Chemistry - Thermodynamics - I Question with Solution | TestHub

Explain the question: Compare property changes for reversible (rev.) versus irreversible (irrev.) adiabatic expansion.

Concept: First Law for adiabatic processes ($\Delta U=w$) and work magnitude relation ($|w_{rev}|>|w_{irrev}|$).

Solution:

In an adiabatic expansion starting from the same initial state to the same final volume, the gas performs work on its surroundings at the expense of its internal energy ($w=\Delta U$).

On a P–V diagram, the reversible expansion curve is higher than the irreversible one, ensuring the area under the curve is larger: $|w_{rev}|>|w_{irrev}|$.

Since more energy is extracted from the gas in the reversible process, its internal energy drop and corresponding temperature drop are greater ($|\Delta U_{rev}|>|\Delta U_{irrev}|$ and $|\Delta T_{rev}|>|\Delta T_{irrev}|$). Because enthalpy change is also proportional to temperature ($\Delta H\propto \Delta T$), it follows that $|\Delta H_{rev}|>|\Delta H_{irrev}|$.

This logic dictates that the final temperature in the reversible process is actually lower than in the irreversible process.

Thus, statements A, C, and D are mathematically and conceptually false.

Final Answer: Options are (A), (C) and (D).