Final answer:
Less work is done during an irreversible process compared to a reversible process in thermodynamics because the system does not remain in equilibrium, leading to energy losses such as friction, which cannot be converted into work. The maximal work obtainable is through reversible processes due to infinitesimal adjustments that maintain equilibrium between the system and the surroundings.
Step-by-step explanation:
When comparing the amount of work done during reversible and irreversible processes in thermodynamics, work done through a reversible path is always greater than that done through an irreversible path for a system undergoing expansion. This is because a reversible process is carried out infinitely slowly, allowing the system to stay in equilibrium with its surroundings at all times. As a result, the external pressure needs to adjust infinitesimally to match the internal pressure, leading to maximum work done as the process expands against the least opposing force. On the other hand, an irreversible process occurs spontaneously and rapidly, not allowing the system and the surroundings to reach equilibrium. The system expands against a greater opposing force due to friction or other dissipative forces, leading to less work output.
In an irreversible process, since equilibrium is not maintained, additional energy is lost to factors like friction or turbulence, and cannot be converted into work. The work done can be visually understood by comparing the area under the pressure-volume (PV) curve in a graph for each process. The area under the curve, representing work, is larger for the reversible process in comparison to the irreversible one, which is typically represented by a smaller area due to a more vertical (steeper) curve.