Maxwell Relations

Why This Matters

Maxwell relations let you convert thermodynamic quantities that are impossible to measure directly into ones you can actually work with in the lab. They come from a straightforward mathematical truth about state functions: the order of differentiation doesn't matter, which reveals deep connections between seemingly unrelated properties.

When you encounter problems involving entropy changes, heat capacities, or equations of state, Maxwell relations are often the key to finding a solution. Don't just memorize the four equations. Understand which thermodynamic potential each one comes from and what physical insight each relation provides. Exam questions frequently ask you to derive a Maxwell relation from scratch or apply one to transform an unmeasurable quantity into something accessible.

---

The Mathematical Foundation: Exact Differentials

Maxwell relations exist because thermodynamic potentials are state functions with exact differentials. That means their mixed second partial derivatives are equal regardless of the order you differentiate.

Equality of Mixed Partial Derivatives

Schwarz's theorem (also called Clairaut's theorem) guarantees that for any well-behaved function $f(x,y)$:

$\frac{\partial^2 f}{\partial x \, \partial y} = \frac{\partial^2 f}{\partial y \, \partial x}$

This mathematical symmetry is the entire foundation of Maxwell relations. No physics is needed to justify it; it's pure calculus. But the physical consequence is significant: it connects thermodynamic properties that seem completely unrelated.

State Functions and Path Independence

State functions like $U$, $H$, $A$, and $G$ depend only on the current state of the system, not on how it got there. Their exact differentials take the form:

$dz = M\,dx + N\,dy \quad \text{where} \quad \left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y$

This exactness condition is precisely what generates each Maxwell relation when you apply it to a thermodynamic potential.

---

The Four Thermodynamic Potentials

Each Maxwell relation derives from one of the four fundamental thermodynamic potentials. The natural variables of each potential determine which Maxwell relation you get.

Internal Energy $U(S,V)$

• Fundamental equation: $dU = T\,dS - P\,dV$, with natural variables entropy and volume
• Coefficients identify: $T = \left(\frac{\partial U}{\partial S}\right)_V$ and $-P = \left(\frac{\partial U}{\partial V}\right)_S$
• Maxwell relation: $\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V$

This connects how temperature changes with volume during an adiabatic process to how pressure changes with entropy at constant volume.

Enthalpy $H(S,P)$

• Fundamental equation: $dH = T\,dS + V\,dP$, with natural variables entropy and pressure
• Coefficients identify: $T = \left(\frac{\partial H}{\partial S}\right)_P$ and $V = \left(\frac{\partial H}{\partial P}\right)_S$
• Maxwell relation: $\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P$

This one is useful for adiabatic processes at constant pressure.

Helmholtz Free Energy $A(T,V)$

• Fundamental equation: $dA = -S\,dT - P\,dV$, with natural variables temperature and volume
• Coefficients identify: $-S = \left(\frac{\partial A}{\partial T}\right)_V$ and $-P = \left(\frac{\partial A}{\partial V}\right)_T$
• Maxwell relation: $\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$

This is arguably the most frequently used Maxwell relation. It connects entropy (unmeasurable directly) to the equation of state (measurable). Any time you need $\left(\frac{\partial S}{\partial V}\right)_T$, you can replace it with $\left(\frac{\partial P}{\partial T}\right)_V$, which you can calculate from any equation of state.

Gibbs Free Energy $G(T,P)$

• Fundamental equation: $dG = -S\,dT + V\,dP$, with natural variables temperature and pressure
• Coefficients identify: $-S = \left(\frac{\partial G}{\partial T}\right)_P$ and $V = \left(\frac{\partial G}{\partial P}\right)_T$
• Maxwell relation: $\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P$

This connects entropy changes to the thermal expansion coefficient. It appears most often in chemistry because lab work typically occurs at constant $T$ and $P$.

---

How to Derive a Maxwell Relation (Step-by-Step)

This process works for any of the four potentials. Here it is using $G$ as an example:

1. Write the fundamental equation for the potential: $dG = -S\,dT + V\,dP$
2. Identify the coefficients. In the general form $dz = M\,dx + N\,dy$, you have $M = -S$ (coefficient of $dT$) and $N = V$ (coefficient of $dP$).
3. Apply the exactness condition: $\left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y$
4. Substitute: $\left(\frac{\partial (-S)}{\partial P}\right)_T = \left(\frac{\partial V}{\partial T}\right)_P$
5. Simplify the sign: $\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P$

Practice this derivation for all four potentials until it feels automatic.

