Ideal Gases: O2 & H2 - Kinetic Energy Comparison

Click For Summary
SUMMARY

The discussion clarifies the kinetic energy comparison between ideal gases O2 and H2 at thermal equilibrium. While both gases exhibit average kinetic energy proportional to temperature, their individual kinetic energies differ due to variations in mass and velocity. The formula for kinetic energy, KE = (3/2) k T, indicates that at the same temperature, average kinetic energies are equal, but individual values vary based on mass and root mean square (RMS) velocity. Thus, the assertion that their kinetic energies are equal is incorrect when considering individual gas properties.

PREREQUISITES
  • Understanding of ideal gas laws
  • Familiarity with kinetic energy formulas
  • Knowledge of root mean square (RMS) velocity
  • Basic principles of thermodynamics
NEXT STEPS
  • Study the ideal gas law and its applications
  • Learn about the derivation of kinetic energy formulas for gases
  • Explore the concept of root mean square (RMS) velocity in detail
  • Investigate the implications of temperature on gas behavior
USEFUL FOR

Students of physics, chemists, and anyone interested in the behavior of gases under thermal equilibrium conditions will benefit from this discussion.

MMS
Messages
146
Reaction score
4
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.
 
Science news on Phys.org
Yes, in fact kinetic energy is proportional to temperature for an ideal gas:

KE = \frac{3}{2} k T
 
  • Like
Likes   Reactions: MMS
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 .
 
  • Like
Likes   Reactions: DuckAmuck and MMS

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
6K
  • · Replies 1 ·
Replies
1
Views
826