Why $pV=\frac13Nm\langle c^2\rangle$ can be applied to diatomic molecules?

Question

For a monoatomic molecule, $$ \frac{1}{2} m \langle c^2\rangle = \frac{3}{2}kT. $$ If I multiply by total number of molecules, $$ \frac{1}{2} Nm \langle c^2\rangle = \frac{3}{2} NkT = \frac{3}{2} pV. $$ Dividing both sides by $3/2$, $$ \frac{1}{3} Nm \langle c^2\rangle = pV\,. $$ However, for diatomic molecules,

$m \langle c^2\rangle/2$ is not $3kT/2$. The energy of the molecules is instead given by equipartition theorem.

So I was expecting a different equation for diatomic molecules, where the coefficient $1/3$ is $1/f$, $f$ being the degree of freedom. However, my notes say that $Nm \langle c^2\rangle/3 = pV$ can apply for diatomic molecules so now I'm confused. What's wrong with my reasoning?

Answer 1

"$\tfrac 12 m\overline{c^2}$ is not $\tfrac 32 kT$"

Yes it is. There is mean energy $\tfrac 12 kT$ per degree of freedom, and a molecule, monatomic or diatomic, has 3 degrees of translational freedom.

"The energy of the molecules is instead given by equipartition theorem."

Yes, but without the "instead". The equipartition theorem tells us that, for a diatomic molecule, rotation gives us 2 more degrees of freedom, each with a mean energy of $\frac 12 kT$.[Quantum considerations tell us that rotation about the line joining the atoms' centres won't usually happen.]

So your equation, $pV =\frac 13 Nm\overline{c^2}$ holds as well for a diatomic gas as a monatomic, under ideal gas conditions (few molecules per unit volume).

What IS different for ideal monatomic and diatomic gases is their internal energies: $\frac 32 NkT$ and $\frac 52 NkT$.