91Ó°ÊÓ

The equation is

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

Here, $∈$is the rotational constant, $k$is the Boltzmann constant, and $T$is the absolute temperature.

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

$Z_{rot}=\frac{kT}{∈}$

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

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

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

The equations are

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

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

$Z_{rot}=1+3exp\left(-\frac{2∈}{kT}\right)+5exp\left(-\frac{6∈}{kT}\right)+7exp\left(-\frac{12∈}{kT}\right)+...101exp\left(-\frac{2550∈}{kT}\right)$

Substitute $107.7$for $\frac{kT}{∈}$in the above equation.

$Z_{rot}=1+3exp\left(-\frac{2}{107.7}\right)+5exp\left(-\frac{6}{107.7}\right)+7exp-\frac{12}{107.7}+...101exp\left(-\frac{2550}{107.7}\right)$

$=108.03$

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