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There are 2 equations to kinetic energy: $\frac{1}{2} mv^2$ and $\frac{3}{2} kT$.

By the first equation kinetic energy depends on mass, while by the second one it doesn't. How comes? So does it depend on mass? Will, for example mol Oxygen and mol Nitrogen have the same kinetic energy in the same temperature? Thank you very much!

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    $\begingroup$ I would suggest looking up what those two expressions are actually for. $\endgroup$ – BioPhysicist Mar 23 '19 at 19:40
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You are mixing two different concepts of kinetic gas theory here (see Wikipedia: Kinetic theory of gases: Temperature and kinetic energy):

  • $\frac{1}{2}mv^2$ is the kinetic energy of a single molecule.
  • $\frac{3}{2}kT$ is the average kinetic energy per molecule, when you have many of them.

Answering your question about Oxygen and Nitrogen: Although Oxygen and Nitrogen molecules have different masses $m$, for a given temperature $T$ they have on average the same kinetic energy.

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  • $\begingroup$ Worth noting here that we're talking about the RMS rather than the arithmetic mean. $\endgroup$ – dmckee --- ex-moderator kitten Mar 23 '19 at 21:22
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In this context $\frac{1}{2}mv^2 = \frac{3}{2}kT$. So the thermal kinetic energy here does not depend on mass. It is pinned to the temperature. At fixed velocity, the kinetic energy of a massive object does depend on mass.

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    $\begingroup$ If you want to make this identification (which is a good idea at times) you should note that the velocity that appears in the equality is the RMS velocity for particles of that mass in the gas: $\frac{1}{2} m v_\text{RMS}^2 = \frac{3}{2} k T$. $\endgroup$ – dmckee --- ex-moderator kitten Mar 23 '19 at 21:21

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