Adiabatic expansion (ideal gas): Which expression correctly gives the work done during a reversible adiabatic expansion from state (p1, v1, T1) to (p2, v2, T2)? (R = gas constant, γ = cp/cv)

Introduction / Context:
Adiabatic processes are central to compressors, turbines, and nozzles. For an ideal gas undergoing a reversible adiabatic (isentropic) expansion, several equivalent formulas exist for work. Selecting the correct sign and arrangement is essential for solving cycle problems accurately.

Given Data / Assumptions:

• Ideal gas with constant specific heats (γ = cp/cv).
• Reversible adiabatic (no heat transfer, Q = 0; isentropic for ideal gas).
• Initial state (p1, v1, T1), final state (p2, v2, T2).

Concept / Approach:
For a reversible adiabatic process: pv^γ = constant and Q = 0. The first law for a closed system gives W = ∫p dv. Using the adiabatic relation, the integral reduces to an expression in end states: W = (p1v1 − p2v2)/(γ − 1). Equivalent temperature-based forms are W = mcv*(T1 − T2) and W = mR(T1 − T2)/(γ − 1). Correct signs reflect that T2 < T1 for an expansion, making W positive (work done by the system).

Step-by-Step Solution:

Verification / Alternative check:
From the first law with Q = 0: ΔU = −W and ΔU = mcv*(T2 − T1) ⇒ W = mcv(T1 − T2), which is consistent with the pressure–volume form above.

Why Other Options Are Wrong:

• (p2v2 − p1v1)/(γ − 1): Sign error.
• mR(T2 − T1)/(γ − 1): Sign error; should be T1 − T2.
• mcv(T2 − T1): Again sign error for expansion.

Common Pitfalls:
Forgetting sign conventions; mixing up γ − 1 with 1 − γ; assuming T2 > T1 in an expansion (true only with heat input, not adiabatic expansion).

Final Answer:
W = (p1v1 - p2v2) / (γ - 1)