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!

Why is the gas energy in a g-field 5/2KNkT ?

  1. Mar 19, 2013 #1
    I am messing around with classical gasses using a demon algorithm, trying to reproduce the behavior of ideal gasses in various situations. Exitingly, this is working very well. Here are my findings so far:

    Classical gas, 3D : E=1.497 NkT - the usual 3/2 NkT result

    Classical gas, 2D : E=0.989 NkT - 2 DoF, so 2/2 NkT

    Relativistic gas, 3D : E = 2.992 NkT - as expected, can be derived from equipartition

    Classical gas, uniform gravitational field, 3D : E = 2.486 NkT

    Now the last one quite surprises me. Why is that he case? Intuitevely I would expect 3 DoF on |p| and 3 DoF on x, so 6*1/2 NkT = 3NkT. Is there a hidden constraint?
     
  2. jcsd
  3. Mar 20, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The equipartition theorem holds true only for phase-space degrees of freedom that enter quadratically in the Hamiltonian. For an ideal gas in a homogeneous gravitational field you have for the single-particle Hamiltonian
    [tex]H=\frac{\vec{p}^2}{2m}+m g z.[/tex]
    The classical probility distribution is thus given by
    [tex]P(\vec{x},\vec{p})=\frac{1}{Z} \exp[-\beta H(\vec{x},\vec{p})].[/tex]
    The mean kinetic energy for particles in a cubic box of length [itex]L[/itex] thus is
    [tex]\langle E_{\text{kin}} \rangle=\frac{3}{2} T[/tex]
    and the mean potential energy
    [tex]\langle m g z \rangle=m g L \left (1+\frac{1}{\exp(m g L/T)-1} \right )-T.[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why is the gas energy in a g-field 5/2KNkT ?
Loading...