1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Ideal Gases

Tags:
  1. Mar 30, 2016 #1

    MMS

    User Avatar

    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. jcsd
  3. Mar 30, 2016 #2
    Yes, in fact kinetic energy is proportional to temperature for an ideal gas:

    [tex] KE = \frac{3}{2} k T [/tex]
     
  4. Mar 30, 2016 #3
    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 .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Ideal Gases
  1. Ke of ideal gases (Replies: 7)

  2. Ideal Gases and Work (Replies: 2)

  3. Ideal Gases (Replies: 4)

Loading...