Discussion
In an adiabatic expansion of an ideal gas
Answer: D
The key's answer is Entropy remains constant.
In an adiabatic process no heat enters or leaves. For a reversible (quasi-static) adiabatic process, dS = dQrev/T = 0, so the entropy stays the same; this is why a reversible adiabatic process is called isentropic. The answer assumes that standard textbook case; a sudden free expansion is also adiabatic but increases entropy.
The other options are wrong. The gas does work while expanding with no heat supplied, so its internal energy and therefore its temperature fall, and the pressure falls too (PVγ is constant).
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