Some advances textbooks define entropy by the formula
$S = - k \sum_{s} P (s) \ln P (s)$
where the sum runs over all microstates accessible to the system and $P (s)$ is the probability of the system being in microstate $s$ .
(a) For an isolated system, role="math" localid="1647056883940" $P (s) = \frac{1}{Ω}$ for all accessible states $s$ . Show that in this case the preceding formula reduces to our familiar definition of entropy.
(b) For a system in thermal equilibrium with a reservoir at temperature $T$ , role="math" localid="1647057328146" $P (s) = \frac{e^{- \frac{E (s)}{k T}}}{Z}$ . Show that in this case as well, the preceding formula agrees with what we already know about entropy.

Question

Some advances textbooks define entropy by the formula
$S = - k \sum_{s} P (s) \ln P (s)$
where the sum runs over all microstates accessible to the system and $P (s)$ is the probability of the system being in microstate $s$ .
(a) For an isolated system, role="math" localid="1647056883940" $P (s) = \frac{1}{Ω}$ for all accessible states $s$ . Show that in this case the preceding formula reduces to our familiar definition of entropy.
(b) For a system in thermal equilibrium with a reservoir at temperature $T$ , role="math" localid="1647057328146" $P (s) = \frac{e^{- \frac{E (s)}{k T}}}{Z}$ . Show that in this case as well, the preceding formula agrees with what we already know about entropy.

Accepted Answer

Part (a): $S = k \ln Ω$

Part (b): $S = \frac{E - F}{T}$

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