91Ó°ÊÓ

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

The rotational partition function of a heterogeneous diatomic molecule is,

$Z_{\text{rot}}=\underset{j=0}{\overset{∞}{∑}}(2j+1)\exp \left(\frac{-j(j+1)∈}{kT}\right)$

At higher temperatures, for $kT≫>∈$, the rotational partition function becomes as follows,

$Z_{\mathrm{rot}}=\frac{kT}{\mathrm{Ï\mu }}$

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

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

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

The rotational partition function of a heterogeneous diatomic molecule is,

$Z_{\mathrm{rot}}=\underset{j=0}{\overset{∞}{∑}}(2j+1)\exp \left(\frac{-j(j+1)∈}{kT}\right)$

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

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

Substitute $107.7$for $\frac{kT}{\mathrm{Ï\mu }}$ 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 $108.03$.