Reversible expansion of an ideal gas under isothermal and adiabatic conditions is shown in the figure ($\mathrm{A}\to \mathrm{B}$ isothermal expansion, $\mathrm{A}\to \mathrm{C}$ adiabatic expansion on a $p$–$V$ plot). Which of the following options is NOT correct?

Solution & Explanation

\textbf{Answer:} (D) $T_C>T_A$ — this is the INCORRECT statement \textbf{Solution:} In a reversible adiabatic expansion ($q=0$), $\Delta U=w<0$, so the gas cools and $\Delta T<0$; hence $T_C<T_A$, making the statement '$T_C>T_A$' incorrect. The other statements are correct: along the isotherm $T_A=T_B$; the isothermal expansion has $\Delta S>0$ while the reversible adiabatic is isentropic ($\Delta S=0$), so $\Delta S_{isothermal}>\Delta S_{adiabatic}$; and the isothermal path lies above the adiabat, so $W_{isothermal}>W_{adiabatic}$. Mapped: source numeric answer '4' → option D.