A gas of molecular hydrogen $\mathrm{H}_{2}$, is originally in equilibrium at a temperature of $1,000 \mathrm{~K}$. It is cooled to $20 \mathrm{~K}$ so quickly that the nuclear spin states of the molecules do not change, although the translational and rotational degrees of freedom do readjust through collisions. What is the approximate internal energy per molecule in terms of temperature units $\mathrm{K}$ ?

Note that the rotational part of the energy for a diatomic molecule is $A l(l+1)$ where $l$ is the rotational quantum number and $A \sim 90 \mathrm{~K}$ for $\mathrm{H}_{2}$. Vibrational motion can be neglected.
Originally the temperature is high and the para- and orthohydrogen molecules are in equilibrium in a ratio of about $1: 3$. When the system is quickly cooled, for a rather long period the nuclear spin states remain the same. The ratio of parahydrogen to orthohydrogen is still $1: 3$. Now the para- and orthohydrogen are no longer in equilibrium but, through collisions, each component is in equilibrium by itself. At the low temperature of $20 \mathrm{~K}$, $\exp (-\beta A) \sim \exp (-90 / 20) \ll 1$, so that each is in its ground state. Thus $\bar{E}_{r, \mathrm{p}}=0, \bar{E}_{\mathrm{r}, \mathrm{o}}=A(1+1) \cdot 1=2 A=180 \mathrm{~K}$, giving

$$
\bar{E}_{\mathrm{r}}=\frac{1}{4} \bar{E}_{\mathrm{r}, \mathrm{p}}+\frac{3}{4} \bar{E}_{\mathrm{r}, \mathrm{O}}=135 \mathrm{~K} .
$$

From equipartition of energy, we have

$$
\bar{E}_{\mathrm{t}}=\frac{3}{2} k T=30 \mathrm{~K} .
$$

The average energy of a molecule is

$$
\bar{E}=\bar{E}_{\mathrm{t}}+\bar{E}_{\mathrm{r}}=165 \mathrm{~K} .
$$