Small Group Activity: Heat capacity of N2

Contemporary Challenges 2021
Students sketch the temperature-dependent heat capacity of molecular nitrogen. They apply the equipartition theorem and compute the temperatures at which degrees of freedom “freeze out.”

In the prompt, explain that 50, 500 and 5000 should be equally spaced, for instance. 2, 5, and 10 are also approximately equally spaced.

As it turns out, the heat capacity is about $\frac52$ for almost the entire temperature range.

For N2 gas molecules in a 10 cm cubic box, the rules of QM dictate the discrete allowed values for:
1. Translational K.E. in one dimension: $\left\{ 1\times 10^{-40}\text{ J}, 4\times 10^{-40}\text{ J}, 9\times 10^{-40}\text{ J}, \ldots \right\}$
2. Rotational K.E.: $\left\{ 0\text{ J}, 0.8\times 10^{-22}\text{ J}, 0.8\times 10^{-22}\text{ J}, 0.8\times 10^{-22}\text{ J}, 2.5\times 10^{-22}\text{ J}, \ldots \right\}$
The bond length of N2 is 1.1 Å. Its mass is $m = 14\text{ amu}\approx 2\times 10^{-26}\text{ kg}$. The moment of inertia is thus \begin{align} I &= 2\cdot 2\times 10^{-26}\text{ kg} \left(0.55\times 10^{-10}\text{ m}\right)^2 \\ &\approx 1.2\times 10^{-46}\text{ kg}\cdot\text{m}^2 \end{align} From which we can find the energy eigenvalues: \begin{align} E_{\ell} &= \frac{\hbar^2}{2I}\ell(\ell+1) \\ &= \frac{\left(10^{-34} \text{ J}\cdot\text{s} \right)^2 }{2\cdot 1.2\times 10^{-46}\text{ kg}\cdot\text{m}^2} \ell(\ell+1) \\ &\approx \ell(\ell+1)\cdot 0.4\times 10^{-22}\text{ J} \end{align}
3. Vibrational energy: $\left\{ 2.3\times 10^{-20}\text{ J}, 6.9\times 10^{-20}\text{ J}, 11.5\times 10^{-20}\text{ J}, \ldots \right\}$
I looked up the experimental vibrational frequency, which is 2358 cm$^{-1}$.
Sketch a graph of $\frac{dU}{dT}$ of $10^{22}$ molecules of N2 gas in the temperature range of 70 K (the temperature at which $N_2$ becomes liquid at 1 atm of pressure) to 5000 K (at which temperature N2 breaks apart).

(use a logarithmic temperature axis)

Keywords
equipartition quantum energy levels
Learning Outcomes