The analysis of this section applies also to linear polyatomic molecules, for which no rotation about the axis of symmetry is possible. An example is CO2, with ϵ=0.000049eV. Estimate the rotational partition function for a CO2 molecule at room temperature. (Note that the arrangement of the atoms is OCO, and the two oxygen atoms are identical.)

Question

Brittany Patton · Accepted Answer

Consider a carbon dioxide CO2 with energy constant ¢ = 0.000049 eV the two oxygen atoms are identical a round C, so we still can use equation 6.33 to find the rotational partition function:    Z→t=kT2ϵ    substitute with the givens at temperature of T = 300 K, (note that the  Boltzmann constant in eV is k=8.617×10−5eVK) to get:    Zrot=(8.617×10−5eVK)(300K)2(0.000049ev)=263.78    Zrot=263.78

The analysis of this section applies also to linear polyatomic molecules, for which no rotation about the axis of symmetry is possible.

Answered question

Answer & Explanation

New Questions in Other