RMS Speed of Gas Molecule for Polyatomic Molecules

Halliday in his book and also many people say that RMS speed, $v_{rms}$ is $\sqrt{\frac{3RT}{M}}$. However, he used this formula in showing that kinetic energy, $K$, is $\frac32kT$. but how about polyatomic gases such as $H_2$, $O_2$, etc. The polyatomic gas by its equipartition energy law or what so called has $K$ that is not $\frac32kT$, it should be $\frac52kT$ or maybe $\frac72kT$ and by that way, its RMS speed should be $\sqrt{\frac{5RT}{M}}$, or $\sqrt{\frac{7RT}{M}}$. But in that book, he showed that the speed of $H_2$ follow the "monoatomic gas" RMS speed. Does the equipartition energy law really exist? The data in Halliday

2 Answers

From the equipartition of energy, energy is equally distributed to every independent degree of freedom of the molecule. (i.e $\frac{kT}{2}$ per degree of freedom.)

For a monoatomic gas particle, there are three translational degrees of freedom, therefore its KE is $\frac{3kT}{2}$, and $v_{rms}$ can quite easily be shown to be $\sqrt{\frac{3RT}{M}}$.

For a polyatomic gas molecule, there are three translational degrees of freedom. The other degrees of freedom are vibrational/rotational.

Only the translational degrees of freedom contribute to the KE (translational) of the gas. The other degrees of freedom contribute to their respective energies. (i.e rotational or vibrational.)

This is why $v_{rms}$ is always equal to $\sqrt{\frac{3RT}{M}}$.

For example: A diatomic gas molecule has total energy $\frac{5kT}{2}$. Only $\frac{3kT}{2}$ corresponds to the translational kinetic energy, whereas $\frac{2kT}{2}$ (sigh) corresponds to the rotational kinetic energy. This gives us $v_{rms}$ as shown above.

The equipartition energy with freedom degree $>$ 3 is showing its vibrational and rotational move, not its translational move. That's why the RMS speed is always $\sqrt\frac{3RT}{M}$