Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures T1 and T2…

🧠 Reversible isothermal work: $w = -nRT\ln(V/V_i)$ — the T controls the steepness of the curve For expansion ($V > V_i$): $w < 0$ (work done by gas). As $V$ increases, $|w|$ increases logarithmically. At higher $T$, the magnitude of work is larger for the same volume ratio.

🗺️ What the correct graph shows

• $w$ starts at 0 when $V = V_i$ (no expansion yet).
• $w$ becomes increasingly negative as $V$ increases — a logarithmic curve going downward.
• The curve for $T_2$ (higher temperature) lies further below the curve for $T_1$, since $|w| \propto T$.
• Both curves are concave (logarithmic shape), not straight lines.

Option (d) correctly shows two logarithmic curves with $T_2$ steeper than $T_1$, both going negative.

⚠️ Trap: A linear $w$–$V$ plot would imply constant-pressure work $w = -P\Delta V$ — not isothermal reversible work. The correct shape is logarithmic.

$\boxed{\text{Answer: (d)}}$