1. Concept:The mechanical work done by a gas expanding from \(V_1\) to \(V_2\) is exactly equal to the area under its curve on a Pressure-Volume (P-V) diagram.
2. Formula:$$W = \int_{V_1}^{V_2} P \, dV$$
3. Solution:- In an Isobaric expansion, pressure remains constantly high at its maximum initial value, generating a rectangular area (\(P \times \Delta V\)).
- In Isothermal expansion, pressure slowly drops as volume increases, creating a curved area smaller than the rectangle.
- In Adiabatic expansion, pressure drops even more steeply, yielding the smallest area.
- Therefore, for given identical volume limits, the Isobaric process yields the maximum mathematical area, and thus maximum work.
4. Why other options are incorrect:Isochoric process produces exactly zero work. Isothermal and adiabatic processes produce progressively less work due to dropping pressures.
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