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Chapter 4 • Theory & Derivations

Thermodynamic Potentials & Cryogenics

Legendre transformations, Internal Energy, Enthalpy, Helmholtz and Gibbs Free Energies, thermodynamic equilibrium criteria, liquid surface interfaces, magnetic work, and cooling by adiabatic demagnetization.

§4.1 Extensive & Intensive Variables & Euler Relations

1. Mathematical Scaling of Thermodynamic Functions

A state variable $X$ is extensive if scaling the system size by a positive parameter $\lambda$ scales the variable by $\lambda$: $X(\lambda N, \lambda V) = \lambda X(N, V)$. A state variable $Y$ is intensive if it remains invariant: $Y(\lambda N, \lambda V) = Y(N, V)$.

Let the fundamental relation for internal energy be $U = U(S, V, N)$. Because $S, V, N$ are all extensive:

$$U(\lambda S, \lambda V, \lambda N) = \lambda U(S, V, N)$$

Differentiating with respect to $\lambda$ via the chain rule:

$$\frac{\partial U}{\partial (\lambda S)} \frac{d(\lambda S)}{d\lambda} + \frac{\partial U}{\partial (\lambda V)} \frac{d(\lambda V)}{d\lambda} + \frac{\partial U}{\partial (\lambda N)} \frac{d(\lambda N)}{d\lambda} = U(S, V, N)$$

Setting $\lambda = 1$:

$$S \left(\frac{\partial U}{\partial S}\right)_{V, N} + V \left(\frac{\partial U}{\partial V}\right)_{S, N} + N \left(\frac{\partial U}{\partial N}\right)_{S, V} = U$$

Recalling definitions of temperature $T = (\partial U/\partial S)_{V, N}$, hydrostatic pressure $P = -(\partial U/\partial V)_{S, N}$, and chemical potential $\mu = (\partial U/\partial N)_{S, V}$, we obtain the fundamental Euler Equation:

$$U = T S - P V + \mu N$$

2. The Gibbs-Duhem Relation

Taking the total differential of the Euler equation:

$$dU = T dS + S dT - P dV - V dP + \mu dN + N d\mu$$

Subtracting the fundamental differential relation $dU = T dS - P dV + \mu dN$:

$$S dT - V dP + N d\mu = 0$$

This is the Gibbs-Duhem Relation. It demonstrates that the intensive parameters $(T, P, \mu)$ of a single-component system cannot all vary independently; varying any two strictly determines the third.

§4.2 The Four Fundamental Potentials (U, H, F, G)

1. Legendre Transformations

In experimental physics, controlling extensive entropy $S$ is difficult; controlling intensive temperature $T$ using a thermal bath is straightforward. Legendre transformations systematically replace an extensive variable with its conjugate intensive derivative without losing any fundamental thermodynamic information.

2. The Four Potentials and Natural Variables

1. Internal Energy $U(S, V)$:

$$dU = T dS - P dV + \mu dN$$

Natural variables: $(S, V, N)$. Conjugate relations: $T = \left(\frac{\partial U}{\partial S}\right)_{V}$, $P = -\left(\frac{\partial U}{\partial V}\right)_{S}$.

2. Enthalpy $H(S, P) \equiv U + P V$:

$$dH = dU + P dV + V dP = (T dS - P dV) + P dV + V dP = T dS + V dP$$

Natural variables: $(S, P, N)$. Conjugate relations: $T = \left(\frac{\partial H}{\partial S}\right)_{P}$, $V = \left(\frac{\partial H}{\partial P}\right)_{S}$. Enthalpy represents heat transferred under constant pressure ($dH = \delta Q_P$).

3. Helmholtz Free Energy $F(T, V) \equiv U - T S$:

$$dF = dU - T dS - S dT = (T dS - P dV) - T dS - S dT = -S dT - P dV$$

Natural variables: $(T, V, N)$. Conjugate relations: $S = -\left(\frac{\partial F}{\partial T}\right)_{V}$, $P = -\left(\frac{\partial F}{\partial V}\right)_{T}$. Helmholtz free energy represents the maximum reversible work extractable from a system during an isothermal process: $W_{\max} = -\Delta F$.

4. Gibbs Free Energy $G(T, P) \equiv H - T S = U + P V - T S$:

$$dG = dH - T dS - S dT = (T dS + V dP) - T dS - S dT = -S dT + V dP$$

Natural variables: $(T, P, N)$. Conjugate relations: $S = -\left(\frac{\partial G}{\partial T}\right)_{P}$, $V = \left(\frac{\partial G}{\partial P}\right)_{T}$. From Euler's relation $U = TS - PV + \mu N$, we obtain $G = \mu N$, proving that chemical potential $\mu = G/N$ is simply molar/molecular Gibbs free energy.

