# Degrees of Freedom of water and $\rm CO_2$ at high temperatures

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?

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.

• 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? Mar 27, 2021 at 6:47
• The temperatures that define the treshold of each stages will be different. Their rotation eigen energies and vibration energies are substantially differemt.
– ytlu
Mar 27, 2021 at 6:52