# Kinetic energy of an ideal gas

I've heard that the kinetic energy of any different gases are the same

All gases at a given temperature have the same average kinetic energy.

But if we have 2 moles of hydrogen and 4 moles of helium in a container at a specific temperature will the kinetic energies become the same? I think it cannot be as the n is different..

As the kinetic energy is 3nrt/2

This might be a small question but this has been a conflict between my friends..

• What has more energy? Two stampeding bulls or four stampeding bulls? If you mixed two brown stampeding bulls and four mottled stampeding bulls, which bulls would have more total energy? The four? Or the two? – Curt F. Dec 4 '15 at 3:42
• Lol then I'm correct.. – Ramith Hettiarachchi Dec 4 '15 at 3:43
• Yeah, but only if you're interested in total energy. If you're interested in per particle energy, then the gases have the same energy. The per-bull energy of the two stampeding bulls is about the same as the per-bull energy of the four. – Curt F. Dec 4 '15 at 3:44
• So what about the average kinetic energies of these gases ? – Ramith Hettiarachchi Dec 4 '15 at 3:48
• The average would be the same because the four mole gas would have more energy in general but if you divide it by four moles, that would give you the average kinetic energy of the gas per particle, and it would be equal to the two mole gas' total kinetic energy divided by two moles. – phi2k Dec 4 '15 at 4:48

The answer is in front of you. $E_v$ is the Total translational kinetic energy not the Average kinetic energy. If you want average translational kinetic energy per mol this is given by: $E_v/n$ or $3/2RT$.