$1 g H_{2}$ gas STP is expanded so that the volume is doubled. Hence, work done is

22.4 L - a t m

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5.6 L - a t m

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11.2 L - a t m

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44.8 L - a t m

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Solution

The correct options are
A $22.4 L - a t m$
C $11.2 L - a t m$
As we know,
2gm hydrogen have volume of 22.4 lit at STP so
$V_{1} (v o l u m e o f 1 g H_{2}) = 11.2 L$ at STP
now,
$V_{2} (v o l u m e o f 1 g H_{2}) = 22.4 L$
so work done is
$∴ W = P Δ V = 11.2 L - a t m$

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