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What is the equation for the speed of a molecule at a specific temperature? I saw two equations $v = \sqrt{\frac{3 k T}{m}}$ and $v =\sqrt{\frac{3RT}{m}}$. What is the difference?

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    $\begingroup$ For ideal gases, average kinetic energy equals to $E=\frac{3}{2}kT=\frac{1}{2}mv^{2}$ where k is equal to the ideal gas constant divided by Avagadro's number $\frac{R}{N_A}$ $\endgroup$
    – Sancol.
    Commented Mar 5 at 17:41
  • $\begingroup$ This link may help physics.stackexchange.com/questions/45838/how-to-deduce-e-3-2kt $\endgroup$
    – Sancol.
    Commented Mar 5 at 17:47

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The speed of a molecule at a specific temperature can be described by the root-mean-square speed equation, which is given by $v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$, where $k$ is the Boltzmann constant, $T$ is the temperature in Kelvin, and $m$ is the mass of the molecule in kilograms.

The equation $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ is another form of the root-mean-square speed, where $R$ is the universal gas constant, $T$ is the temperature in Kelvin, and $M$ is the molar mass of the gas in kilograms per mole (kg/mol).

The difference between the two equations lies in the use of $k$ instead of $R$ and $m$ instead of $M$. I think you assumed that both equations have mass $m$ in the denominator! But if you use $R$ you need to use the molar mass $M$!

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A molecule does not have a temperature, only a gas of molecules in thermal equilibrium has. The molecules in such a gas can have a wide range of speeds. The speeds obey a Maxwell-Boltzmann distribution with average speed determined by $<v^2>=3kT/m$.

Distribution of particle speed for 1 million oxygen particles at -100, 20 and 600 °C

Distribution of particle speed for 1 million oxygen particles at -100, 20 and 600 °C. From https://en.m.wikipedia.org/wiki/Maxwell–Boltzmann_statistics.

Note on ‘oxygen particles’ added to the Wikipedia talk page:

“In the figure caption ‘oxygen particles’ are mentioned. It should be specified whether atoms, molecules, ozone or pieces of frozen oxygen 😀 are meant. If these are not atoms, rotation and vibration come into play. If they are, what’s keeping them from forming dimers?”

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