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It is a common textbook question to treat the root mean square velocities of an ideal gas as given by the following equation: $$v_\text{rms}=\sqrt{\frac{3RT}{M}}$$ I was wondering about the validity of this equation, particularly as it applies to gases that are not monatomic. For example, in the following question from Zumdahl et al., it says:

Calculate the root mean square velocities of $\ce{CH4}$ and $\ce{N2}$ molecules at $\pu{273 K}$ and $\pu{546 K}$.

The answer key gives $\pu{652 m/s}$ for methane at $\pu{273 K}$, $\pu{493 m/s}$ for nitrogen at $\pu{273 K}$, $\pu{922 m/s}$ for methane at $\pu{546 K}$, and $\pu{697 m/s}$ for nitrogen at $\pu{546 K}$. These were all calculated with the equation $$v_\text{rms}=\sqrt{\frac{3RT}{M}}$$

Why is this equation valid for all ideal gases? Shouldn't it be $$v_\text{rms}=\sqrt{\frac{6RT}{M}} \text{ for methane}$$ (because there are $3$ rotational degrees of freedom in addition to $3$ translational degrees of freedom), and $$v_\text{rms}=\sqrt{\frac{5RT}{M}} \text{ for nitrogen}$$ (because there are $2$ rotational degrees of freedom)?

If my thoughts are not true, then why is it commonly reported that the average molar specific heat of an ideal monatomic gas (at constant volume) is $3/2RT$, that of an ideal diatomic gas is $5/2RT$, and that of a nonlinear gas is $3RT$?

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  • $\begingroup$ By velocity i think it's meant to be velocity of centre of mass , so the rotatory velocity does not impart any velocity to centre of mass. $\endgroup$
    – aryan bansal
    Commented Jan 29, 2020 at 19:17

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$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.

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  • $\begingroup$ I am not too sure about your statement. Consider an ideal gas; the total internal energy is defined as $U = \text{PE} + \text{KE}$, where $\text{PE}$ and $\text{KE}$ are the potential and kinetic energies, respectively. Since in an ideal gas, there are no attractive or repulsive forces, $\text{PE} = 0 $. Are you saying that $U = \text{PE} + \text{KE} = 0 + 3/2kt$ for all gases? Surely this can't be true.... $\endgroup$
    – Cyclopropane
    Commented Jan 27, 2020 at 14:10
  • $\begingroup$ @DrPepper No, I do not say that. KE >= translational KE =3/2kT. ( Al as mean energy) $\endgroup$
    – Poutnik
    Commented Jan 27, 2020 at 14:24
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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.

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  • $\begingroup$ so the kinetic energy only takes into account the translational modes? $\endgroup$
    – Cyclopropane
    Commented Jan 27, 2020 at 5:34
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    $\begingroup$ @DrPepper No, the kinetic energy is the sum of the translational, vibrational, and rotational kinetic energy. The speed, however, is due only to the translational (as opposed to rotational or vibrational) kinetic energy. This is because translational kinetic energy is the energy the molecules have as a result of how fast they move (i.e., translate) through space (and their mass). $\endgroup$
    – theorist
    Commented Jan 27, 2020 at 5:46
  • $\begingroup$ Thank you so much! I understand it now :) Your comment was very helpful. $\endgroup$
    – Cyclopropane
    Commented Jan 27, 2020 at 14:14

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