8.3 Second and Third Laws of Thermodynamics: Entropy, Carnot Engine, and Absolute Entropies

Second and Third Laws of Thermodynamics: Entropy, Carnot Engine, and Absolute Entropies

While the First Law governs energy conservation, it cannot predict whether a physical or chemical process occurs spontaneously. The Second and Third Laws of Thermodynamics introduce entropy ($S$) as the fundamental state function establishing the arrow of time, the theoretical efficiency limit of thermodynamic engines, and the quantification of absolute standard entropies.

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1. Spontaneity and Statements of the Second Law

A spontaneous process is a transformation that proceeds in a definite direction without continuous input of external work. Natural processes disperse energy and matter, moving isolated systems toward macroscopic equilibrium.

Classic Formulations of the Second Law

1. Clausius Statement: It is impossible to construct a cyclic thermodynamic device whose sole effect is the continuous transfer of heat from a cooler body to a hotter body without an input of external work.
2. Kelvin-Planck Statement: It is impossible to construct a cyclic heat engine that extracts heat from a single thermal reservoir and converts it completely into mechanical work without rejecting a portion of heat to a colder reservoir.
3. Entropy Formulation: The entropy of the universe (system plus surroundings) increases during any spontaneous, irreversible process and remains constant during a reversible process: $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} > 0 \quad \text{[Irreversible / Spontaneous]}$ $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} = 0 \quad \text{[Reversible / Equilibrium]}$

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2. Thermodynamic Definition of Entropy and the Clausius Inequality

Entropy is defined differentially along a reversible thermal path:

$dS \equiv \frac{\delta q_{rev}}{T}$

The Clausius Inequality

For any real, irreversible change between two states, the actual heat absorbed ($\delta q$) is less than the heat that would be absorbed along a reversible path ($\delta q_{rev}$) connecting the exact same states because internal dissipation generates entropy:

$dS \ge \frac{\delta q}{T} \iff \oint \frac{\delta q}{T} \le 0$ where the equality holds strictly for reversible transformations.

Entropy Change of the Surroundings ($\Delta S_{surr}$)

Because the surroundings act as an essentially infinite, constant-temperature heat reservoir operating quasi-statically, any heat exchanged by the system ($q_{sys}$) enters or leaves the surroundings reversibly:

$\Delta S_{surr} = -\frac{q_{sys}}{T_{surr}}$

Worked Step-by-Step Calculation: Irreversible Heat Transfer

Scenario: A block of copper transfers $6000\text{ J}$ of heat directly and irreversibly from a high-temperature thermal reservoir at $T_H = 600.0\text{ K}$ to a low-temperature reservoir at $T_C = 300.0\text{ K}$. Calculate $\Delta S_{hot}$, $\Delta S_{cold}$, and $\Delta S_{univ}$.

Step 1: Compute entropy changes of individual reservoirs: $\Delta S_{hot} = \frac{q_{hot}}{T_H} = \frac{-6000\text{ J}}{600.0\text{ K}} = -10.00\text{ J/K}$ $\Delta S_{cold} = \frac{q_{cold}}{T_C} = \frac{+6000\text{ J}}{300.0\text{ K}} = +20.00\text{ J/K}$

Step 2: Compute net entropy change of the universe: $\Delta S_{univ} = \Delta S_{hot} + \Delta S_{cold} = -10.00\text{ J/K} + 20.00\text{ J/K} = +10.00\text{ J/K}$ Thermodynamic Deduction: Because $\Delta S_{univ} = +10.00\text{ J/K} > 0$, the Second Law is strictly satisfied. The spontaneous unidirectional flow of heat down a temperature gradient generates net entropy.

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3. The Carnot Cycle and Theoretical Efficiency

The Carnot cycle represents an idealized, fully reversible four-stroke thermodynamic engine operating between a hot reservoir at $T_H$ and a cold reservoir at $T_C$.

