How do I calculate the degrees of freedom for $\rm CO_2$ and water at high temperatures? I'm confused because I know $\rm CO_2$ is linear while H2O is nonlinear, and am wondering how this would affect the total degrees of freedom.

I know for $\rm CO_2$ it would be 3 translational + 2 rotational + 4 vibrational+ 4 bending = 13.

For steam at high temperatures I am guessing it would it be 3 translational + 3 rotational + 3 vibrational = 9?


1 Answer 1


For rotation, bending, and vidration, they are different combinations of motions of the constituent atoms. These eigen modes are not of degree of freedom.

For $CO_2$ at very low temperature, while rotation modes are still frozen, the whole molecule moves rigidly as one single point particle, the activate degree of freedom is 3, reflecting a $C_v = \frac{3}{2} K T$.

As temperature rising to activate the rotational modes, there is additional 2 degree of freedom $3+2 = 5$, reflecting a $C_v = \frac{5}{2} K T$.

As higher temperature to activate the vibration modes, there is additional 2 degree of freedom $5 + 2 = 7$, reflecting a $C_v = \frac{7}{2} K T$. This is how we typically count the degree of freedom from tri-atomic molecule, $3\times 3 - 2 = 7$. Each atom hae 3 degrees of freedom, with two binding constrains.

As temperature high enough to break the molecule, each atom has 3 degree of freedom, $3 \times 3 = 9$, reflecting a $C_v = \frac{9}{2} K T$.

Similar argument can be applied to $H_2O$ molecule.

  • $\begingroup$ So would the degrees of freedom always be the same for water and carbon dioxide? Since a water molecule is a bent molecule, won't it have fewer symmetries, so more rotational degrees of freedom? $\endgroup$
    – pctree
    Mar 27, 2021 at 6:47
  • $\begingroup$ The temperatures that define the treshold of each stages will be different. Their rotation eigen energies and vibration energies are substantially differemt. $\endgroup$
    – ytlu
    Mar 27, 2021 at 6:52

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