One mole of a monatomic ideal gas is initially at pressure P0, volume V0 and temperature T0. It is taken to a volume 2V0 by three different…

With $\gamma=5/3$. (i) $W_i=Q_i=RT_0\ln2\approx0.693RT_0$. (ii) $T_f=T_02^{-2/3}$, $W_{ii}=C_V(T_0-T_f)=1.5RT_0(1-2^{-2/3})\approx0.555RT_0$ and $P_f=P_02^{-5/3}\approx0.315P_0$. (iii) No work, no heat, so $\Delta U=0$, $T_f=T_0$ and $P_f=P_0/2$. (A) is incorrect: $0.555RT_0<0.693RT_0$ (the isothermal curve lies above the adiabat, so it encloses more area). (B) is incorrect: free expansion is irreversible and $\Delta S=R\ln2>0$ (compute along a reversible isothermal path between the same end states). (C) is incorrect: $0.5P_0\ne0.315P_0$. (D) is correct: $Q_i\approx0.693RT_0>W_{ii}\approx0.555RT_0$. Remember that the entropy change must be evaluated along a reversible path connecting the same end states, which for an ideal gas in free expansion is an isothermal expansion, giving $\Delta S=R\ln2$. The adiabat in a $P$-$V$ diagram falls more steeply than the isotherm, which is why both its final pressure and its work are smaller than the isothermal values, and why (iii) ends at a pressure between the two.