Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetic energy for a mole of a gas discrepancy

  1. Jul 22, 2011 #1
    Exam krackers says that the average kinetic energy for a mole of a gas is K.E.avg = 3/2 RT

    but my chemistry book says that the total kinetic energy for a mole of gas is = 3/2 RT
    meaning that
    NA (average K.E. of one molecule) = 3/2 RT


    These 2 statements are different, for the bottom one says that the total K.E. for a mole of gas is = 3/2 RT, yet the top one says that the average K.E. for a mole is =3/2RT

    Which is correct?
     
  2. jcsd
  3. Jul 22, 2011 #2

    Ygggdrasil

    User Avatar
    Science Advisor

    What units does the expression 3/2 RT have?
     
  4. Jul 22, 2011 #3
    R = 8.314 J/(K x mol)

    so 3/2 RT units are J/ mol


    Average kinetic energy for one molecule is: 1/2 m (mean square speed)...
    so the units for that are: (kg m2) / (s2 molecule)

    so this means my chemistry book is correct because

    (average kinetic energy for one molecule) (NA) works because the units work:

    since NA = molecules/ mol.... the multiplication equals 1 J/mol....


    right??

    So then why does my Exam Krackers book say that the average kinetic energy for a mole of gas = 3/2 RT...?
     
  5. Jul 22, 2011 #4

    Ygggdrasil

    User Avatar
    Science Advisor

    You are correct that (3/2)RT gives the total energy per mole of a monatomic ideal gas. The average kinetic energy of a molecule of that gas is given by a very similar formula, (3/2)kBT, where kB is the Boltzman constant. As you can probably guess, kB = R/NA.

    I'm not sure why your other book says that (3/2)RT is the average kinetic energy for a mole of gas.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinetic energy for a mole of a gas discrepancy
  1. Kinetic gas equation (Replies: 3)

Loading...