mean vs rms hard

Question

For any Maxwell gas: v_rms compared to v_avg is:

RankGuru · Accepted Answer

As established in KTG-008:
- $v_{rms} = \sqrt{3} \sqrt{\frac{k_B T}{m}} \approx 1.732 \dots$
- $v_{avg} = \sqrt{8/\pi} \sqrt{\frac{k_B T}{m}} \approx 1.595 \dots$
Mathematically, since $\sqrt{3} > \sqrt{8/\pi} \approx \sqrt{2.54}$, $v_{rms}$ is always greater than $v_{avg}$ for any gas following the Maxwell distribution.

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mean vs rms hard

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