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Calculating temperature from entropy

  1. Mar 10, 2012 #1
    Consider N >> 1 particles that can either be in groundstate [itex]\epsilon_0[/itex] or excited state [itex]\epsilon_1[/itex] and are thermally isolated, so the internal energy is fixed at [itex]U = (N - n) \epsilon_0 + n \epsilon_1[/itex]. We want to calculate the temperature of this system.

    This is how I attempt it: First calculate the multiplicity. For two particles, the possible microstates are -2, 0, 0, 2 and for 3 particles it's -3, -1, 1, 3, etc, so that the multiplicity becomes:
    [tex]\Omega(n) = \frac{N!}{\frac{N + n}{2}! \frac{N - n}{2}!}[/tex]
    The entropy is, using the Stirling approximation, [itex]S = k_B \ln{\Omega(n)} = k_B(N \ln N - \frac{N + n}{2} \ln{\frac{N + n}{2}} - \frac{N - n}{2} \ln{\frac{N - n}{2}})[/itex].

    The temperature we can calculate from [itex]T = \left(\frac{\partial S}{\partial U}\right)^{-1}[/itex]. This is where I don't know how to continue (not sure I am on the right track in the first place).
  2. jcsd
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