# What is the equation for the speed of a molecule at a specific temperature?

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?

• 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 $\frac{R}{N_A}$ Commented Mar 5 at 17:41
• This link may help physics.stackexchange.com/questions/45838/how-to-deduce-e-3-2kt Commented Mar 5 at 17:47

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 $$M$$!

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 $$=3kT/m$$.

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?”