Maxwell Relations

Why This Matters

Maxwell relations are the Swiss Army knife of thermodynamics—they let you convert quantities that are impossible to measure directly into ones you can actually work with in the lab. You're being tested on your ability to derive these relations from thermodynamic potentials, recognize which relation to apply in a given situation, and use them to simplify complex calculations. These relations demonstrate a fundamental mathematical truth about thermodynamic 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 unlocking 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.

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The Mathematical Foundation: Exact Differentials

Maxwell relations exist because thermodynamic potentials are state functions with exact differentials—meaning the mixed second partial derivatives are equal regardless of the order of differentiation.

Equality of Mixed Partial Derivatives

• Schwarz's theorem guarantees that for any well-behaved function $f(x,y)$, we have $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$
• This mathematical symmetry is the entire foundation of Maxwell relations—no physics required, just calculus
• The physical consequence is profound: it reveals hidden connections between thermodynamic properties that appear unrelated

State Functions and Path Independence

• State functions like $U$, $H$, $A$, and $G$ depend only on the current state, not how the system got there
• Their exact differentials can be written as $dz = M\,dx + N\,dy$ where $\left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y$
• This exactness condition is precisely what generates each Maxwell relation when applied to thermodynamic potentials

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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: $dU = TdS - PdV$, giving 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: $\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V$—connects adiabatic temperature-volume behavior to entropy-pressure relationships

Enthalpy $H(S,P)$

• Fundamental equation: $dH = TdS + VdP$, 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: $\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P$—useful for adiabatic processes at constant pressure

Helmholtz Free Energy $A(T,V)$

• Fundamental equation: $dA = -SdT - PdV$, 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: $\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$—the most frequently used relation, connects entropy to the equation of state

Gibbs Free Energy $G(T,P)$

• Fundamental equation: $dG = -SdT + VdP$, 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: $\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P$—connects entropy changes to thermal expansion, essential for chemical applications

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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 problem: $\left(\frac{\partial S}{\partial V}\right)_T$ and $\left(\frac{\partial S}{\partial P}\right)_T$ cannot be measured directly—you can't "see" entropy
• The solution: Use Maxwell relations to convert these to $\left(\frac{\partial P}{\partial T}\right)_V$ and $-\left(\frac{\partial V}{\partial T}\right)_P$, both measurable from equations of state
• Exam application: When asked to find $\Delta S$ for an isothermal process, immediately look for the appropriate Maxwell relation

Deriving Heat Capacity Relationships

• $C_P - C_V$ derivation relies critically on Maxwell relations to eliminate entropy derivatives
• The final result $C_P - C_V = \frac{TV\alpha^2}{\kappa_T}$ uses $\alpha$ (thermal expansion) and $\kappa_T$ (isothermal compressibility)—both measurable
• Why it matters: This equation explains why $C_P > C_V$ for all substances and quantifies the difference

Joule-Thomson Coefficient

• The coefficient $\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$ describes temperature change during throttling
• Maxwell relations help express this in terms of $C_P$, $V$, and $\alpha$: $\mu_{JT} = \frac{V}{C_P}(T\alpha - 1)$
• Physical insight: Predicts whether a gas cools or warms upon expansion—critical for refrigeration and gas liquefaction

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Special Systems and Phase Transitions

Maxwell relations become particularly powerful when applied to specific systems or phase boundaries. They reveal universal behavior that transcends the details of any particular substance.

Ideal Gas Verification

• Test the relations: For an ideal gas with $PV = nRT$, verify that $\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V}$ and $\left(\frac{\partial S}{\partial V}\right)_T = \frac{nR}{V}$
• Physical meaning: Entropy increases with volume at constant temperature because more microstates become accessible
• Exam strategy: Ideal gas problems are excellent for checking your Maxwell relation work—if the relation doesn't hold, you made an error

Clapeyron Equation and Phase Boundaries

• Phase equilibrium requires $dG^\alpha = dG^\beta$ along the coexistence curve
• The Clapeyron equation $\frac{dP}{dT} = \frac{\Delta S}{\Delta V} = \frac{\Delta H}{T\Delta V}$ follows from Maxwell-type reasoning
• Applications: Predicts how melting points change with pressure, explains why ice skating works, describes vapor pressure curves

Critical Point Behavior

• Near the critical point, $\left(\frac{\partial P}{\partial V}\right)_T \to 0$ and $\left(\frac{\partial^2 P}{\partial V^2}\right)_T \to 0$
• Maxwell relations predict diverging response functions like $\kappa_T$ and $C_P$ at criticality
• Significance: Universal behavior near critical points—all fluids behave similarly regardless of chemical identity

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Quick Reference Table

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Self-Check Questions

1. Starting from $dG = -SdT + VdP$, 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 and contrast the Maxwell relations derived from $U(S,V)$ and $H(S,P)$—what physical situations favor each one?

4. An FRQ asks you 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?

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.