 Solution

Entropy is a thermodynamic state function. For a reversible heat transfer, $dS=\delta Q_{\mathrm{rev}}/T$; the Clausius inequality generalizes this to irreversible processes. The combined entropy of a system and its surroundings does not decrease. A subsystem can lose entropy by exporting heat or matter, so the second-law statement must specify the whole isolated combination. In a simple compressible system,which identifies temperature, pressure and chemical potential as derivatives of the energy representation of the state function. Thermodynamic equilibrium maximizes entropy under fixed total energy, volume and particle number. A heat reservoir instead leads to minimizing the Helmholtz free energy at fixed $T,V,N$, or the Gibbs free energy at fixed $T,P,N$.

Statistical mechanics explains the state count. In a microcanonical ensemble, equally probable compatible microstates give Boltzmann's entropy formula, $S=k_B\ln \Omega$. With general discrete probabilities, the Gibbs entropy isThe microcanonical formula follows by setting every nonzero probability to $1/\Omega$. For a canonical ensemble, $p_j=e^{-\beta E_j}/Z$, soThis turns microscopic energy levels and their populations into a measurable thermodynamic quantity. Classical phase-space counting needs a consistent microscopic normalization and indistinguishability convention; quantum state counting supplies that normalization. Independent systems have multiplicative state counts and therefore additive entropies.

For an ideal gas freely expanding from volume $V$ to $2V$ in an insulated enclosure, energy remains fixed while each particle gains twice the accessible position volume. The state count increases by $2^N$, giving $\boxed{\Delta S=N{k}_B\ln 2}$. The actual irreversible process transfers no heat. A reversible isothermal comparison path transfers heat $N{k}_{B}T\ln 2$ and computes the same entropy change because the initial and final thermodynamic equilibrium states are the same.

Calorimetry supplies another connection to experiment: at fixed pressure and within one phase, $S(T)-S(0)={\int }_0^T{C}_P(T^{\prime })d{T}^{\prime }/T^{\prime }$, with latent-heat contributions $\mathcal{L}/T$ at first-order transitions. The third law of thermodynamics fixes $S(0)=0$ for an ideal perfect crystal with a unique ground state. Degeneracy or frozen disorder can produce residual entropy. Entropy thus links heat-capacity measurements, phase thermodynamic equilibrium and the number of accessible microscopic configurations.