Ideal-gas entropy dependence: For an ideal gas under standard thermodynamics, the specific entropy s is a function of which state variables? Select the most complete statement.

Introduction / Context:
Understanding what entropy depends on for an ideal gas is foundational for compressors, turbines, and nozzles. In thermodynamic modeling, we often need to predict how entropy changes when pressure or temperature changes along a process path. This question checks the correct functional dependence of ideal-gas entropy.

Given Data / Assumptions:

• Ideal-gas behavior is assumed.
• Specific heats may be taken as constant over a small range, or treated as temperature-dependent without changing the qualitative dependence.
• We are concerned with state-variable dependence, not path details.

Concept / Approach:

For an ideal gas, entropy is a state function that can be expressed relative to a reference state using temperature and pressure. A common expression is s2 − s1 = ∫(Cp(T) dT / T) − R ln(p2/p1). This clearly shows separate contributions: a temperature term through Cp/T and a pressure term through the logarithm. Hence, entropy depends on both temperature and pressure for an ideal gas state description.

Step-by-Step Solution:

Verification / Alternative check:

Alternate expression using density shows the same dependence: s2 − s1 = ∫(Cv(T) dT / T) + R ln(v2/v1); via ideal-gas law, v and p are linked at fixed T, reconfirming dependence on T and p.

Why Other Options Are Wrong:

Pressure only or temperature only omits one of the two controlling variables. 'Neither' contradicts state equations. Molecular weight affects R and Cp values but not the qualitative dependence on T and p.

Common Pitfalls:

Confusing entropy with enthalpy (which for ideal gases depends on T alone) and forgetting the logarithmic pressure term in entropy.

Final Answer:

both (a) & (b)