Maxwell’s Thermodynamic Relations

1. Mathematical Foundation of Maxwell's Relations

James Clerk Maxwell (1871) realized that because the four thermodynamic potentials ($U, H, F, G$) are exact state functions, their mixed second partial derivatives must commute by Schwarz's theorem:

2. Systematic Derivations from Fundamental Differentials

1. First Maxwell Relation (from Internal Energy $U(S, V)$):

Since $dU$ is exact:

Equating mixed second derivatives $\frac{\partial^2 U}{\partial V \partial S} = \frac{\partial^2 U}{\partial S \partial V}$:

2. Second Maxwell Relation (from Enthalpy $H(S, P)$):

Equating $\frac{\partial^2 H}{\partial P \partial S} = \frac{\partial^2 H}{\partial S \partial P}$:

3. Third Maxwell Relation (from Helmholtz Free Energy $F(T, V)$):

Equating $\frac{\partial^2 F}{\partial V \partial T} = \frac{\partial^2 F}{\partial T \partial V}$:

4. Fourth Maxwell Relation (from Gibbs Free Energy $G(T, P)$):

Equating $\frac{\partial^2 G}{\partial P \partial T} = \frac{\partial^2 G}{\partial T \partial P}$:

3. Max Born's Mnemonic Square

To quickly reconstruct these relations, Max Born proposed the mnemonic square:

$$\begin{array}{ccc} -V & & P \\ F & & H \\ -T & & S \end{array} \quad \text{or mnemonic: "Good Physicists Have Studied Under Very Fine Teachers"}$$

The potentials flank the edges between their natural variables. Arrow directions encode negative signs.

1. Phase Coexistence Equilibrium

Consider two distinct macroscopic phases of a pure substance (e.g., liquid 1 and vapor 2) coexisting in dynamic phase equilibrium at temperature $T$ and pressure $P$. Equilibrium requires equality of their specific Gibbs free energies (chemical potentials):

If temperature shifts by an infinitesimal $dT$, the equilibrium coexistence pressure must shift along the phase line by $dP$ such that:

Expanding in total differentials $dg = -s dT + v dP$:

Rearranging terms:

2. The Latent Heat Formula

The entropy change during a reversible phase transformation at constant temperature $T$ is related to the latent heat $L$ per mole by $\Delta s = L / T$:

This is the Clausius-Clapeyron Equation.

3. Physical Applications

1. Boiling Point Elevation with Pressure: For vaporization, $v_{\text{vapor}} \gg v_{\text{liquid}} \implies \Delta v > 0$. Since vaporization is endothermic ($L > 0$), $\frac{dP}{dT} > 0$. Increasing pressure elevates the boiling point (utilized in pressure cookers and autoclave sterilization).

2. Depression of Freezing Point of Water: For water, ice has a lower density than liquid water at $0^\circ\text{C}$ ($v_{\text{water}} < v_{\text{ice}} \implies \Delta v = v_{\text{water}} - v_{\text{ice}} < 0$). Because latent heat of fusion $L_f > 0$:

Applying pressure lowers the melting temperature of ice (facilitating glacier regelation and ice skating).

1. Derivation of the First TdS Equation

Let entropy be expressed as a function of temperature and volume, $S = S(T, V)$:

Multiplying by absolute temperature $T$:

Using the definition of isochoric heat capacity $C_v = T (\partial S/\partial T)_V$ and the Third Maxwell Relation $(\partial S/\partial V)_T = (\partial P/\partial T)_V$:

This is the First $T dS$ Equation.

2. Derivation of the Second TdS Equation

Now let entropy be expressed as a function of temperature and pressure, $S = S(T, P)$:

Multiplying by $T$:

Using isobaric heat capacity $C_p = T (\partial S/\partial T)_P$ and the Fourth Maxwell Relation $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$:

This is the Second $T dS$ Equation.

3. Rigorous Derivation of $C_p - C_v$

Equating the First and Second $T dS$ equations:

Dividing by $dT$ at constant pressure ($dP = 0$):

Expressing in terms of thermal expansion coefficient $\alpha = \frac{1}{V}(\partial V/\partial T)_P$ and isothermal compressibility $\kappa_T = -\frac{1}{V}(\partial V/\partial P)_T$, with the cyclic identity $(\partial P/\partial T)_V = \alpha / \kappa_T$:

1. The First Energy Equation (Volume Dependence of U)

From the First Law, $dU = T dS - P dV$. Dividing by $dV$ at constant temperature $T$:

Substituting the Third Maxwell Relation $(\partial S/\partial V)_T = (\partial P/\partial T)_V$:

This is the First Energy Equation. The term $(\partial U/\partial V)_T$ is called the internal pressure ($P_i$), representing the internal force per unit area exerted by intermolecular attractive forces.

Examples:

1. Ideal Gas: $P = nRT / V \implies (\partial P/\partial T)_V = nR/V$.

proving Joule's law theoretically without approximations.

2. Van der Waals Gas: $P = \frac{n R T}{V - n b} - \frac{n^2 a}{V^2} \implies (\partial P/\partial T)_V = \frac{n R}{V - n b}$.

Internal energy increases during isothermal expansion as intermolecular bonds stretch.

2. The Second Energy Equation (Pressure Dependence of H)

From enthalpy $dH = T dS + V dP$, dividing by $dP$ at constant temperature $T$:

Substituting the Fourth Maxwell Relation $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$:

This is the Second Energy Equation. For an ideal gas, $(\partial V/\partial T)_P = V/T$, so $(\partial H/\partial P)_T = 0$.

1. The Porous Plug Throttling Experiment

In the Joule-Thomson (Joule-Kelvin) experiment (1852), a continuous gas stream under steady high pressure $P_1$ is forced through an insulated porous plug or throttling valve to a lower pressure $P_2$.

• Upstream work done on the gas by piston: $W_1 = -P_1 V_1$.
• Downstream work done by the gas on piston: $W_2 = +P_2 V_2$.
• Net work done by gas: $W = P_2 V_2 - P_1 V_1$.
• Because the tube is thermally insulated, $Q = 0$.

By the First Law:

A throttling process is strictly isenthalpic ($\Delta H = 0$).

2. Derivation of the Joule-Thomson Coefficient

The Joule-Thomson Coefficient $\mu_{\text{JT}}$ is the rate of temperature change with pressure under constant enthalpy:

By the cyclic permutation identity for $(T, P, H)$:

Substituting the Second Energy Equation $(\partial H/\partial P)_T = V - T(\partial V/\partial T)_P$:

3. Joule-Thomson Cooling for Real Gases

1. Ideal Gas: $(\partial V/\partial T)_P = V/T \implies \mu_{\text{JT}} = 0$ (no temperature change).

2. Van der Waals Gas: Using $(P + a/V_m^2)(V_m - b) \approx RT$, we find:

• Cooling Zone ($\mu_{\text{JT}} > 0$): Since $dP < 0$ across the throttle, $dT = \mu_{\text{JT}} dP < 0$. Gas cools if $T < T_i$.
• Heating Zone ($\mu_{\text{JT}} < 0$): Gas heats up upon throttling if $T > T_i$.
• Temperature of Inversion $T_i$: The boundary where $\mu_{\text{JT}} = 0$:

For air, $T_i \approx 600\text{ K}$ (cools at room temperature). For hydrogen ($T_i \approx 200\text{ K}$) and helium ($T_i \approx 40\text{ K}$), pre-cooling below their inversion temperatures is mandatory before throttling can produce liquefaction.