I Ideal Gases: O2 & H2 - Kinetic Energy Comparison

AI Thread Summary
In the discussion about the kinetic energy of ideal gases O2 and H2 at thermal equilibrium, it is clarified that while both gases can have the same average kinetic energy due to being at the same temperature, their individual kinetic energies differ due to their distinct masses and velocities. The average kinetic energy for ideal gases is proportional to temperature, expressed as KE = (3/2) k T. The confusion arises from the relationship between mass and velocity, where the kinetic energy formula incorporates velocity squared. Thus, while the average kinetic energy is equal, the individual kinetic energies of O2 and H2 are not the same. Understanding these distinctions is crucial for accurate comparisons of kinetic energy in different gases.
MMS
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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.
 
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Yes, in fact kinetic energy is proportional to temperature for an ideal gas:

KE = \frac{3}{2} k T
 
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MMS said:
multiplication of both terms (since the energy is given by half m(gas)*v(gas)) gives an equal number.
Did I get this right?

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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