Starting with the same initial conditions, an ideal gas expands from volume $$V_1$$ to $$V_2$$ in three different ways. The work done by the gas is $$W_1$$ if the process is purely isothermal, $$W_2$$, if the process is purely adiabatic and $$W_3$$ if the process is purely isobaric. Then, choose the correct option

An ideal gas expands from $$V_1$$ to $$V_2$$ via three processes starting from the same initial conditions. We compare the work done: $$W_1$$ (isothermal), $$W_2$$ (adiabatic), $$W_3$$ (isobaric).

All three processes start from the same point $$(V_1, P_1)$$ on the P-V diagram and end at $$V_2$$.

For expansion from $$V_1$$ to $$V_2$$:

- Isobaric process: Pressure stays at $$P_1$$ throughout. The P-V curve is a horizontal line at $$P = P_1$$.

- Isothermal process: $$PV = P_1V_1 = \text{const}$$, so $$P = P_1V_1/V$$. The pressure drops as $$1/V$$ — the curve lies below the isobaric line.

- Adiabatic process: $$PV^\gamma = P_1V_1^\gamma = \text{const}$$, with $$\gamma > 1$$. The pressure drops faster than in the isothermal case, so this curve lies below the isothermal curve.

Work done equals the area under the P-V curve from $$V_1$$ to $$V_2$$. Since the isobaric curve is highest, isothermal is in the middle, and adiabatic is lowest:

$$W_{\text{adiabatic}} < W_{\text{isothermal}} < W_{\text{isobaric}}$$

$$W_2 < W_1 < W_3$$

This ordering can be understood physically: in an isobaric process, the gas maintains constant pressure (maximum area under curve). In an isothermal process, the gas absorbs heat to maintain temperature, keeping the pressure relatively high. In an adiabatic process, no heat is absorbed, so the temperature and pressure drop the most, giving the smallest area under the curve.

Hence, the correct answer is Option A: $$W_2 < W_1 < W_3$$.