What is the degree of freedom for O2?

1 Answer
Apr 5, 2018

If you mean, what degrees of freedom #N# there are for #"O"_2#, it is a linear diatomic molecule, which can clearly move around in three dimensions, so #N_(tr) = 3#.

Also, its rotation can be described by two angles (#theta,phi#) since it is linear, so #N_(rot) = 2#. (A nonlinear polyatomic molecule would need three angles #theta, phi, gamma#.)

There is only one vibrational mode, so there is one vibrational degree of freedom. But at ordinary temperatures, it is in the ground vibrational state (#Theta_(vib) = (tildeomega_e)/(0.695 "cm"^(-1)"/K") = "2273.7 K"#), where #tildeomega_e = "1580.193 cm"^(-1)#, so we ignore the vibrational contribution if we wanted the heat capacity #C_V#.

Therefore, to a good approximation, by omitting #N_(vib)#,

#N = N_(tr) + N_(rot) + N_(vib) ~~ 3 + 2 + 0 = color(blue)(5)#

For example, in actuality, #C_V = N/2R = (5/2 + 0.0283) R = "21.02 J/mol"cdot"K"# for #"O"_2#, and not #6/2R = "24.94 J/mol"cdot"K"# at #"298 K"#, so it is a good thing we omitted #N_(vib)#. Now we are much closer to the actual value of #"21.07 J/mol"cdot"K"#.