One mole of an ideal monatomic gas at 300 K and 10 L expands to 20 L by three different routes: (a) reversible isothermal expansion, (b) reversible…

Option (A) is false: a reversible adiabatic process has $q_{rev}=0$, so $\Delta S_{gas}=0$, even though the volume doubles (the temperature falls to compensate). Option (B) is true: in the isothermal reversible case the surroundings lose exactly what the gas gains, so $\Delta S_{univ}=0$, and for an ideal gas at constant $T$, $\Delta G=\Delta H-T\Delta S=-RT\ln 2$. Option (C) is false: the adiabatic route ends at $T_2=300(1/2)^{2/3}\approx189\ \mathrm{K}$ and does $|w|=\tfrac32R(300-189)\approx1.38\ \mathrm{kJ}$, which is less than the isothermal $RT\ln2\approx1.73\ \mathrm{kJ}$. Option (D) is true: free expansion of an ideal gas has $q=w=0$ and the same end state as route (a), so $\Delta S_{gas}=R\ln2=5.76\ \mathrm{J\,K^{-1}}$ and nothing happens to the insulated surroundings, so $\Delta S_{univ}=5.76\ \mathrm{J\,K^{-1}}$ and the process is irreversible.