Thermodynamics — Maxwell’s relations are valid for which type of thermodynamic processes/systems?

Introduction:
Maxwell’s thermodynamic relations are elegant identities linking partial derivatives of state variables. They arise from the equality of mixed second derivatives of thermodynamic potentials and require conditions under which those potentials are exact differentials. Understanding when they apply is foundational for property calculations.

Given Data / Assumptions:

• Thermodynamic potentials (U, H, F, G) are functions of state.
• Systems considered are in thermodynamic equilibrium and undergo reversible, quasi-static changes.
• No explicit non-equilibrium transport gradients (finite rate irreversibilities) are present.

Concept / Approach:

Maxwell relations stem from the fundamental equations (e.g., dU = T dS − p dV for simple compressible systems) and Legendre transforms that define H, F, and G. Because these differentials are exact in reversible, equilibrium settings, mixed partial derivatives commute, yielding the Maxwell identities that connect measurable properties like (∂T/∂V)_S and (∂p/∂S)_V, etc. Irreversible processes break the exact differential form due to entropy production terms, so the identities no longer strictly hold without additional assumptions or corrections.

Step-by-Step Solution:

Verification / Alternative check:

Deriving one Maxwell relation from Helmholtz free energy F(T,V): dF = −S dT − p dV → (∂S/∂V)_T = (∂p/∂T)_V confirms the equilibrium, reversible context.

Why Other Options Are Wrong:

A/B: While such systems may be in equilibrium, the relations are about reversible thermodynamic paths; the phrasing is too narrow or misleading. C/E: Irreversible/non-equilibrium processes violate the assumptions behind exact differentials and mixed-partial symmetry.

Common Pitfalls:

Applying Maxwell relations directly to fast, dissipative processes or to systems far from equilibrium where local state variables are ill-defined.

Final Answer:

reversible thermodynamic processes