Volume Relationships and State Function Proof

For the two adiabatic stages: $\frac{T_H}{T_C} = \left(\frac{V_3}{V_2}\right)^{\gamma - 1} \quad \text{and} \quad \frac{T_H}{T_C} = \left(\frac{V_4}{V_1}\right)^{\gamma - 1} \implies \frac{V_3}{V_2} = \frac{V_4}{V_1} \iff \frac{V_2}{V_1} = \frac{V_3}{V_4}$ Substituting this equality into the cyclic entropy sum: $\oint \frac{\delta q_{rev}}{T} = \frac{q_H}{T_H} + \frac{q_C}{T_C} = nR \ln\left(\frac{V_2}{V_1}\right) + nR \ln\left(\frac{V_4}{V_3}\right) = nR \ln\left(\frac{V_2}{V_1}\right) - nR \ln\left(\frac{V_3}{V_4}\right) = 0$ This proves mathematically that $\frac{\delta q_{rev}}{T}$ is an exact differential, establishing that entropy ($S$) is a true state function.

Carnot Thermal Efficiency ($\eta$)

The thermodynamic efficiency is defined as net mechanical work delivered per unit heat absorbed from the hot reservoir: $\eta \equiv \frac{-w_{net}}{q_H} = \frac{q_H + q_C}{q_H} = 1 + \frac{q_C}{q_H}$ Because $\frac{q_C}{q_H} = -\frac{T_C}{T_H}$:

$\eta_{Carnot} = 1 - \frac{T_C}{T_H} = \frac{T_H - T_C}{T_H}$

Carnot's Principles:

1. No heat engine operating between two specified heat reservoirs can be more efficient than a reversible Carnot engine operating between the same two reservoirs ($\eta_{irr} < \eta_{Carnot}$).
2. All reversible heat engines operating between the same two temperatures possess the identical efficiency, completely independent of the working fluid (ideal gas, steam, chemical solution).

Worked Carnot Calculation

Scenario: A high-temperature geothermal powerplant extracts thermal steam at $227.0^\circ\text{C}$ ($500.15\text{ K}$) and discharges rejected heat to a cooling pond at $27.0^\circ\text{C}$ ($300.15\text{ K}$). If the plant extracts $120.0\text{ MW}$ of thermal heat from the underground steam field:

• (a) What is the maximum theoretical efficiency?
• (b) What is the minimum rate of waste heat rejected to the pond?

Step 1: Compute efficiency: $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300.15\text{ K}}{500.15\text{ K}} = 1 - 0.60012 = 0.39988 \approx 40.0\%$

Step 2: Compute net work rate (electric power) and rejected heat: $\dot{W}_{net} = \eta \cdot \dot{Q}_H = (0.39988)(120.0\text{ MW}) = 47.99\text{ MW} \approx 48.0\text{ MW}$ $\dot{Q}_C = \dot{Q}_H - \dot{W}_{net} = 120.0\text{ MW} - 48.0\text{ MW} = 72.0\text{ MW}$

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4. Entropy Calculations Across Thermodynamic Pathways

1. Isothermal Ideal Gas Expansion / Compression

$\Delta S = \int_{1}^{2} \frac{\delta q_{rev}}{T} = \frac{1}{T} \int_{1}^{2} P\,dV = nR \int_{V_1}^{V_2} \frac{dV}{V} = nR \ln\left(\frac{V_2}{V_1}\right) = -nR \ln\left(\frac{P_2}{P_1}\right)$

2. Isobaric Heating / Cooling ($P = \text{const}$)

$\Delta S = \int_{T_1}^{T_2} \frac{n C_{p,m}}{T}\,dT = n C_{p,m} \ln\left(\frac{T_2}{T_1}\right) \quad \text{[if } C_p \text{ is constant]}$

3. Isochoric Heating / Cooling ($V = \text{const}$)

$\Delta S = \int_{T_1}^{T_2} \frac{n C_{v,m}}{T}\,dT = n C_{v,m} \ln\left(\frac{T_2}{T_1}\right) \quad \text{[if } C_v \text{ is constant]}$

4. General State Change of an Ideal Gas ($T_1, P_1 \rightarrow T_2, P_2$)

$\Delta S = n C_{p,m} \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right)$

5. Phase Transitions at Equilibrium ($T_{tr}, P = 1\text{ bar}$)

At the normal transition temperature (e.g., normal boiling point $T_b$, normal melting point $T_m$), coexistence of two phases is in thermodynamic equilibrium ($\Delta G = 0$): $\Delta S_{tr} = \frac{\Delta H_{tr}}{T_{tr}}$

• Normal fusion: $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_m}$
• Normal vaporization: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}$

Trouton's Rule and Structural Anomalies

Trouton's Rule states that for a wide variety of non-associated, normal liquids, the standard molar entropy of vaporization at the normal boiling point is nearly constant:

$\Delta S_{vap}^\circ = \frac{\Delta H_{vap}}{T_b} \approx 85 - 88\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ Physical Origin: In all normal liquids, the molar volume expansion upon vaporization is approximately uniform ($\approx 1000\times$), leading to a roughly identical increase in translational positional disorder.

Common Deviations from Trouton's Rule:

• Positive Deviations ($\Delta S_{vap} \gg 88\text{ J/mol}\cdot\text{K}$): Strongly hydrogen-bonded liquids like water ($\Delta S_{vap} = 109.1\text{ J/mol}\cdot\text{K}$) and ethanol ($\Delta S_{vap} = 110.0\text{ J/mol}\cdot\text{K}$). The liquid phase possesses substantial three-dimensional structural ordering. Vaporization shatters this extensive H-bonded network, resulting in a dramatically larger entropy jump.
• Negative Deviations ($\Delta S_{vap} \ll 85\text{ J/mol}\cdot\text{K}$): Carboxylic acids such as acetic acid ($\Delta S_{vap} \approx 62\text{ J/mol}\cdot\text{K}$) undergo vapor-phase dimerization via double hydrogen bonds ($2\text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_2$), retaining significant order in the gaseous state.

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5. Statistical Entropy and the Third Law

Boltzmann's Statistical Formulation

Ludwig Boltzmann established the microscopic foundation of entropy connecting thermodynamic state functions to quantum microstate multiplicity ($W$):

$S = k_B \ln W$ where $k_B = \frac{R}{N_A} = 1.38065 \times 10^{-23}\text{ J/K}$ is Boltzmann's constant, and $W$ is the thermodynamic probability (the total number of accessible microstates consistent with the macroscopic energy, volume, and composition).

The Third Law of Thermodynamics

1. Nernst Heat Theorem (1906): In any chemical or physical transformation between pure crystalline substances in internal equilibrium, the entropy change approaches zero as temperature approaches absolute zero: $\lim_{T \to 0\text{ K}} \Delta S = 0$
2. Planck Formulation (1911): The absolute entropy of all pure, perfectly crystalline substances at absolute zero ($0\text{ K}$) is identically zero: $S(0\text{ K}) = 0$ Microscopic Basis: In a perfect crystalline lattice at $0\text{ K}$, thermal motion ceases and every atom occupies its unique ground-state lattice position. Hence, only one unique microstate exists ($W = 1$), yielding $S = k_B \ln(1) = 0$.

Experimental Determination of Absolute Standard Entropies ($S_{298}^\circ$)

Because $S(0\text{ K}) = 0$, absolute entropies can be determined experimentally via calorimetric heat capacity measurements:

$S^\circ(T) = \int_{0}^{T_m} \frac{C_p(s)}{T}\,dT + \frac{\Delta H_{fus}}{T_m} + \int_{T_m}^{T_b} \frac{C_p(l)}{T}\,dT + \frac{\Delta H_{vap}}{T_b} + \int_{T_b}^{T} \frac{C_p(g)}{T}\,dT$ Near $0\text{ K}$ ($0\text{ to } \sim 15\text{ K}$), experimental heat capacities are extrapolated using Debye's $T^3$ Law for non-metallic crystalline solids: $C_p \approx a T^3 \implies S(T) = \int_{0}^{T} \frac{a T^3}{T}\,dT = \frac{a T^3}{3} = \frac{C_p(T)}{3}$

Residual Entropy at $0\text{ K}$

When a real substance is cooled, kinetic barriers often freeze molecules in random orientations before true thermodynamic equilibrium is attained, resulting in residual entropy at $0\text{ K}$ ($S_{0\text{ K}} > 0$):