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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


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

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