Two moles of a gas is expanded to double its volume by two different processors. One is isobaric and the other is isothermal. If $w_{1}$ and $w_{2}$ are the works done respectively, then

w_{2} = \frac{w_{1}}{ln 2}

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w_{2} = w_{1}

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w_{2} = w_{1} ln 2

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w_{1}^{2} = w_{2} ln 2

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Solution

The correct option is C $w_{2} = w_{1} ln 2$
Volume is changed from $V_{1} = V_{o}$ to $V_{2} = 2 V_{o}$ .
Temperature and pressure of gas at $V_{o}$ are $T_{o}$ and $P_{o}$ respectively.
Thus, from ideal gas equation $P_{o} V_{o} = n R T_{o}$ .....(1)
Work done in isobaric process $w_{1} = P_{o} (V_{2} - V_{1}) = P_{o} (2 V_{o} - V_{o}) = P_{o} V_{o}$ .....(2)
Work done in isothermal process $w_{2} = n R T_{o} ln \frac{V_{2}}{V_{1}}$
$∴$ $w_{2} = n R T_{o} ln \frac{2 V_{o}}{V_{o}} = n R T_{o} ln 2$
Or $w_{2} = P_{o} V_{o} ln 2$ (from 1)
$⟹ w_{2} = w_{1} ln 2$

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