Work and heat in thermodynamic processes – select the correct statement Which of the following statements is correct regarding reversible/irreversible work and the nature of thermodynamic quantities?

Introduction / Context:
Comparing different processes for ideal gases (isothermal vs. adiabatic) and understanding state vs. path functions are foundational for cycle analysis and equipment design (compressors, expanders, engines).

Given Data / Assumptions:

• Ideal gas behavior for qualitative comparisons.
• Same initial state and same final volume are considered for work comparison.
• Reversible paths maximize work magnitude for a given change.

Concept / Approach:
For a reversible expansion from the same initial state to the same final volume, the isothermal path has higher pressures at any intermediate volume than the adiabatic path; thus the area under the P–V curve (work by the gas) is larger for the isothermal case. Therefore, reversible adiabatic work (by the gas) is less than reversible isothermal work to the same V2. Also, reversible work magnitude exceeds irreversible work for the same endpoints. Finally, heat and work are path functions, whereas enthalpy and entropy are state functions.

Step-by-Step Solution:

Verification / Alternative check:
Using ideal-gas formulas, W_rev,isothermal = nRT ln(V2/V1), whereas W_rev,adiabatic = (P1V1 − P2V2)/(γ − 1). For the same V2/V1, the isothermal work is larger.

Why Other Options Are Wrong:

• Irreversible work more than reversible: False; reversible is maximum (for work output).
• All four are state functions: False; heat and work are not state functions.
• Closed system cannot exchange energy: False; it exchanges energy (heat/work) but not matter.

Common Pitfalls:
Comparing to the same final pressure instead of the same final volume; mixing up sign conventions for work.

Final Answer:
Under reversible conditions, the adiabatic work is less than isothermal work.