Answer
(a) stay the same
(b) I
Work Step by Step
(a) We know that for an adiabatic process, the heat added or extracted from the gas is zero. As $\Delta S=\frac{\Delta Q}{\Delta T}$
$\implies \Delta S=\frac{0}{\Delta T}=0$. Hence, the entropy of the gas stays the same.
(b) The best explanation is option (I) -- that is, the process is reversible, and no heat is added to the gas. Therefore, the entropy of the gas remains the same.
