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A thermodynamic system is taken through the cycle abcda. The work done by the gas along the path bc is:
Path bc is a vertical line on P-V diagram -> Volume is strictly constant (dV = 0) -> Work W = ∫ P dV = 0 J!
In any quasi-static thermodynamic process, the mechanical work done by a gas during transition from an initial state to a final state is represented by the area under the process curve on a P-V diagram:
From the given P-V diagram: - Volume at state b: V_b = 400 cm³ - Volume at state c: V_c = 400 cm³ The process bc is represented by a vertical straight line perpendicular to the volume axis. The volume remains strictly constant throughout the entire transformation.
Substituting dV = 0 into the work integral:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Incorrect; corresponds to work along path cd (isobaric compression: -300 kPa × 300 cm³ = -90 J).
Incorrect; represents the total net work of the full cycle abcda (-60 J).
Work along an isochoric path is identically zero (dV = 0).
Corresponds to work along path ab (isobaric expansion: +100 kPa × 300 cm³ = +30 J).
Students frequently see a cyclic P-V diagram and automatically calculate the enclosed loop area (300 - 100) kPa × (400 - 100) cm³ = 60 J, picking -60 J.
"In an isochoric (constant volume) process, the system performs no mechanical boundary work because displacement of the boundary is zero (W = 0)."
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