Given: An ideal gas starts from the same initial state and expands from volume $V_1$ to $V_2$ by three different processes:

• Isothermal: work $W_1$
• Adiabatic: work $W_2$
• Isobaric: work $W_3$

We must compare $W_1, W_2, W_3$.

Work done in each process

Work done by a gas is $W = \int_{V_1}^{V_2} P\,dV$

So the comparison depends on the pressure-volume curve for each process.

Isobaric process

For isobaric expansion, pressure remains constant at the initial pressure $P_1$: $W_3 = P_1 (V_2 - V_1)$

Isothermal process

For isothermal expansion of an ideal gas, $P = \frac{nRT}{V} = \frac{P_1V_1}{V}$ Hence, $W_1 = \int_{V_1}^{V_2} \frac{P_1V_1}{V} dV = P_1V_1 \ln\left(\frac{V_2}{V_1}\right)$

Adiabatic process

For adiabatic expansion, $PV^\gamma = \text{constant}$ so pressure falls faster with volume than in the isothermal case. Therefore, for every intermediate volume between $V_1$ and $V_2$, $P_{\text{adiabatic}} < P_{\text{isothermal}} < P_{\text{isobaric}}$

Since work is the area under the $P$–$V$ curve from $V_1$ to $V_2$, $W_2 < W_1 < W_3$

Direct mathematical check

Let $x = \frac{V_2}{V_1} > 1$

Then $W_3 = P_1V_1(x-1)$ $W_1 = P_1V_1\ln x$

Since for $x>1$, $\ln x < x-1$ we get $W_1 < W_3$

Also, adiabatic curve lies below isothermal during expansion, so $W_2 < W_1$

Therefore, $W_2 < W_1 < W_3$

Matching with options

This ordering means:

• smallest: $W_2$
• middle: $W_1$
• largest: $W_3$

So the correct option is D.

Comparison with stored answer

Stored correct answer: D

This matches our derived answer.