Reversible vs Irreversible Expansion: Why Maximum Work is Reversible

Work done by a gas = external pressure x change in volume (area under the p-V curve). In a reversible expansion, the external pressure is kept only INFINITELY smaller than the gas pressure at every single step. So the gas is always pushing against the highest possible opposing pressure. In an irreversible expansion the external pressure is a fixed, much smaller value from the start. Since work = p_ext x deltaV, a bigger opposing pressure over the whole path means more work. That is why reversible expansion gives the MAXIMUM possible work.

For a reversible isothermal expansion of an ideal gas, w = -2.303 nRT log(V2/V1), which you can also write with pressures as w = -2.303 nRT log(P1/P2). The minus sign means the gas does work ON the surroundings (work by the gas is negative in the IUPAC sign convention). For irreversible expansion against a constant pressure you instead use w = -p_ext x (V2 - V1). NEET gives you both types, so know which formula matches which process.

In a reversible expansion the external pressure keeps changing at every step (it always tracks the gas pressure p = nRT/V). Because p changes continuously, you must integrate, and that integration gives the natural log (ln) term. In an irreversible expansion the external pressure is CONSTANT, so the pressure does not change during the push and you simply multiply p_ext by the total volume change. Constant pressure = simple multiplication; changing pressure = log.

Work is a PATH FUNCTION. This concept proves it: expanding a gas from the same start to the same end gives DIFFERENT work for reversible vs irreversible paths. Reversible gives the most work, irreversible gives less. Since the answer depends on HOW you go (the path), not just the start and end points, work cannot be a state function. Internal energy (U) and enthalpy (H) are state functions; heat (q) and work (w) are path functions.

Work equals the AREA under the p-V curve between the start and end volumes. The reversible path is a smooth curve that stays HIGHER on the graph (larger area), while an irreversible path against a low constant pressure is a flat, low line (smaller area). So the curve that encloses the largest area from start volume to end volume represents the maximum work. This is exactly what NEET 2022 asked.

In an isothermal expansion, temperature is constant, so for an ideal gas deltaU = 0. By the first law, deltaU = q + w, so q = -w. This means all the work done by the gas is exactly replaced by heat absorbed from the surroundings. In reversible isothermal expansion the gas does the MOST work, so it also absorbs the MOST heat. The internal energy itself does not change because T is fixed.

Because the gas pushes against the highest possible external pressure (just below its own) at every step, and work = pressure x volume change, so a bigger opposing pressure over the whole path gives the most work.

Use w = -2.303 nRT log(V2/V1) = -2.303 nRT log(P1/P2). The log appears because the external pressure changes continuously during a reversible process.

Use w = -p_ext x (V2 - V1) when the external pressure is constant. This always gives a smaller magnitude of work than the reversible case for the same start and end states.

For expansion, yes: reversible expansion does the maximum work. For compression it is reversed: reversible compression needs the minimum work done on the gas. Reversible is always the 'best case' path.

Because different paths (reversible vs irreversible) between the same two states give different amounts of work. A state function would give the same value regardless of path.