---

Practical Applications

Maxwell relations transform unmeasurable quantities into experimentally accessible ones. The key is recognizing when a derivative you need can be replaced by one you can measure.

Calculating Entropy Changes

The derivatives $\left(\frac{\partial S}{\partial V}\right)_T$ and $\left(\frac{\partial S}{\partial P}\right)_T$ cannot be measured directly. You can't stick a probe into a system and read out entropy. But Maxwell relations convert these into $\left(\frac{\partial P}{\partial T}\right)_V$ and $-\left(\frac{\partial V}{\partial T}\right)_P$, both of which you can calculate from an equation of state.

When asked to find $\Delta S$ for an isothermal process, immediately look for the appropriate Maxwell relation.

Deriving the $C_P - C_V$ Relationship

The derivation of $C_P - C_V$ relies critically on Maxwell relations to eliminate entropy derivatives. The final result is:

$C_P - C_V = \frac{TV\alpha^2}{\kappa_T}$

where $\alpha$ is the thermal expansion coefficient and $\kappa_T$ is the isothermal compressibility. Both are measurable. This equation explains why $C_P > C_V$ for all substances (since $T$, $V$, and $\kappa_T$ are always positive, and $\alpha^2$ is non-negative) and quantifies the difference.

Joule-Thomson Coefficient

The Joule-Thomson coefficient $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$ describes how temperature changes during a throttling process. Using Maxwell relations and some manipulation, you can express it as:

$\mu_{JT} = \frac{V}{C_P}(T\alpha - 1)$

This predicts whether a gas cools ($\mu_{JT} > 0$) or warms ($\mu_{JT} < 0$) upon expansion, which is critical for refrigeration and gas liquefaction.

---

Special Systems and Phase Transitions

Maxwell relations become particularly powerful when applied to specific systems or phase boundaries.

Ideal Gas Verification

For an ideal gas ($PV = nRT$), you can verify the Helmholtz Maxwell relation directly:

$\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V}$

The Maxwell relation says this must equal $\left(\frac{\partial S}{\partial V}\right)_T$, and it does. The physical meaning: entropy increases with volume at constant temperature because more spatial microstates become accessible.

Ideal gas problems are excellent for checking your work. If the Maxwell relation doesn't hold for an ideal gas, you've made an error somewhere.

Clapeyron Equation and Phase Boundaries

At a phase boundary, two phases are in equilibrium, so $dG^\alpha = dG^\beta$ along the coexistence curve. This leads to the Clapeyron equation:

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

This follows from the same kind of reasoning that underlies Maxwell relations (equating derivatives of $G$). It predicts how melting points and boiling points shift with pressure and describes vapor pressure curves.

Critical Point Behavior

Near the critical point, $\left(\frac{\partial P}{\partial V}\right)_T \to 0$ and $\left(\frac{\partial^2 P}{\partial V^2}\right)_T \to 0$. Through Maxwell relations, this leads to diverging response functions like $\kappa_T$ and $C_P$ at criticality. All fluids show this same universal behavior near their critical points, regardless of chemical identity.

---

Quick Reference Table

---

Self-Check Questions

1. Starting from $dG = -S\,dT + V\,dP$, derive the Maxwell relation $\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P$ and explain why this relation is useful for calculating entropy changes at constant temperature.

2. Which two thermodynamic potentials have temperature as a natural variable, and how do you decide which one to use for a given problem?

3. Compare the Maxwell relations derived from $U(S,V)$ and $H(S,P)$. What physical situations favor each one?

4. You need to express $\left(\frac{\partial S}{\partial V}\right)_T$ in terms of measurable quantities for a van der Waals gas. Which Maxwell relation do you use, and what equation of state derivative do you need to evaluate?

5. Explain why Maxwell relations cannot be applied to heat ($q$) or work ($w$), and identify what mathematical property distinguishes state functions from path functions in this context.