§4.3 Equilibrium Criteria & Extremum Principles

1. Thermodynamic Equilibrium Criteria

From the Clausius inequality for a non-isolated system in contact with surroundings at temperature $T_0$ and pressure $P_0$:

$$dU \le T_0 dS - P_0 dV$$

Applying this inequality to constrained systems generates specific extremum principles:

1. Isolated System ($U, V$ constant):

$$dS_{U, V} \ge 0 \implies S \text{ is a maximum at equilibrium.}$$

2. System at Constant Entropy and Volume ($S, V$ constant):

$$dU_{S, V} \le 0 \implies U \text{ is a minimum at equilibrium.}$$

3. System at Constant Temperature and Volume ($T = T_0, V$ constant):

$$d(U - T S)_{T, V} \le 0 \implies dF_{T, V} \le 0 \implies F \text{ is a minimum at equilibrium.}$$

4. System at Constant Temperature and Pressure ($T = T_0, P = P_0$):

$$d(U - T S + P V)_{T, P} \le 0 \implies dG_{T, P} \le 0 \implies G \text{ is a minimum at equilibrium.}$$

2. Practical Importance of Gibbs Energy Minimum

Most real-world chemical reactions, phase transitions, and biological processes occur at constant atmospheric pressure and ambient temperature. Therefore:

$$\Delta G < 0 \implies \text{Process is thermodynamically spontaneous.}$$
$$\Delta G = 0 \implies \text{System is in dynamic phase/chemical equilibrium.}$$
$$\Delta G > 0 \implies \text{Process is non-spontaneous (requires external work input).}$$

§4.4 Thermodynamics of Liquid Surface Films

1. Interfacial Work and Surface Free Energy

Creating an interface of area $A$ requires pulling molecules from the bulk liquid against attractive intermolecular cohesive forces to the surface. The work performed to expand surface area by $dA$ at surface tension $\gamma$ is:

$$\delta W = -\gamma dA$$

Including interfacial work in the fundamental First Law relation:

$$dU = T dS - P dV + \gamma dA$$

For the Helmholtz free energy $F = U - TS$:

$$dF = -S dT - P dV + \gamma dA$$

At constant temperature and volume:

$$\gamma = \left(\frac{\partial F}{\partial A}\right)_{T, V} = \left(\frac{\partial G}{\partial A}\right)_{T, P}$$

Thus, surface tension is the surface Helmholtz/Gibbs free energy per unit area.

2. Temperature Dependence of Surface Tension & Total Surface Energy

Applying Maxwell's cross-derivative reciprocity to $dF$:

$$\left(\frac{\partial S}{\partial A}\right)_{T, V} = -\left(\frac{\partial \gamma}{\partial T}\right)_A$$

Since surface tension decreases monotonically with temperature ($d\gamma/dT < 0$), surface entropy per unit area $s_A = -d\gamma/dT$ is positive; creating new surface area absorbs heat isothermally ($q_A = T s_A = -T d\gamma/dT$).

The Total Surface Internal Energy Density $u_A = U_A / A$ is:

$$u_A = f_A + T s_A = \gamma - T \left(\frac{d\gamma}{dT}\right)$$

Because $d\gamma/dT < 0$, total surface energy $u_A$ is strictly greater than surface tension $\gamma$.

§4.5 Magnetic Thermodynamics & Adiabatic Demagnetization

1. Thermodynamics of Magnetic Systems

For a paramagnetic material placed inside an external magnetic field $\vec{H}$, work done by the magnetic field generator to alter total sample magnetic dipole moment $\vec{M}$ is:

$$\delta W_{\text{mag}} = -\mu_0 \vec{H} \cdot d\vec{M}$$

The fundamental thermodynamic relation becomes:

$$dU = T dS - P dV + \mu_0 H dM$$

Neglecting small magnetostrictive volume changes ($dV \approx 0$):

$$dU = T dS + \mu_0 H dM$$

The magnetic Gibbs free energy is defined as $G_{\text{mag}} = U - TS - \mu_0 HM$:

$$dG_{\text{mag}} = -S dT - \mu_0 M dH$$

2. Curie's Law and Spin Entropy

For an ideal paramagnetic salt (e.g., cerium magnesium nitrate, gadolinium sulfate), the magnetic dipoles follow Curie's law:

$$M = \frac{C H}{T}$$

Applying Maxwell's cross-derivative to $dG_{\text{mag}}$:

$$\left(\frac{\partial S}{\partial H}\right)_T = \mu_0 \left(\frac{\partial M}{\partial T}\right)_H = -\mu_0 \frac{C H}{T^2} < 0$$

Applying an external magnetic field forces randomly tumbling atomic spins to align parallel to $\vec{H}$, drastically reducing spin orientation entropy ($S_{\text{spin}}$).

