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$$S=k\ln \Omega$$

What assumptions are being made about the system/process that allow for using those expressions?

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In your first formula, you cannot use a derivative $dQ$, but a variation $\delta Q$. And this is only true for a reversible process, that is $dS = \frac{\delta Q_{Reversible}}{T}$. Second formula is correct. – Trimok Jun 11 '13 at 18:30
I'm interested in the precise ramifications - for example I've read something about the second formula being a result of equal probabilities for all microstates, else it is replaced by Gibbs entropy. I have a feeling that the second one, being sort of 'thermodynamic' (and so, inherently "macroscopic") has an even tighter region of usability. – DepeHb Jun 11 '13 at 18:38
Not at all, this is a fundamental formula, which states that the entropy is proportionnal to the logarithm of the number of configurations. See the grave of Boltzmann – Trimok Jun 11 '13 at 18:44
Each microscopic configuration is equiprobable, the entropy itself is a macroscopic quantity – Trimok Jun 11 '13 at 18:48
Then I don't understand what is meant here:… (admittedly its a wikipedia article so I know it should be taken with a grain of salt, but still something's got to give?) – DepeHb Jun 11 '13 at 19:17

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