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So I know we have the Gibbs Entropy Formula $$S=k\sum_i p_i \ln p_i$$ And I've seen the probabilities for the microstates be said to be the Boltzmann probabilities $$p_i\propto e^{-\frac{E_i}{kT}}$$ But I've also seen that to obtain Boltzmann's entropy formula you assume all states have the same probability so, if $\Omega$ is the number of accessible microstates then we have $$S=k \ln \Omega$$ So either the probability for all microstates is the same, it is proportional to the Boltzmann Factor, or I am missing a big chunk of information.

Please help.

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The Boltzmann's entropy formula is valid in the microcanical ensemble, where the number of particles $N$, the volume $V$ and the energy $E$ are fixed. In this case we postulate that all microstates are equally probable, and define $\Omega$ as the number of microstates that have an energy comprised between $E$ and $E + \Delta$, where $\Delta \ll E$.

If we are in the canonical ensemble, in which $N$, $V$ and temperature $T$ are fixed, things are a bit different. In particular, the Boltzmann relation for $S$ is not valid, as the entropy is now given by

$$ S = -\left( \frac{\partial F}{\partial T} \right)_{N,V} $$

where $F$ is the Helmholtz free energy. In this ensemble, the probability of a microstate, as you wrote, is proportional to the Boltzmann factor:

$$ p_i\propto e^{-\frac{E_i}{kT}} $$

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