3. Cooling by Adiabatic Demagnetization

Adiabatic demagnetization (proposed by Debye and Giauque in 1926) achieves sub-millikelvin temperatures through a two-stage thermodynamic sequence:

1. Isothermal Magnetization ($A \to B$): The paramagnetic salt is immersed in a liquid helium bath at $T_i \approx 1.0\text{ K}$. A powerful magnetic field ($H_i \sim 2\text{ to } 5\text{ T}$) is applied. The spins align, releasing heat of magnetization $Q = T_i \Delta S_{\text{mag}}$ to the helium bath via helium exchange gas.

2. Adiabatic Demagnetization ($B \to C$): The exchange gas is pumped out to thermally isolate the salt. The external magnetic field is reduced slowly to zero ($H \to 0$).

Because the process is isentropic ($\Delta S_{\text{total}} = 0$):

$$S_{\text{spin}}(H_i, T_i) + S_{\text{lattice}}(T_i) = S_{\text{spin}}(0, T_f) + S_{\text{lattice}}(T_f)$$

As $H \to 0$, the magnetic spins disorder thermally, absorbing entropy and heat from the crystal lattice vibrations, causing the lattice temperature to plummet:

$$T_f = T_i \frac{\sqrt{h_{\text{int}}^2}}{H_i} = T_i \frac{h_{\text{int}}}{H_i}$$

where $h_{\text{int}} \sim 0.01\text{ T}$ is the small internal local dipole field of the crystal. By this technique, temperatures down to $10^{-3}\text{ K}$ (electronic demagnetization) and $10^{-6}\text{ K}$ (nuclear demagnetization of copper nuclei) are routinely achieved.

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 4.1: Surface Film Cooling During Reversible Adiabatic Expansion

A soap film stretched across a rectangular wire frame has a surface area of $A = 0.0200\text{ m}^2$ at temperature $T_0 = 293.15\text{ K}$. Its surface tension follows the linear empirical relation $\gamma(T) = \gamma_0 - b(T - T_0)$, where $\gamma_0 = 0.0280\text{ N/m}$ and $b = 1.40\times 10^{-4}\text{ N/(m}\cdot\text{K)}$. The total heat capacity of the liquid film is $C = 8.40\text{ J/K}$. (a) Calculate the total surface energy density $u_A$ at $T_0$. (b) If the film area is stretched reversibly and adiabatically from $A_1 = 0.0200\text{ m}^2$ to $A_2 = 0.0600\text{ m}^2$, derive and calculate the resulting temperature drop $\Delta T$.

Step 1: Calculate Total Surface Energy Density $u_A$

The surface internal energy density is:

$$u_A = \gamma - T \left(\frac{d\gamma}{dT}\right)$$

Given $\frac{d\gamma}{dT} = -b = -1.40 \times 10^{-4}\text{ N/(m}\cdot\text{K)}$:

$$u_A = 0.0280 - (293.15)(-1.40 \times 10^{-4}) = 0.0280 + 0.04104 = 0.06904\text{ J/m}^2$$

Notice that total surface internal energy is nearly $2.5\times$ greater than surface tension $\gamma$, because creating surface absorbs latent heat of orientation.

Step 2: Formulate Differential Equation for Adiabatic Stretching

For a reversible adiabatic process ($dS = 0$), the total entropy is a function of temperature and area, $S = S(T, A)$:

$$dS = \left(\frac{\partial S}{\partial T}\right)_A dT + \left(\frac{\partial S}{\partial A}\right)_T dA = 0$$

Recalling that $\left(\frac{\partial S}{\partial T}\right)_A = \frac{C}{T}$ and from Maxwell's relation $\left(\frac{\partial S}{\partial A}\right)_T = -\frac{d\gamma}{dT} = b$:

$$\frac{C}{T} dT + b dA = 0 \implies dT = -\frac{b T}{C} dA$$

Since $\Delta T \ll T_0$, we can approximate $T \approx T_0$:

$$\Delta T = -\frac{b T_0}{C} \Delta A$$
Step 3: Numerical Evaluation of Temperature Drop

