We can't really look up the average molecular kinetic energy, but we can check by looking at the constant-pressure molecular heat capacity:

${\mathbb{C}}_{P} \equiv {C}_{P} / \left(n {N}_{A}\right)$

$\approx \frac{N + 2}{2} {k}_{B}$ $\text{J/molecule"cdot"K}$,

or the constant-pressure molar heat capacity:

${\overline{C}}_{P} \equiv {C}_{P} / n$

$\approx \frac{N + 2}{2} R$ $\text{J/mol"cdot"K}$.

From the $5$ degrees of freedom we found, we can approximate:

${\overline{C}}_{P} \approx \frac{5 + 2}{2} R = \frac{7}{2} R \approx \underline{\text{29.10 J/mol"cdot "K}}$

From NIST, we have the graph for ${\overline{C}}_{P}$: which shows $\underline{\text{29.124 J/mol"cdot"K}}$ at $\text{299 K}$, around 0.08% error.