An ideal gas expands from an initial volume V to a final volume 2V...

The magnitude of work done by a gas during expansion is equal to the area under the P-V curve for that process.

Starting from the same initial state $(P, V)$ and expanding to the same final volume $2V$:

In an isobaric process, the pressure remains constant, so its P-V curve is a horizontal line.

In an isothermal process, the pressure decreases as volume increases. Its curve falls below the isobaric line.

In an adiabatic process, the pressure decreases more rapidly than in an isothermal process because its curve is steeper (slope is $\gamma$ times the isothermal slope). Its curve falls below the isothermal curve.

Therefore, the area under the curves follows the order: Area(isobaric) > Area(isothermal) > Area(adiabatic).

This means the work done follows the sequence: $W_{1} > W_{2} > W_{3}$.

Answer: $W_{1} > W_{2} > W_{3}$