6 g of $O_{2}$ gas at STP is expanded against atmosphere so that the volume is doubled. Thus, the work done is:

22.4 L atm

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- 4.2 L atm

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5.6 L atm

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11.2 L atm

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Solution

The correct option is B - 4.2 L atm
Expansion work,
$W = - p_{e x t} △ V = - P_{e x t} (V_{f i n a l} - V_{i n i t i a l})$

At STP we have:
$V = 22.4 L$
$p_{e x t} = 1 a t m$
Moles of $O_{2}$ $= \frac{6}{32} = 0.1875 m o l$
$V_{i n t i a l} = 22.4 \times 0.1875 = 4.2 L$
$V_{i n i t i a l} = 4.2 L$
$V_{f i n a l} = 2 \times 4.2 = 8.4 L$
Thus, $△ V = 4.2 L$
So, $W = - (1 a t m \times 4.2 L) = - 4.2 L a t m$

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