Entropy

Remarks

By the Clausius theorem, $\oint \delta Q / T = 0$ around every reversible cycle, so $S(A)$ does not depend on the choice of $\gamma$. So $S$ is a state function, and $dS = \delta Q_{\mathrm{rev}} / T$. For $n$ moles of ideal gas expanded reversibly at constant $T$ from $V_1$ to $V_2$, $Q = nRT \ln(V_2 / V_1)$, so $\Delta S = nR \ln(V_2 / V_1)$. The free expansion of the same gas into a vacuum is not reversible, so the formula does not apply to it. There $Q = 0$, but $\Delta S = nR \ln(V_2 / V_1) > 0$, as computed along a reversible path. In statistical mechanics, $S = k_B \ln \Omega$, where $\Omega$ counts the microstates and $k_B$ is the Boltzmann constant.