# I Ideal Gases

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1. Mar 30, 2016

### MMS

Hello,

I want to make sure I understand the following considering ideal gases.
Assuming I have two different types of gases, say, O2 and H2 (each at thermal equilibrium), is it correct to say that the kinetic energy of the O2 gas equals to the kinetic energy of the H2 gas since they're both ideal gases?
I see this as follows: Each of their masses is of course different and so is the (mean) velocity. However, the multiplication of both terms (since the energy is given by half m(gas)*v(gas)) gives an equal number.
Did I get this right?

Thanks in advance.

2. Mar 30, 2016

### DuckAmuck

Yes, in fact kinetic energy is proportional to temperature for an ideal gas:

$$KE = \frac{3}{2} k T$$

3. Mar 30, 2016

### drvrm

i think it should be velocity square- may be a typo.
if you are talking of averge K.E.(due to R.M.S.velocity) then the energy comes out to be (3/2) kT and if two gases are at same temp its average K.E. will be equal but individual kinetic energy depends on mass as well as velocity and will be different .

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