The Maxwell Relations and Their Uses
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You must be comfortable with the four thermodynamic potentials (internal energy, enthalpy, Helmholtz free energy, Gibbs free energy) and their total differentials. You should also understand the concept of exact differentials and the mathematical condition for a differential to be integrable.
Why this matters
The Maxwell relations allow us to replace difficult-to-measure thermodynamic quantities with easily measurable ones.
In experimental thermodynamics, we rarely have direct access to entropy or internal energy. We can, however, measure pressure, volume, and temperature with high precision. The Maxwell relations, derived from the symmetry of second derivatives of thermodynamic potentials, provide the crucial bridge between these measurable variables and the abstract state functions. They are the workhorses of thermodynamic analysis, enabling the calculation of heat capacities, equation of state corrections, and the behaviour of real gases without ever needing to count molecules directly.
The Symmetry of Mixed Derivatives
Imagine a smooth landscape defined by a function of two variables, such as the height of a hill depending on your latitude and longitude. If you walk north and then east, the change in height is the same as if you walked east and then north. This is a fundamental property of smooth functions: the order in which you take partial derivatives does not matter. In calculus, this is known as the equality of mixed partial derivatives. Thermodynamic potentials like the Helmholtz free energy are smooth functions of their natural variables. Therefore, if we differentiate the potential with respect to temperature and then volume, we must get the same result as differentiating with respect to volume and then temperature. This mathematical symmetry, when applied to the differentials of the thermodynamic potentials, yields the Maxwell relations. They are not new physical laws; they are consequences of the mathematical consistency of the state functions we have already defined.
Deriving the Relations from Potentials
To obtain the Maxwell relations, we examine the total differential of each thermodynamic potential. For the internal energy \( U \), the differential is \( \mathrm{d}U = T\mathrm{d}S - P\mathrm{d}V \). Since \( U \) is a state function, its differential is exact. This implies that the cross-derivatives of its coefficients must be equal. Specifically, the derivative of \( T \) with respect to \( V \) at constant \( S \) must equal the negative of the derivative of \( P \) with respect to \( S \) at constant \( V \). This gives us the first Maxwell relation: \( \left( \frac{\partial T}{\partial V} \right)_S = - \left( \frac{\partial P}{\partial S} \right)_V \). Similarly, by considering the other potentials, we can derive the other three relations. For the Helmholtz free energy \( F = U - TS \), the differential is \( \mathrm{d}F = -S\mathrm{d}T - P\mathrm{d}V \). Applying the same logic, we find \( \left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V \). This particular relation is often the most useful in practice because it relates entropy changes to pressure-temperature behaviour, both of which are measurable.
Practical Applications
The primary utility of the Maxwell relations is in calculating entropy changes. Since entropy is not directly measurable, we use these relations to express \( \mathrm{d}S \) in terms of \( \mathrm{d}T \) and \( \mathrm{d}V \) or \( \mathrm{d}P \). For example, using the Helmholtz relation above, we can write the change in entropy for an ideal gas. For an ideal gas, \( P = \frac{nRT}{V} \), so \( \left( \frac{\partial P}{\partial T} \right)_V = \frac{nR}{V} \). Thus, \( \left( \frac{\partial S}{\partial V} \right)_T = \frac{nR}{V} \). Integrating this with respect to volume gives the familiar logarithmic dependence of entropy on volume. This allows us to calculate the entropy change for any process involving an ideal gas, provided we know the initial and final states. The relations also allow us to calculate the Joule-Thomson coefficient and other response functions that describe how real gases deviate from ideal behaviour.
Consistency and Constraints
The Maxwell relations also serve as a consistency check for equations of state. If a proposed equation of state for a substance does not satisfy the Maxwell relations, it is thermodynamically inconsistent and therefore physically impossible. For instance, if we propose a new equation of state for a fluid, we can use the Maxwell relations to predict how its entropy should vary with volume and temperature. If experimental measurements of heat capacity or thermal expansion do not match these predictions, the proposed equation of state is invalid. This makes the Maxwell relations a powerful tool for validating theoretical models of matter. They ensure that our macroscopic descriptions of thermodynamics are internally consistent and compatible with the second law.
The derivation
The Maxwell Relations
Equality of mixed second partials of the potentials yields the four Maxwell relations linking entropy derivatives to measurable mechanical quantities.
A common pitfall is assuming that the Maxwell relations apply to non-equilibrium states. They are derived from the assumption that the system is in internal equilibrium and that the thermodynamic potentials are well-defined state functions. If the system is undergoing a rapid, irreversible process, the local variables may not satisfy the equilibrium relations, and the Maxwell relations may not hold locally. Always ensure that the system is in a quasi-static equilibrium state before applying these relations. C
Worked example
Problem. Consider one mole of an ideal gas undergoing an isothermal expansion from volume \( V_1 = 10\ \mathrm{L} \) to \( V_2 = 20\ \mathrm{L} \) at a constant temperature \( T = 300\ \mathrm{K} \). Calculate the change in entropy \( \Delta S \) using the Maxwell relation derived from the Helmholtz free energy.
- Identify the relevant Maxwell relation: \( \left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V \).
- Use the ideal gas law \( P = \frac{nRT}{V} \) to find \( \left( \frac{\partial P}{\partial T} \right)_V = \frac{nR}{V} \).
- Substitute this into the Maxwell relation: \( \left( \frac{\partial S}{\partial V} \right)_T = \frac{nR}{V} \).
- Integrate with respect to volume from \( V_1 \) to \( V_2 \): \( \Delta S = \int_{V_1}^{V_2} \frac{nR}{V} \mathrm{d}V = nR \ln\left( \frac{V_2}{V_1} \right) \).
- Substitute the numerical values: \( n = 1\ \mathrm{mol} \), \( R = 8.314\ \mathrm{J/(mol \cdot K)} \), \( V_1 = 10\ \mathrm{L} \), \( V_2 = 20\ \mathrm{L} \).
- Calculate \( \Delta S = 1 \times 8.314 \times \ln(2) \approx 8.314 \times 0.693 \approx 5.76\ \mathrm{J/K} \).
Units check. The gas constant \( R \) has units of \( \mathrm{J/(mol \cdot K)} \). The natural logarithm is dimensionless. The number of moles \( n \) is in moles. Thus, the units are \( \mathrm{mol} \times \mathrm{J/(mol \cdot K)} = \mathrm{J/K} \), which is the correct unit for entropy.
Three quick tests
- What is the Maxwell relation derived from the Gibbs free energy that relates entropy and pressure?
Answer
\( \left( \frac{\partial S}{\partial P} \right)_T = - \left( \frac{\partial V}{\partial T} \right)_P \) - If \( \left( \frac{\partial P}{\partial T} \right)_V \) is positive, what can you say about \( \left( \frac{\partial S}{\partial V} \right)_T \)?
Answer
It is also positive, meaning entropy increases with volume at constant temperature. - Why are the Maxwell relations considered a consequence of the second law?
Answer
Because they rely on the existence of well-defined thermodynamic potentials, which are constructed using entropy, a quantity defined by the second law.
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.