Reversible expansion of an ideal gas under isothermal and adiabatic conditions is shown in the figure, where $AB$ is the isothermal expansion and $AC$ is the adiabatic expansion, both starting from the common state $A$. Which of the following options is NOT correct?

Solution & Explanation

\textbf{Answer:} (d) $T_C>T_A$ \textbf{Solution:} Analyse each statement for the two reversible expansions starting from $A$: $\bullet$ (a) In a reversible isothermal expansion the gas absorbs heat, so entropy increases ($\Delta S>0$); in a reversible adiabatic expansion $\Delta S=0$. Hence the entropy change in the isothermal case is greater: \textbf{correct}. $\bullet$ (b) $AB$ is isothermal, so the temperature is constant: $T_A=T_B$: \textbf{correct}. $\bullet$ (c) On the $P$-$V$ diagram the adiabat $AC$ is steeper and lies below the isotherm $AB$, so the area under $AB$ (work in isothermal) is greater than the area under $AC$ (work in adiabatic): $W_{\text{isothermal}}>W_{\text{adiabatic}}$: \textbf{correct}. $\bullet$ (d) During an adiabatic \emph{expansion}, the gas does work at the expense of its internal energy, so its temperature \emph{falls}: $T_C<T_A$. Therefore $T_C>T_A$ is \textbf{incorrect}. The statement that is NOT correct is (d).