Question

The first excited vibrational energy level of diatomic chlo- rine (Cl2) is 558 cm^-1 above the ground state. Wave- numbers, the units in which vibrational frequencies are usually recorded, are effectively units of energy, with 1 cm ^-1 = 1.986445 X 10^-23 J. If every vibrational energy level is equally spaced, and has a degeneracy of 1, sum over the lowest 4 vibrational levels to obtain a vibrational partition function for chlorine.
Required:
Determine the populations of each of the four levels at 298 K and the average molar vibrational energy (Em.vib) for chlorine at 298 K.

Answer 1

Answer:

The answer is "0.0000190 and 2.7 J".

Explanation:

$\to P_4=\frac{e^{-\beta}(4+\frac{1}{2}) hev}{2vb}$

         $=\frac{e^{-9(1.35)}}{0.278v}\\\\=\frac{e^{-12.5}}{0.278}\\\\=\frac{0.000005285}{0.278}\\\\=0.0000190$

Given:

$h=6.626 \times 10^{-34}\\\\c=3\times 10^{10}\\\\v=558\ cm^{-1}\\\\K=1.38 \times 10^{-23}\\\\ T=298$

$E=\frac{hcv}{KT}$

by putting the value into the above formula so, the value is 2.7 J