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With same ini...

Question

With 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 isothermal, $W_{2}$ if isobaric and $W_{3}$ if adiabatic, then

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

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

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

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

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Solution

The correct option is A $W_{2} > W_{1} > W_{3}$
Isobaric process is that which occurs at constant pressure. So, in figure, $W_{2}$ denotes work done during isobaric process. While for isothermal or adiabatic curve slope is give by $\frac{d p}{d V}$ , also slope $(\frac{d p}{d V})$ of an adiabatic curve is $γ -$ times the slope $(\frac{d p}{d V})$ of an isothermal curve. As $γ > 1$ , therefore adiabatic curve at any point is steeper than the isothermal curve at that point. Isothermal curve lies above the adiabatic curve in case of expansion, work done also depends on area of $p - V$ curve. Here, for isobaric process, work done is largest, next for isothermal and then for adiabatic process,
$∴ W_{2} > W_{1} > W_{3}$

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