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

22.4 L atm

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5.6 L atm

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111.2 L atm

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1.135 kJ

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Solution

The correct option is D
1.135 kJ

At STP 1 gr of $H_{2} = \frac{1}{2} m o l e$ of $H_{2}$ , will occupy $11.2 l i t$

Thus if volume is doubled= $V_{2} - V_{1} = 22.4 - 11.2 = 11.2 l i t$

$W = p (V_{2} - V_{1}) = 1 \times 11.2 l i t a t m (1 l i t a t m = 101.33) = 1.135 k J$

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