Second Law and Entropy §

Equivalence proof sketch. Assume Kelvin–Planck is false: a device converts $Q_C$ from a single reservoir at $T_C$ entirely into work $W = Q_C$. Use that work to drive a Carnot refrigerator between $T_C$ and $T_H$, pumping heat $Q_C'$ from $T_C$ to $T_H$. Net effect: heat flows from $T_C$ to $T_H$ with no other change — violating Clausius. The reverse argument works similarly.

Practical meaning. A heat engine must reject waste heat to a cold sink. A refrigerator must consume work (or a higher-temperature heat input) to pump heat upward. These are not engineering limitations; they are fundamental.

State function. Although $\delta Q$ is path-dependent, $\delta Q_{\text{rev}}/T$ is an exact differential. The entropy change between two states is the same regardless of the actual path; you compute it along any convenient reversible path connecting the same states.

Units. $[S] = \text{J/K}$. Molar entropy: J/(mol·K). Specific entropy: J/(kg·K).

Absolute entropy (third law). The third law states $S \to 0$ as $T \to 0$ for a perfect crystal. This gives entropy an absolute zero, unlike internal energy.

Statistical meaning. $S = k_B \ln W$ (Boltzmann) or $S = -k_B \sum_i p_i \ln p_i$ (Gibbs/von Neumann). See Statistical Foundations of Thermodynamics for the microstate counting.

Derivation sketch. Consider a system undergoing an irreversible process from state 1 to state 2. Connect state 2 back to state 1 by a reversible path. The total cycle is irreversible, so $\oint \delta Q/T < 0$. Splitting the integral:

The second integral is $-(S_2 - S_1) = -\Delta S$. Therefore $\Delta S > \int_1^2 \delta Q_{\text{irr}}/T$.

Isolated system. If $Q = 0$ (adiabatic, isolated), then $\Delta S \geq 0$. Entropy of an isolated system never decreases. This is the most common form used to judge spontaneity.

Connection to lost work. In any real (irreversible) process, some work potential is destroyed. The Gouy–Stodola theorem quantifies this: the exergy destroyed equals the environment temperature times the entropy produced. For a reversible process, $\sigma = 0$ and no exergy is destroyed. For an irreversible process, $\sigma > 0$ and the lost work is proportional to the irreversibility.

Engineering use. In cycle analysis, the entropy production rate identifies where in a power plant or refrigeration system the largest losses occur (combustion, throttling, friction, heat transfer). See Thermodynamic Cycles and Machines for the cycle-by-cycle analysis.

Isothermal expansion ($T_f = T_i$): $\Delta S = nR\ln(V_f/V_i) > 0$ for expansion.

Adiabatic reversible ($Q = 0$, reversible): $\Delta S = 0$ (isentropic).

Free expansion (irreversible, adiabatic, ideal gas): $\Delta S = nR\ln(V_f/V_i) > 0$. Even though $Q = 0$, entropy increases because the process is irreversible.

Mixing. When two different ideal gases mix at the same $T, P$, the entropy of mixing is $\Delta S_{\text{mix}} = -nR\sum_i x_i \ln x_i > 0$. This is a purely entropic driving force (no enthalpy change for ideal gases).

No engine can exceed Carnot. Suppose an engine E has efficiency $\eta_E > \eta_{\text{Carnot}}$. Run E forward and a Carnot refrigerator (the reverse of a Carnot engine) backward, both between the same reservoirs, sized so they exchange the same heat with the cold reservoir. The net effect: heat is drawn from the hot reservoir and fully converted to work, with no other change — a perpetual motion machine of the second kind, forbidden by the second law.

See Thermodynamic Cycles and Machines for the full Carnot cycle analysis.

Part 1: Entropy change.

Free expansion: $Q = 0$, $W = 0$, so $\Delta U = 0$ and $T_f = T_i = 300\text{ K}$ (ideal gas).

Entropy change (compute along a reversible isothermal path between the same states):

The surroundings exchange no heat ($Q = 0$), so $\Delta S_{\text{surr}} = 0$.

The process is irreversible (as expected for a free expansion), and the total entropy increases.

Part 2: Minimum work to restore the original state.

To compress the gas back isothermally at 300 K from $V_f = 2V_i$ to $V_i$, the minimum work is the reversible isothermal compression work:

This equals $T_0\,\Delta S_{\text{universe}} = 300 \times 5.76 \approx 1728 \text{ J}$ (rounding). The entropy increase quantifies the lost work potential (exergy): the free expansion destroyed 1729 J of useful work that can never be recovered without expending at least that much work to undo the process.