The increase in surface area is $\Delta A = A_2 - A_1 = 0.0600 - 0.0200 = 0.0400\text{ m}^2$.

$$\Delta T = -\frac{(1.40 \times 10^{-4}\text{ N/(m}\cdot\text{K)})(293.15\text{ K})}{8.40\text{ J/K}} (0.0400\text{ m}^2)$$
$$\Delta T = -\frac{0.04104}{8.40} \times 0.0400 = - (0.004886)(0.0400) = -1.954 \times 10^{-4}\text{ K}$$

For double-sided soap film (two interfaces, area $2\Delta A$):

$$\Delta T_{\text{film}} = 2 \times (-1.954 \times 10^{-4}\text{ K}) = -3.91 \times 10^{-4}\text{ K} = -0.391\text{ mK}$$
Final Answer & Physical Insight

(a) Total surface energy density $u_A = 0.0690\text{ J/m}^2$. (b) Adiabatic stretching produces a cooling effect of $\Delta T = -0.391\text{ mK}$ for a double-sided soap film.

Standard Example 4.2: Cooling of Paramagnetic Salt by Adiabatic Demagnetization

A crystal of gadolinium sulfate of mass $m = 0.150\text{ kg}$ is magnetized isothermally at initial temperature $T_i = 1.20\text{ K}$ in a magnetic field of $H_i = 2.50\text{ T}$. The effective internal local dipole field of the crystal lattice is $h_{\text{int}} = 0.0350\text{ T}$. (a) Calculate the final temperature $T_f$ when the external field is reduced adiabatically to zero ($H_f = 0$). (b) Calculate the final temperature if the field is reduced to a residual value of $H_f = 0.200\text{ T}$. (c) Determine the heat absorbed from a cryogenic sample when warmed back up to $1.20\text{ K}$ if the total heat capacity is approximated as $C = \alpha T^3$ with $\alpha = 0.0450\text{ J/K}^4$.

Step 1: Calculate Final Temperature for Complete Demagnetization ($H_f = 0$)

During isentropic demagnetization of an ideal paramagnetic dipole system, total entropy depends on the effective field-to-temperature ratio $\frac{\sqrt{H^2 + h_{\text{int}}^2}}{T} = \text{constant}$.

$$T_f = T_i \sqrt{\frac{H_f^2 + h_{\text{int}}^2}{H_i^2 + h_{\text{int}}^2}}$$

For $H_f = 0$:

$$T_f = T_i \frac{h_{\text{int}}}{\sqrt{H_i^2 + h_{\text{int}}^2}}$$

Since $H_i = 2.50\text{ T} \gg h_{\text{int}} = 0.0350\text{ T}$, $\sqrt{2.50^2 + 0.0350^2} \approx 2.5002\text{ T}$:

$$T_f = (1.20\text{ K}) \left(\frac{0.0350}{2.5002}\right) = (1.20)(0.01400) = 0.0168\text{ K} = 16.8\text{ mK}$$
Step 2: Calculate Final Temperature for Partial Demagnetization ($H_f = 0.200\text{ T}$)

With residual external field $H_f = 0.200\text{ T}$:

$$\sqrt{H_f^2 + h_{\text{int}}^2} = \sqrt{(0.200)^2 + (0.0350)^2} = \sqrt{0.0400 + 0.001225} = \sqrt{0.041225} = 0.2030\text{ T}$$
$$T_f = (1.20\text{ K}) \left(\frac{0.2030}{2.5002}\right) = (1.20)(0.08119) = 0.0974\text{ K} = 97.4\text{ mK}$$
Step 3: Calculate Cryogenic Heat Absorbed During Reheating

The heat capacity of the lattice and spin reservoir is $C = \alpha T^3$ with $\alpha = 0.0450\text{ J/K}^4$. Reheating from $T_f = 0.0168\text{ K}$ to $T_i = 1.20\text{ K}$:

$$Q = \int_{T_f}^{T_i} \alpha T^3 dT = \frac{\alpha}{4} [T_i^4 - T_f^4]$$
$$(1.20)^4 = 2.0736, \quad (0.0168)^4 \approx 7.96 \times 10^{-8} \approx 0$$
$$Q = \frac{0.0450}{4} (2.0736) = (0.01125)(2.0736) = 0.02333\text{ J} = 23.33\text{ mJ}$$
Final Answer & Physical Insight

(a) Complete demagnetization yields $T_f = 16.8\text{ mK}$. (b) Partial demagnetization to $0.20\text{ T}$ yields $T_f = 97.4\text{ mK}$. (c) Cryogenic cooling capacity absorbed upon warming to $1.20\text{ K}$ is $Q = 23.3\text{ mJ}$.

