Final answer:
The adiabatic expansion of an ideal gas in a vacuum is irreversible because there are no equilibrium states during the expansion, no work is performed, and no heat is exchanged.
Step-by-step explanation:
The expansion of an ideal gas adiabatically in a vacuum is an example of an irreversible process. An irreversible process is one where the system does not pass through equilibrium states, and once the process occurs, it cannot return to its original state without leaving changes in the surroundings. For the expansion into a vacuum, the external pressure is zero, meaning no work is done by the gas (Pext = 0, so PΔV = 0), and since it is adiabatic, no heat is transferred (Q = 0). The first law of thermodynamics confirms that the internal energy of the system remains unchanged (dEint = Q - W = 0).
However, because the gas expands spontaneously and the external pressure is zero, there are no opposing forces to allow for a quasi-static (equilibrium) process. This spontaneity and lack of equilibrium states entail that the process cannot be reversed by an infinitesimal change, making it irreversible. Furthermore, since the process is adiabatic but not quasi-static, the resulting P-V curve for the expansion would be steeper than that of the corresponding isotherm, reflecting the lack of heat exchange with the surrounding environment.