$W_{1}, W_{2}, W_{3}$ are work done in adiabatic, isobaric and iso-thermal process respectively during an expansion process. Arrange them in increasing order.

W_{1} > W_{2} > W_{3}

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W_{2} > W_{1} > W_{3}

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W_{3} > W_{1} > W_{2}

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W_{2} > W_{3} > W_{1}

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Solution

The correct option is B $W_{2} > W_{3} > W_{1}$
Let the state be $(P_{i}, V_{i}, T_{i})$ and final volume be $V_{f}$
For adiabatic process, $P V^{γ} = c o n s t a n t \Rightarrow$ work done = $W_{1} = \frac{P_{i} V_{i} - P_{f} V_{f}}{γ - 1} = \frac{P_{i} V_{i} (1 - (\frac{V_{i}}{V_{f}})^{γ - 1})}{γ - 1}$

For isobar, $d P = 0 \Rightarrow W_{2} = P_{i} (V_{f} - V_{i})$

For isotherm, $d T = 0 \Rightarrow W_{3} = P_{i} V_{i} ln (V_{f} / V_{i})$
Expanding the log term yields $W_{3} < W_{2}$
Also, $W_{3} = (\frac{P_{i} V_{i}}{γ - 1}) ln (\frac{V_{f}}{V_{i}})^{γ - 1} < (\frac{P_{i} V_{i}}{γ - 1}) (1 - (\frac{V_{i}}{V_{f}})^{γ - 1}) \Rightarrow W_{3} > W_{1}$

Hence $W_{2} > W_{3} > W_{1}$

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