Degrees of freedom of carbon dioxide based on quadratic energy terms

Question

We all learned that by theorem of equipartition of energy, the quadratic terms in position or velocity components contribute to the number of degrees of freedom. Therefore, for a diatomic molecule, the vibration degree of freedom is 2. As I understand it and in any text, the vibrational degrees of freedom in diatomic molecule is counted like this:

1. The first vibrational freedom comes from $\frac{1}{2}\mu v^2$ which contribute $\frac{1}{2}kT$ to the total energy, coming from the kinetic energy term.
2. The second vibrational freedom comes from $\frac{1}{2}k x^2$ which contribute $\frac{1}{2}kT$ to the total energy, coming from the potential energy of the spring.

However, there is a general formula for linear molecules to count the number of vibration degrees of freedom, which is $3N-5$, using this equation, the vibration degree of freedom comes out to be only $(3\times 2-5 =)\ 1$ which is confusing and does not equate to the above deduction.

Also, this is the same case with $CO_2$, if we look up number of vibrational degrees of freedom of $CO_2$, there are 4, corresponding to symmetric stretching, asymmetric stretching and two bending. If there are four modes of vibration, there must be 8 different quadratic components that contribute to the number of degrees of freedom coming from the quadratic potential and kinetic energy terms. Why this is not so? What am I missing? Please help!

Answer 1

You need to be consistent with your terminology. A diatomic molecule has 2 vibration degrees of freedom, but only one mode (it can only stretch/squeeze the bond). In general, a linear molecule with $N>1$ atoms has $3N-5$ vibration modes, but $2(3N-5)$ vibration degrees of freedom. Setting $N=2$, you do recover the consistency with the previous case. There is a factor $2$ between modes and degrees of freedom. This comes from the fact that each harmonic mode has a kinetic term and a potential term, each contributing for one degree of freedom. In terms of heat capacity, by the equipartition theorem, each term contributes to $\frac12k_BT$, so each mode contributes to $k_BT$.

For $CO_2$, just set $N=3$. You therefore have $4$ vibration modes. Two for each bond that elongates, two for bending the molecule in any direction. You therefore have $8$ vibration degrees of freedom to which you can add the $5$ degrees of freedom due to translation (3) and rotation (2), giving a total of $13$. For example, at very high temperature where all the degrees of freedom have unfrozen (the vibrational degrees of freedom unfreeze last), you would therefore expect a heat capacity of $\frac{13}2k_BT$. In practice, this line of reasoning works well for mono/di-atomic molecules, but starts to deviate at triatomic molecules like $CO_2$.