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6.26

Expert-verifiedFound in: Page 236

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

For a $CO$ molecule, the constant $\u20ac$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

The rotational partition function of a heterogeneous diatomic molecule

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 ${Z}_{rot}=\frac{kT}{\in}$

${Z}_{rot}=\frac{(8.617\times {10}^{-5}eV/K)(300K)}{0.00024eV}\phantom{\rule{0ex}{0ex}}=107.7$

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

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 $108.03$

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