One mole of a monoatomic ideal gas is taken along two cyclic processes $E \to F \to G \to E$ and $E \to F \to H \to E$ as shown on given $P - V$ diagram.
The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

Which of the following is/are correct ?

W_{E F H E} > W_{E F G E}

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W_{E F} = 0

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W_{E F H E} < W_{E F G E}

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W_{F H} < W_{F G}

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Solution

The correct option is D $W_{F H} < W_{F G}$
As work done is area under PV curve,

From graph, it can be deduced that,
$E F \to isochoric$
$W_{E F} = 0$

$F H \to adiabatic$
$F G \to isothermal$
Area under $F - H <$ Area under $F - G$
$W_{F H} < W_{F G}$

Area of cyclic loop $E - F - H - E <$ Area of cyclic loop $E - F - G - H - E$
$W_{E F H E} < W_{E F G E}$

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