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Ke of ideal gases

  1. May 5, 2003 #1
    when you equate the two formulas for ideal gases, one is evetually left with a formula to calculate the ke. of the ideal gas (3/2kt i think) how come the ke is independent of the mass of the molecule ?
  2. jcsd
  3. May 5, 2003 #2
    If you mix many balls with different mass in a box and shake the box, each ball will have the same average kinetic energy regardless difference in their mass.
  4. May 5, 2003 #3
    wait.... i got an idea, does the formula mean that although the ke. will be what the formula spits out, the velocities for heavier moleules (lets say of another ideal gas) would be lower, and thus the formula would be indepndent of mass.. ?
  5. May 5, 2003 #4
  6. May 5, 2003 #5
    A simple way to see it.

    The gas properties are defined by the pressure and temperature. The temperature is basically the mean energy per unit quantum, in this case atoms or molecules are the "quanta" involved, (this isn't the standard definition of quantum) and the pressure is the mean energy per unit volume. The heavier molecules move slower to have the same energy as the lighter ones.

    In non-ideal gasses some of the energy is tied up in rotational modes, which is why they have differing ratios of specific heat.
  7. May 6, 2003 #6
    thanks guys,
    i think i finally got it !
  8. May 6, 2003 #7
    Mobility of elastic gases

    Hi apache,
    I am attaching the table referred to below with the hope that the narrow field of this posting causes scrambling.


    The Mobilities according to Graham’s law are listed in the table above as referenced to the mobility of N2, the major molecular constituent of the atmosphere. Positive factors indicate propensities of given gases to rise while negative factors indicate falling tendencies. Those labeled “gas” are usually Brownian gases in that their boiling points are above ambient atmospheric temperatures. For example when solid Iodine sublimes, a maroon cloud hovers nearby until dispersed and/or condensed. It should be remembered that volatility is a function of boiling point temperature as contrasted with mobility, which is the inverse square root of the molecular mass.

    Attached Files:

    Last edited: May 9, 2003
  9. May 6, 2003 #8
    Re: A simple way to see it.

    Standard postulate here is "In thermal equilibrium energy is equally distributed among all available degrees of freedom (equipartition)", so temperature T is defined in such way that each degree has in the average kT/2 amount of energy. If a molecule is monoatomic, it has 3 degrees only(x,y,z), thus <E>=3kT/2, if diatomic then it has two more (rotational) degrees, thus <E>=5kT/2, and so on.
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