Difficulty: Easy
Correct Answer: volume
Explanation:
Introduction / Context:
In classical thermodynamics for closed systems doing only boundary work, mechanical work arises from volume change against an external pressure. Many reaction systems are designed to eliminate this contribution so that energy accounting focuses on heat effects and changes in internal energy or enthalpy. Understanding when the PΔV term vanishes is fundamental.
Given Data / Assumptions:
Concept / Approach:
Boundary work for a quasi-static step is W = ∫ P_ext dV. If the process is conducted at constant volume, dV = 0 at every instant, so W = 0 regardless of pressure or temperature changes. At constant pressure, work generally does not vanish if volume changes; at constant temperature, work may be nonzero (e.g., ideal-gas expansions/compressions).
Step-by-Step Solution:
Write boundary work: W = ∫ P_ext dV.Impose constraint: constant volume → dV = 0 for the entire process.Therefore, W = 0 exactly; no mechanical work is performed by the system.Other constraints (constant T or constant P) do not inherently force dV = 0.
Verification / Alternative check:
Bomb calorimetry is conducted at constant volume specifically so that the measured heat equals the change in internal energy (ΔU) of reaction for the system, confirming that boundary work is zero in such setups.
Why Other Options Are Wrong:
Constant temperature: systems can expand or contract, giving nonzero work. Constant pressure: volume usually changes during reaction, so work is typically nonzero. 'None of these' contradicts the constant-volume case.
Common Pitfalls:
Final Answer:
volume
Discussion & Comments