![]() ![]() We can compare the irreversible process to a reversible adiabatic expansion where the final pressure ends up being the same as the final pressure of the irreversible expansion. The work done is simply the external pressure times the change in volume. Using the ideal gas laws we can determine the final volume and temperature (point 2a) when equilibrium is reached, assuming the initial volumes and temperatures are as indicated for point 1. This results in a non quasi-static (irreversible) adiabatic expansion against a constant external pressure. If the cycle is reversible (Carnot Cycle) On the other hand, if a heat engine operates between two thermal reservoirs in a reversible cycle (Carnot cycle) the heat engine will not create entropy andįor any heat engine operating between two thermal reservoirs the cycle efficiency is the net work done divided by the gross heat added, or If a heat engine creates entropy, the cycle is irreversible, and Consequently, any entropy created by the system is transferred to the surroundings. The entropy change of the system is always zero for any complete cycle (reversible or not) because entropy is a state function of the system and does not depend on the path. ![]() The equation is the total change in entropy (system + surroundings) and does account for the possibility that the engine creates entropy. I'm assuming that something is wrong with my reasoning, but what? I don't buy that this is the total change in entropy because he doesn't account for any entropy created due to work being done by the engine. ![]()
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