Work from gas expansion over a fixed pressure range: For a given initial state and the same final pressure, which reversible path yields the maximum work from the expanding gas?

Introduction / Context:Mechanical work from a gas depends on both endpoints and the path on a P–V diagram. For ideal gases, comparing common reversible paths (isothermal, adiabatic/isentropic) reveals which produces the greatest area under the curve (integral of P dV) for a specified pressure drop.

Given Data / Assumptions:

• Ideal-gas behavior for clarity of comparison.
• Reversible (quasi-static) paths to avoid additional entropy production.
• Initial state fixed; final pressure is specified and identical across paths.

Concept / Approach:Reversible isothermal expansion follows P = nRT/V with T constant, producing a relatively high pressure at any volume compared with a reversible adiabatic whose pressure falls faster with increasing volume (P ∝ V^−γ). The larger average pressure during isothermal expansion yields greater work over the same pressure range.

Step-by-Step Solution:Work (reversible) = ∫ P dV.Isothermal: P_iso = nRT/V → W_iso = nRT ln(V2/V1).Adiabatic/isentropic: P_ad drops more steeply; area under curve is smaller to the same final pressure.Therefore, reversible isothermal expansion gives maximum work for ideal gases across the same pressure drop.

Verification / Alternative check:Graphical P–V comparison shows the isothermal path lies above the adiabatic between the same endpoints in pressure, confirming a larger integral of P dV.

Why Other Options Are Wrong:Adiabatic and isentropic (same for ideal gases) yield less work than isothermal for the specified condition. 'None of these' contradicts known results.

Common Pitfalls:

• Comparing at the same final volume rather than same final pressure; the result depends on constraint.
• Ignoring real-gas deviations; however, the teaching result holds for ideal-gas comparisons.

Final Answer:isothermal