Standard Example 4.3: Derivation of Gibbs-Helmholtz Equation & Chemical Reaction Spontaneity

Starting from the definition of Gibbs free energy $G = H - TS$ and the fundamental differential $dG = -S dT + V dP$: (a) derive the Gibbs-Helmholtz equation in both the form $\left(\frac{\partial (G/T)}{\partial T}\right)_P = -\frac{H}{T^2}$ and $\left(\frac{\partial (\Delta G/T)}{\partial (1/T)}\right)_P = \Delta H$. (b) A chemical reaction has an enthalpy change of $\Delta H^\circ = -84.2\text{ kJ/mol}$ and an entropy change of $\Delta S^\circ = -165.0\text{ J/(mol}\cdot\text{K)}$. Calculate $\Delta G^\circ$ at $298.15\text{ K}$ and determine the inversion temperature above which the reaction ceases to be spontaneous.

Step 1: Derivation of the Gibbs-Helmholtz Relation

From $dG = -S dT + V dP$, at constant pressure ($dP = 0$):

$$\left(\frac{\partial G}{\partial T}\right)_P = -S$$

Substituting $-S = \frac{G - H}{T}$ from $G = H - TS$:

$$\left(\frac{\partial G}{\partial T}\right)_P = \frac{G - H}{T} \implies G - T \left(\frac{\partial G}{\partial T}\right)_P = H$$

Now evaluate the derivative of the quotient $G / T$ with respect to $T$:

$$\left(\frac{\partial (G/T)}{\partial T}\right)_P = \frac{T \left(\frac{\partial G}{\partial T}\right)_P - G}{T^2} = -\frac{G - T \left(\frac{\partial G}{\partial T}\right)_P}{T^2} = -\frac{H}{T^2}$$

Using the chain rule with $u = 1/T$, so $du = -dT/T^2$ or $\frac{d}{d(1/T)} = -T^2 \frac{d}{dT}$:

$$\left(\frac{\partial (G/T)}{\partial (1/T)}\right)_P = -T^2 \left(\frac{\partial (G/T)}{\partial T}\right)_P = -T^2 \left(-\frac{H}{T^2}\right) = H$$

For a finite reaction transformation: $\left(\frac{\partial (\Delta G/T)}{\partial (1/T)}\right)_P = \Delta H$.

Step 2: Calculate $\Delta G^\circ$ at $298.15\text{ K}$

Using $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$:

$$\Delta H^\circ = -84200\text{ J/mol}$$
$$T \Delta S^\circ = (298.15\text{ K})(-165.0\text{ J/(mol}\cdot\text{K)}) = -49194.75\text{ J/mol}$$
$$\Delta G^\circ = -84200 - (-49194.75) = -84200 + 49194.75 = -35005.25\text{ J/mol} = -35.01\text{ kJ/mol}$$

Since $\Delta G^\circ < 0$, the reaction is thermodynamically spontaneous at $298.15\text{ K}$.

Step 3: Calculate Inversion Temperature for Spontaneity

Spontaneity boundary occurs where $\Delta G^\circ = 0$:

$$\Delta H^\circ - T_{\text{inv}} \Delta S^\circ = 0 \implies T_{\text{inv}} = \frac{\Delta H^\circ}{\Delta S^\circ}$$
$$T_{\text{inv}} = \frac{-84200\text{ J/mol}}{-165.0\text{ J/(mol}\cdot\text{K)}} = 510.30\text{ K} = 237.15^\circ\text{C}$$

For $T < 510.3\text{ K}$, $\Delta G^\circ < 0$ (spontaneous). For $T > 510.3\text{ K}$, the unfavorable entropy loss ($-T\Delta S > 0$) overcomes the favorable exothermic enthalpy ($\Delta H < 0$), causing $\Delta G^\circ > 0$ and terminating spontaneity.

Final Answer & Physical Insight

(a) Proof yields $\left(\frac{\partial (\Delta G/T)}{\partial (1/T)}\right)_P = \Delta H$. (b) At $298.15\text{ K}$, $\Delta G^\circ = -35.01\text{ kJ/mol}$ (spontaneous); inversion temperature is $T_{\text{inv}} = 510.3\text{ K}$ ($237.2^\circ\text{C}$).