Entropy as a state function — choosing appropriate independent variables: For a simple compressible substance, entropy can be expressed as a function of which independent variables?

Introduction / Context:
Entropy s is a property (state function). For simple compressible substances (including ideal gases), the thermodynamic state is fully specified by two independent intensive variables. Choosing the proper pair is foundational for property calculations and tables.

Given Data / Assumptions:

• Single-phase, simple compressible substance.
• Equilibrium conditions (uniform properties).
• Fixed composition.

Concept / Approach:
Because s is a state property, it can be expressed as s = s(p, T) or s = s(T, v) (and other valid pairs like s(p, v) with constitutive closure). “Heat” and “work” are path-dependent interactions, not properties; they cannot serve as state coordinates. The choice of independent variables affects convenience: gases often use (T, p), while tables for water may use (T, v) or (T, p).

Step-by-Step Solution:
Recognize that two independent variables determine the state for a simple compressible substance.Identify valid pairs: (p, T) and (T, v) are both sufficient to fix state and hence s.Eliminate invalid pair: (Q, W) are not properties and depend on the process path.Therefore, the most complete correct option is “either (p, T) or (T, v).”

Verification / Alternative check:
Property relations (e.g., Maxwell relations) and equations of state allow expressing s in different coordinate pairs; steam tables frequently tabulate s against T and p.

Why Other Options Are Wrong:
Options A and B are each partially correct but not uniquely most complete; option C is incorrect because Q and W are not state variables.

Common Pitfalls:
Assuming a unique pair is required; in reality, any independent pair plus an equation of state suffices for a simple compressible substance.

Final Answer:
either (p, T) or (T, v) for a simple compressible substance