Question

For a $CO$molecule, the constant $€$is approximately $0.00024eV$.(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 $CO$molecule at room temperature $\left(300K\right)$, first using the exact formula 6.30 and then using the approximate formula 6.31

Short answer

The rotational partition function of a heterogeneous diatomic molecule

Step-by-step solution

Rotational partition function:

The equation is

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

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

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

$Z_{rot}=\frac{kT}{\in }$

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

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

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

The rotational partition function of a heterogeneous diatomic molecule:

The equations are

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

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

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

Substitute $107.7$for $\frac{kT}{\in }$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$108.03$