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

Box containing particles and be left alone

  1. Feb 8, 2006 #1
    Suppose I have a box containing particles and be left alone, with a total amount of energy, E. Hence, at equilibrium, these particles will obey the Maxwell-Boltzmann distribution. The entropy of the system is,

    [tex]S=NkT lnV[/tex]

    where T is the temperature of the system. V = volume of box.

    Suppose now, I divide this box equally into two, hence, now, each new box's volume is V/2, and the number of particles in each box is just N/2. But the temperature of these two new systems will stay the same, since they were in equilibrium initially.
    Now, the new entropy of both systems would be:

    [tex] S_{new}=\frac{N}{2}kTln V/2 + \frac{N}{2}kTln V/2 [/tex]
    [tex] S_{new} = NkT ln V - NkTln 2 [/tex]

    But, [tex]S_{new} < S[/tex]

    Where did I go wrong?
    Last edited: Feb 8, 2006
  2. jcsd
  3. Feb 9, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    You're adressing the famous "Gibbs paradox" in statistical mechanics (there's actually no paradox), but you're using an incorrect formula for entropy...

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook