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Chapter 3 • Theory & Derivations

Entropy & The Third Law of Thermodynamics

Macroscopic entropy, Clausius theorem and inequality, entropy of ideal gases and universal entropy growth, T-S diagrams, the Third Law (Nernst-Planck theorem), and Ehrenfest classification of phase transitions.

§3.1 Concept of Entropy & Clausius Theorem

1. Clausius Theorem for Cyclic Reversible Processes

In 1854, Rudolf Clausius discovered that the Second Law implies the existence of a fundamental thermodynamic state function. Consider an arbitrary reversible cyclic process operating on a state plane $(P, V)$.

Any arbitrary reversible cycle $\mathcal{C}$ can be decomposed into an infinite grid of infinitesimal Carnot cycles operating between adjacent adiabats. For each infinitesimal Carnot cycle operating between reservoirs at temperatures $T_{1, i}$ and $T_{2, i}$:

$$\frac{\delta Q_{1, i}}{T_{1, i}} + \frac{\delta Q_{2, i}}{T_{2, i}} = 0$$

Summing over all sub-cycles, the internal adiabatic and isothermal paths cancel pairwise because each internal segment is traversed in opposite directions by adjacent cycles. The only uncompensated heat exchanges are along the outer closed boundary:

$$\sum_{i} \frac{\delta Q_{\text{rev}, i}}{T_i} = 0$$

Taking the continuum limit as the number of sub-cycles $N \to \infty$:

$$\oint_{\mathcal{C}} \frac{\delta Q_{\text{rev}}}{T} = 0$$

This is Clausius's Theorem.

2. Definition of Entropy as a State Function

Because the cyclic integral of $\frac{\delta Q_{\text{rev}}}{T}$ vanishes for every conceivable reversible loop, the line integral between any two equilibrium states $A$ and $B$ is strictly independent of the reversible path:

$$\int_{\text{Path 1}}^{A \to B} \frac{\delta Q_{\text{rev}}}{T} = \int_{\text{Path 2}}^{A \to B} \frac{\delta Q_{\text{rev}}}{T}$$

Consequently, $\frac{\delta Q_{\text{rev}}}{T}$ is an exact differential of an extensive state function, which Clausius named Entropy ($S$):

$$dS \equiv \frac{\delta Q_{\text{rev}}}{T}$$

The finite entropy difference between states $A$ and $B$ is:

$$\Delta S = S(B) - S(A) = \int_A^B \frac{\delta Q_{\text{rev}}}{T}$$

The absolute value of entropy is defined relative to an arbitrary reference state ($S_0$), until anchored by the Third Law.

§3.2 Clausius Inequality & Principle of Increase of Entropy

1. The Clausius Inequality

Now consider a real, irreversible cyclic process $\mathcal{C}$. Let the system absorb heat $\delta Q$ at temperature $T$ during an irreversible forward path $A \to B$, and return to initial state $A$ via an idealized reversible path $B \to A$. By Carnot's theorem, an irreversible engine is less efficient than a reversible engine:

$$\eta_{\text{irrev}} < \eta_{\text{rev}} \implies \oint \frac{\delta Q}{T} < 0$$

Combining reversible ($= 0$) and irreversible ($< 0$) cases yields the universal Clausius Inequality:

$$\oint \frac{\delta Q}{T} \le 0$$

where equality holds if and only if the entire cycle is strictly reversible.

2. The Second Law in Differential Form

Splitting the cyclic integral into the irreversible path $I$ ($A \to B$) and reversible path $R$ ($B \to A$):

$$\int_{A, I}^B \frac{\delta Q}{T} + \int_{B, R}^A \frac{\delta Q_{\text{rev}}}{T} \le 0$$

Since the second path is reversible, $\int_{B, R}^A \frac{\delta Q_{\text{rev}}}{T} = S_A - S_B = -\Delta S$:

$$\int_{A, I}^B \frac{\delta Q}{T} - \Delta S \le 0 \implies \Delta S \ge \int_{A, I}^B \frac{\delta Q}{T}$$

In differential form:

$$dS \ge \frac{\delta Q}{T}$$

For an isolated system, which cannot exchange heat with its surroundings ($\delta Q = 0$):

$$dS_{\text{isolated}} \ge 0$$

This is the Principle of Increase of Entropy:

$$\Delta S_{\text{isolated}} > 0 \quad (\text{spontaneous natural processes}), \quad \Delta S_{\text{isolated}} = 0 \quad (\text{reversible equilibrium})$$

3. Entropy of the Universe

Any system interacting with its environment can be embedded within an overall isolated composite universe:

$$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \ge 0$$

While the entropy of a localized open subsystem can decrease (e.g., crystal freezing, living organism biological growth), it does so only by ejecting an even greater quantity of entropy into its surrounding environment, ensuring that total universal entropy relentlessly increases.

§3.3 Entropy of Ideal Gases & Irreversible Transformations

1. Analytical Expressions for Ideal Gas Entropy

From the combined First and Second Laws for a reversible process in a simple fluid:

$$T dS = dU + P dV = C_v dT + P dV$$

For an ideal gas, substituting $dU = n C_{v, m} dT$ and $P = n R T / V$:

$$dS = n C_{v, m} \frac{dT}{T} + n R \frac{dV}{V}$$

Integrating from state $(T_1, V_1)$ to $(T_2, V_2)$:

$$S(T_2, V_2) - S(T_1, V_1) = n C_{v, m} \ln\left(\frac{T_2}{T_1}\right) + n R \ln\left(\frac{V_2}{V_1}\right)$$

Alternatively, using enthalpy $T dS = dH - V dP = n C_{p, m} dT - V dP$:

$$dS = n C_{p, m} \frac{dT}{T} - n R \frac{dP}{P}$$
$$S(T_2, P_2) - S(T_1, P_1) = n C_{p, m} \ln\left(\frac{T_2}{T_1}\right) - n R \ln\left(\frac{P_2}{P_1}\right)$$

2. Free Adiabatic Expansion (Joule Expansion)

Consider $n$ moles of an ideal gas expanding freely into an evacuated volume from $V_1$ to $V_2 = 2 V_1$ inside an adiabatic container:

  • $Q = 0, W = 0 \implies \Delta U = 0 \implies T_2 = T_1$.
  • Because the expansion is irreversible, we cannot integrate $\delta Q / T$ along the actual path.
  • But because $S$ is a state function, $\Delta S$ depends solely on initial and final states. We construct an imaginary reversible isothermal expansion connecting $(T_1, V_1)$ to $(T_1, V_2)$:
$$\Delta S_{\text{system}} = n R \ln\left(\frac{V_2}{V_1}\right) = n R \ln 2 > 0$$
  • Surroundings experience zero heat transfer: $\Delta S_{\text{surroundings}} = 0$.
  • Total universal entropy generated:
$$\Delta S_{\text{universe}} = n R \ln 2 > 0$$

This positive entropy change quantifies the intrinsic irreversibility of free expansion.

§3.4 Temperature-Entropy (T-S) Diagrams

1. Representation of Thermodynamic Cycles on T-S Coordinates

The Temperature-Entropy ($T$-$S$) diagram provides deep geometric insight into thermodynamic transformations because the area under a reversible curve directly represents heat exchanged:

$$\delta Q_{\text{rev}} = T dS \implies Q_{\text{rev}} = \int_1^2 T dS$$
  • An isothermal process ($T = \text{const}$) maps to a horizontal straight line.
  • A reversible adiabatic process ($dS = 0$) maps to a vertical straight line (an isentropic process).

2. The Carnot Cycle on a T-S Diagram

On a $T$-$S$ plane, the Carnot cycle maps to an exact rectangle:

  1. Stage $A \to B$ (Isothermal expansion at $T_H$): Horizontal line from $S_A$ to $S_B$. Heat absorbed:
$$Q_H = \text{Area under } AB = T_H (S_B - S_A) = T_H \Delta S$$
  1. Stage $B \to C$ (Adiabatic expansion): Vertical line downward at constant entropy $S_B$ from $T_H$ to $T_C$ ($Q = 0$).
  2. Stage $C \to D$ (Isothermal compression at $T_C$): Horizontal line from $S_B$ back to $S_A$. Heat rejected:
$$Q_C = \text{Area under } CD = T_C (S_B - S_A) = T_C \Delta S$$
  1. Stage $D \to A$ (Adiabatic compression): Vertical line upward at constant entropy $S_A$ from $T_C$ back to $T_H$ ($Q = 0$).

3. Net Mechanical Work and Efficiency

The net mechanical work done per cycle is the enclosed area of the rectangle:

$$W_{\text{net}} = Q_H - Q_C = (T_H - T_C)(S_B - S_A) = (T_H - T_C) \Delta S$$

The thermal efficiency is the ratio of enclosed area to total area under the upper horizontal curve:

$$\eta = \frac{W_{\text{net}}}{Q_H} = \frac{(T_H - T_C) \Delta S}{T_H \Delta S} = 1 - \frac{T_C}{T_H}$$

This offers an elegant geometric proof of Carnot efficiency.

§3.5 The Third Law of Thermodynamics & Unattainability of Absolute Zero

1. Nernst Heat Theorem & Planck Formulation

In 1906, Walther Nernst observed that during chemical reactions between condensed phases, the entropy change $\Delta S$ approaches zero as temperature approaches absolute zero:

$$\lim_{T \to 0} \Delta S = 0$$

Max Planck (1911) generalized this into the Third Law of Thermodynamics:

$$\text{The entropy of all pure, perfectly crystalline substances approaches a universal constant (which can be chosen as zero) as absolute temperature approaches zero:}$$
$$\lim_{T \to 0} S = 0$$

2. Physical Consequences of the Third Law

1. Vanishing Heat Capacities: The entropy of a substance at temperature $T$ is $S(T) = \int_0^T \frac{C_p(T')}{T'} dT'$. For this integral to converge at $T = 0$, $C_p$ and $C_v$ must vanish:

$$\lim_{T \to 0} C_p = 0, \quad \lim_{T \to 0} C_v = 0$$

In quantum physics, Debye's law confirms $C_v \propto T^3$ for non-metallic crystals and $C_v \propto T$ for conduction electrons as $T \to 0$.

2. Vanishing Thermal Expansion: Using Maxwell's relation $(\partial V/\partial T)_P = -(\partial S/\partial P)_T$, and since $S \to 0$ independent of pressure:

$$\lim_{T \to 0} \left(\frac{\partial V}{\partial T}\right)_P = 0 \implies \lim_{T \to 0} \alpha = 0$$

3. Principle of Unattainability of Absolute Zero

The Third Law is equivalent to the proposition: It is physically impossible to reduce the temperature of any macroscopic system to absolute zero ($T = 0\text{ K}$) in a finite number of thermodynamic operations.

To cool a system, one alternates between isothermal parameter changes (e.g., magnetization $M$) and adiabatic changes (e.g., demagnetization). Because all isentropic curves converge to $S = 0$ at $T = 0\text{ K}$, the temperature steps $\Delta T$ achieved per cycle diminish asymptotically to zero as $T \to 0$, requiring an infinite sequence of operations to reach $0\text{ K}$.

§3.6 First and Second Order Phase Transitions

1. Ehrenfest Classification Scheme

Paul Ehrenfest (1933) classified phase transitions according to the lowest order derivative of the Gibbs Free Energy $G(T, P)$ that displays a mathematical discontinuity:

2. First-Order Phase Transitions

A transition is first-order if the first partial derivatives of $G$ exhibit finite discontinuities across the phase boundary:

  • Entropy discontinuity:
$$\Delta S = S_2 - S_1 = -\left[\left(\frac{\partial G_2}{\partial T}\right)_P - \left(\frac{\partial G_1}{\partial T}\right)_P\right] = \frac{L}{T} \neq 0$$

where $L$ is the latent heat of transformation.

  • Volume discontinuity:
$$\Delta V = V_2 - V_1 = \left(\frac{\partial G_2}{\partial P}\right)_T - \left(\frac{\partial G_1}{\partial P}\right)_T \neq 0$$

Examples include solid-liquid melting, liquid-vapor boiling, and solid-vapor sublimation. Along the coexistence curve, equality of chemical potentials $G_1(T, P) = G_2(T, P)$ leads directly to the Clausius-Clapeyron equation:

$$\frac{dP}{dT} = \frac{\Delta S}{\Delta V} = \frac{L}{T \Delta V}$$

3. Second-Order Phase Transitions

A transition is second-order (continuous) if the first derivatives of $G$ are continuous ($\Delta S = 0, \Delta V = 0, L = 0$), but the second partial derivatives exhibit finite jump discontinuities or power-law divergences:

1. Specific Heat Discontinuity:

$$\Delta C_p = -T \left[ \left(\frac{\partial^2 G_2}{\partial T^2}\right)_P - \left(\frac{\partial^2 G_1}{\partial T^2}\right)_P \right]$$

2. Thermal Expansion Discontinuity:

$$\Delta \alpha = \frac{1}{V} \left[ \frac{\partial^2 G_2}{\partial T \partial P} - \frac{\partial^2 G_1}{\partial T \partial P} \right]$$

3. Isothermal Compressibility Discontinuity:

$$\Delta \kappa_T = -\frac{1}{V} \left[ \left(\frac{\partial^2 G_2}{\partial P^2}\right)_T - \left(\frac{\partial^2 G_1}{\partial P^2}\right)_T \right]$$

Examples include:

  • Ferromagnetic to paramagnetic transition at the Curie temperature $T_C$.
  • Superconducting transition in zero external magnetic field.
  • Order-disorder transitions in binary metal alloys (e.g., $\beta$-brass).
  • Liquid Helium-4 normal to superfluid lambda transition at $T_\lambda = 2.17\text{ K}$.
Standard University Exam Examination Problems

Rigorous Analytical & Numerical Solved Problems

Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.

Standard Example 3.1: Thermal Equilibration of Two Identical Solid Blocks

Two identical blocks of copper, each of mass $m = 2.50\text{ kg}$ and constant specific heat capacity $c = 385\text{ J/(kg}\cdot\text{K)}$, are initially at temperatures $T_1 = 373.15\text{ K}$ ($100^\circ\text{C}$) and $T_2 = 273.15\text{ K}$ ($0^\circ\text{C}$). The blocks are brought into thermal contact inside a thermally insulated calorimeter until they reach mutual equilibrium. (a) Calculate the final equilibrium temperature $T_f$. (b) Calculate the entropy change of Block 1, Block 2, and the universe. (c) If a Carnot engine had been operated reversibly between the two blocks to extract the maximum mechanical work while bringing them to equilibrium, calculate the final common temperature $T_f'$ and the maximum work delivered $W_{\max}$.

Step 1: Final Temperature under Direct Irreversible Contact

By energy conservation in an insulated system ($Q_1 + Q_2 = 0$):

$$m c (T_f - T_1) + m c (T_f - T_2) = 0 \implies 2 T_f = T_1 + T_2$$
$$T_f = \frac{T_1 + T_2}{2} = \frac{373.15 + 273.15}{2} = 323.15\text{ K} = 50.0^\circ\text{C}$$

The final temperature is the arithmetic mean.

Step 2: Entropy Changes in Irreversible Thermal Contact

For each block, entropy change is integrated via $\Delta S = \int \frac{m c dT}{T} = m c \ln\left(\frac{T_f}{T_i}\right)$: Total heat capacity $C = m c = (2.50)(385) = 962.5\text{ J/K}$.

$$\Delta S_1 = C \ln\left(\frac{323.15}{373.15}\right) = 962.5 \ln(0.8660) = 962.5(-0.14387) = -138.48\text{ J/K}$$
$$\Delta S_2 = C \ln\left(\frac{323.15}{273.15}\right) = 962.5 \ln(1.1830) = 962.5(+0.16809) = +161.79\text{ J/K}$$

Net universal entropy production is:

$$\Delta S_{\text{univ}} = \Delta S_1 + \Delta S_2 = -138.48 + 161.79 = +23.31\text{ J/K}$$

Since $\Delta S_{\text{univ}} > 0$, the process is spontaneously irreversible.

Step 3: Reversible Extraction of Maximum Work via Carnot Engine

For a reversible process operating between the two finite blocks, the total entropy change of the universe must be zero:

$$\Delta S_{\text{univ}} = \Delta S_1 + \Delta S_2 = C \ln\left(\frac{T_f'}{T_1}\right) + C \ln\left(\frac{T_f'}{T_2}\right) = 0$$
$$C \ln\left(\frac{(T_f')^2}{T_1 T_2}\right) = 0 \implies (T_f')^2 = T_1 T_2 \implies T_f' = \sqrt{T_1 T_2}$$

The reversible final temperature is the geometric mean:

$$T_f' = \sqrt{(373.15)(273.15)} = \sqrt{101915.9} = 319.24\text{ K} = 46.09^\circ\text{C}$$

Notice $T_f' < T_f$ because energy has been extracted from the blocks as useful mechanical work.

Step 4: Maximum Work Delivered $W_{\max}$

By the First Law, the work delivered is the decrease in total internal energy of both blocks:

$$W_{\max} = -(\Delta U_1 + \Delta U_2) = C (T_1 - T_f') + C (T_2 - T_f') = C (T_1 + T_2 - 2 T_f')$$
$$W_{\max} = 962.5 [ 373.15 + 273.15 - 2(319.24) ] = 962.5 [ 646.30 - 638.48 ] = 962.5 (7.82) = 7526.8\text{ J} = 7.53\text{ kJ}$$
Final Answer & Physical Insight

(a) Direct thermal contact: $T_f = 323.15\text{ K}$ ($50.0^\circ\text{C}$). (b) $\Delta S_1 = -138.5\text{ J/K}$, $\Delta S_2 = +161.8\text{ J/K}$, $\Delta S_{\text{univ}} = +23.3\text{ J/K}$. (c) Reversible work extraction: $T_f' = 319.24\text{ K}$ ($46.1^\circ\text{C}$) and $W_{\max} = 7.53\text{ kJ}$.

Standard Example 3.2: Entropy of Free Gas Expansion vs Irreversible Heat Transfer

An insulated rigid tank is divided into two equal compartments, each of volume $V_0 = 0.050\text{ m}^3$, by a partition. One compartment contains $n = 3.00\text{ mol}$ of an ideal gas at $T_0 = 400.0\text{ K}$, and the second compartment is evacuated. The partition ruptures, and the gas expands freely to fill the entire container. Subsequently, the gas is placed in thermal contact with a large heat reservoir at $T_R = 300.0\text{ K}$ until thermal equilibrium is attained. Calculate: (a) the entropy change of the gas during the free expansion stage, (b) the entropy change of the gas during cooling, (c) the entropy change of the reservoir during cooling, and (d) the total entropy generation of the universe across both stages.

Step 1: Free Expansion Stage Entropy Change

During free expansion into vacuum, $Q = 0$ and $W = 0$, so $\Delta U = 0$ and $T_1 = T_0 = 400.0\text{ K}$. The volume doubles: $V_1 = 2 V_0 = 0.100\text{ m}^3$.

$$\Delta S_{\text{gas, 1}} = n R \ln\left(\frac{V_1}{V_0}\right) = (3.00\text{ mol})(8.314\text{ J/(mol}\cdot\text{K)}) \ln(2) = (24.942)(0.69315) = +17.29\text{ J/K}$$

The isolated surroundings experience zero change: $\Delta S_{\text{surr, 1}} = 0$.

$$\Delta S_{\text{univ, 1}} = +17.29\text{ J/K}$$
Step 2: Cooling Stage Entropy Change of the Gas

During cooling at constant volume $V_1$ from $T_1 = 400.0\text{ K}$ to $T_2 = 300.0\text{ K}$ for an ideal gas with $C_v = \frac{5}{2} R$ (diatomic gas):

$$C_v = \frac{5}{2} (8.314) = 20.785\text{ J/(mol}\cdot\text{K)}$$
$$\Delta S_{\text{gas, 2}} = n C_v \ln\left(\frac{T_2}{T_1}\right) = (3.00)(20.785) \ln\left(\frac{300.0}{400.0}\right) = (62.355)(-0.28768) = -17.94\text{ J/K}$$
Step 3: Entropy Change of the Reservoir

The heat released by the gas during isochoric cooling is:

$$Q_{\text{gas}} = n C_v (T_2 - T_1) = (3.00)(20.785)(300 - 400) = -6235.5\text{ J}$$

The reservoir absorbs heat $Q_{\text{res}} = -Q_{\text{gas}} = +6235.5\text{ J}$ at constant temperature $T_R = 300.0\text{ K}$:

$$\Delta S_{\text{res}} = \frac{Q_{\text{res}}}{T_R} = \frac{+6235.5\text{ J}}{300.0\text{ K}} = +20.785\text{ J/K}$$
Step 4: Total Entropy Generation of the Universe

For stage 2 (thermal cooling across finite temperature difference):

$$\Delta S_{\text{univ, 2}} = \Delta S_{\text{gas, 2}} + \Delta S_{\text{res}} = -17.94 + 20.79 = +2.85\text{ J/K}$$

Summing both irreversible stages:

$$\Delta S_{\text{univ, total}} = \Delta S_{\text{univ, 1}} + \Delta S_{\text{univ, 2}} = +17.29 + 2.85 = +20.14\text{ J/K}$$
Final Answer & Physical Insight

(a) Free expansion gas entropy $\Delta S_{\text{gas, 1}} = +17.29\text{ J/K}$. (b) Cooling gas entropy $\Delta S_{\text{gas, 2}} = -17.94\text{ J/K}$. (c) Reservoir entropy $\Delta S_{\text{res}} = +20.79\text{ J/K}$. (d) Total universal entropy produced $\Delta S_{\text{univ}} = +20.14\text{ J/K}$.

Standard Example 3.3: Third Law Low-Temperature Debye Entropy Integration

At low temperatures ($T < 20\text{ K}$), the molar heat capacity of solid silver obeys the Debye cubic power law $C_{v, m}(T) = a T^3$, where $a = 1.70\times 10^{-4}\text{ J/(mol}\cdot\text{K}^4)$. (a) Verify that this relation satisfies the Third Law of Thermodynamics. (b) Calculate the absolute molar entropy of silver at $T = 15.0\text{ K}$. (c) Calculate the heat absorbed per mole when warming from $0\text{ K}$ to $15.0\text{ K}$ and compare the heat absorbed with the product $T S(T)$.

Step 1: Verify Third Law Compliance

According to the Third Law (Nernst-Planck theorem), $C_v \to 0$ as $T \to 0$.

$$\lim_{T \to 0} C_{v, m}(T) = \lim_{T \to 0} (a T^3) = 0$$

Furthermore, the entropy integral at the lower limit is:

$$S(T) = \int_0^T \frac{C_v(T')}{T'} dT' = \int_0^T a (T')^2 dT' = \frac{a T^3}{3}$$

Since $\lim_{T \to 0} S(T) = 0$, the expression strictly obeys the Third Law without logarithmic divergences.

Step 2: Calculate Absolute Molar Entropy at $15.0\text{ K}$

Evaluating at $T = 15.0\text{ K}$:

$$S(15.0) = \frac{a (15.0)^3}{3} = \frac{(1.70 \times 10^{-4})(3375)}{3} = \frac{0.57375}{3} = 0.19125\text{ J/(mol}\cdot\text{K)}$$

Notice that the molar heat capacity at this temperature is:

$$C_v(15.0) = a (15.0)^3 = 0.57375\text{ J/(mol}\cdot\text{K)}$$

Remarkably, $S(T) = \frac{1}{3} C_v(T)$ for any substance obeying Debye's $T^3$ law.

Step 3: Calculate Heat Absorbed and Compare with $T S(T)$

The total heat absorbed per mole from $0\text{ K}$ to $T$ is:

$$Q = \int_0^T C_v(T') dT' = \int_0^T a (T')^3 dT' = \frac{a T^4}{4}$$

Evaluating at $T = 15.0\text{ K}$ ($15^4 = 50625$):

$$Q = \frac{(1.70 \times 10^{-4})(50625)}{4} = \frac{8.60625}{4} = 2.1516\text{ J/mol}$$

Comparing with the product $T S(T)$:

$$T S(T) = T \left(\frac{a T^3}{3}\right) = \frac{a T^4}{3} = \frac{8.60625}{3} = 2.8688\text{ J/mol}$$
$$Q = \frac{3}{4} T S(T)$$

This shows that $25\\%$ of the thermal energy $T S$ represents bound unavailable energy locked within the low-frequency phonon modes.

Final Answer & Physical Insight

(a) $C_v \to 0$ and $S \to 0$ as $T \to 0$, rigorously satisfying the Third Law. (b) Absolute molar entropy $S(15.0\text{ K}) = 0.191\text{ J/(mol}\cdot\text{K)}$. (c) Heat absorbed $Q = 2.15\text{ J/mol}$, satisfying the exact scaling $Q = \frac{3}{4} T S$.