The pressure-volume work for an ideal gas can be calculated by using the expression given as,
$w = \int_{V_{i}}^{V_{f}} p_{ex} d V$ . The work can also be calculated from the pV - plot by using the area under the curve within the specified limits. When an ideal gas is expanded (a) reversibly or (b) irreversibly from volume $V_{i}$ to $V_{f}$ , choose the correct option regarding the magnitude of work done.

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$w (reversible) > w (irreversible)$

$w (reversible) = w (irreversible)$

$w (reversible) < w (irreversible)$

$w (reversible) > w (irreversible) + p_{ex} . Δ V$

answer is C.

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Detailed Solution

$W = - P_{ext} Δ V$

For expansion to occur $P_{e x t} < P_{g a s}$ for expansion to occur maximum value of $P_{e x t} = P_{g a s}$

In a reversible process at each and every stage the $P_{e x t}$ attains its maximum value $P_{g a s}$ .

$(∴ P_{gas} ≃ P_{gas})$ so maximum work is done

In Irreversible expansion $P_{gas} > > P_{ext}$ so work done is not as much as in reversible process

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