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Molecular nitrogen is heated in an electric arc, and it is found spectroscopically that the relative populations of excited vibrational levels is

$f_0/f_0=1.0; f_1/f_0=0.2; f_2/f_0=0.04; f_3/f_0=0.008; f_4/f_0=0.002$.

Is the nitrogen in thermodynamic equilibrium with respect to vibrational energy? What is the vibrational temperature of the gas? Is this necessarily the same as the translational temperature?

I found this problem in Mcquarrie's Statistical Mechanics and didn't understand the role of the electric arc for the heating of the gas. Knowing that the vibrational temperature for $N_2$ is 3374 K and the relative population of all excited states are given by

$$ f_{n>0}=\sum_{n=1}^{\infty}\frac{e^{-\beta h\nu(n+1/2)}}{q_{vib}}=1-f_0=e^{-\beta h\nu}=e^{-\Theta_v/T} $$

When a molecule is heated, the ratio $\Theta_v/T$ is lower at high temperatures and the population of the vibrational excited states is augmented.

So, if $N_2$ is placed in an electric arc for heating, there exists some perturbation on vibrational energy levels, like a Stark effect, that makes a perturbation on the thermodynamic equilibrium?

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  • $\begingroup$ No. You checked what you were supposed to check. The arc is not heating the gas, the function of the electric discharge is to make electronic excitations. The vibrational spectrum does not need to be thermal: the distribution would start out as Franck-Condon factors. $\endgroup$
    – user137289
    Dec 24 '16 at 20:33
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One can see by inspection that the vibrational occupancy numbers are an arithmetic series. A harmonic oscillator has evenly spaced energy levels, so one can conclude that the occupancy probability is given by the Boltzmann factor for some vibrational temperature. To calculate that temperature, one would need to know the vibrational frequency of the excited state.

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