For a CO molecule, the constant $ϵ$is approximately $0.00024\mathrm{eV}$. (This number is measured using microwave spectroscopy, that is, by measuring the microwave frequencies needed to excite the molecules into higher rotational states.) Calculate the rotational partition function for a $\mathrm{CO}$molecule at room temperature $(300\mathrm{K})$, first using the exact formula $6.30$ and then using the approximate formula $6.31$.

We are given that the value of constant$\epsilon$is $0.00024\mathrm{eV}$.

The rotational partition function of a heterogeneous diatomic molecule is,

$Z_{\text{rot}}=\sum _{j=0}^{\infty }(2j+1)\exp \left(\frac{-j(j+1)\in }{kT}\right)$

At higher temperatures, for $kT\gg >\in$, the rotational partition function becomes as follows,

$Z_{\mathrm{rot}}=\frac{kT}{ϵ}$

Substitute $8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}$ for k, $300K$ for T, and $0.00024\mathrm{eV}$in the equation $Z_{\text{rot}}=\frac{kT}{ϵ}$, we get

$$\begin{gathered} Z_{\mathrm{rot}}=\frac{\left(8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)(300\mathrm{K})}{0.00024\mathrm{eV}} \\ =107.7 \end{gathered}$$

Therefore, the rotational partition function of a CO molecule is $107.7$.

The rotational partition function of a heterogeneous diatomic molecule is,

$Z_{\mathrm{rot}}=\sum _{j=0}^{\infty }(2j+1)\exp \left(\frac{-j(j+1)\in }{kT}\right)$

Expand the above summation from $j=0$to $j=50$,

role="math" $Z_{\mathrm{rot}}=1+3\exp \left(-\frac{2ϵ}{kT}\right)+5\exp \left(-\frac{6ϵ}{kT}\right)+7\exp \left(-\frac{12ϵ}{kT}\right)+\dots 101\exp \left(-\frac{2550ϵ}{kT}\right)$

Substitute $107.7$for $\frac{kT}{ϵ}$ in the above equation,

$$\begin{gathered} Z_{\text{rot}}=1+3\exp \left(-\frac{2}{107.7}\right)+5\exp \left(-\frac{6}{107.7}\right)+7\exp \left(-\frac{12}{107.7}\right) \\ =108.03 \end{gathered}$$

Hence, the exact value of rotational partition function of a CO molecule is $108.03$.