Chemistry - According to KMT, is the velocity of an ideal gas always sqrt(3RT/M)?

Solution 1:

$E=\frac 12mv^2 \implies v=\sqrt{\frac{2E}{m}} $ is valid for translational kinetic energy and the speed of the centre of mass.

Vibrational or rotational energy does not count. An object may vibrate or rotate even if it's centre of mass has zero speed.

As each available degree of freedom has the mean energy $E=\frac 12kT$, and as there are 3 independent translational degrees of freedom, the mean translational kinetic energy is $E_\mathrm{transl}=\frac 32kT \le E_\mathrm{kin}$

and the respective $v_\mathrm{RMS}=\sqrt{\frac{2E_{transl}}{m}}= \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$

even for ideal gas consisting of multiatom molecules.

Solution 2:

According to the equipartition theorem, energy is shared equally among all accessible degress of freedom.

A monatomic ideal gas has three translational degrees of freedom, so each translational degree of freedom has an energy of $\frac12 k T$, for a total translational energy of $\frac32 k T$.

A diatomic ideal gas has three translational degrees of freedom and two rotational degrees of freedom. At the same temperature, it thus has the same translational kinetic energy as a monatomoic ideal gas ($\frac32 k T$) and thus, for the same mass, the same average speed.

However, its two rotational degrees of freedom give it an additional $2\times\frac12 k T = k T$ of rotational kinetic energy. As a result, the heat capacity of a diatomic ideal gas is higher, because to heat it to the same temperature requires more energy, because you have to put energy not only into the three translational modes, but also into the two rotational modes.