$160 g$ of oxygen gas expand at STP to occupy double of its original volume. The work during the process is:
( $1 L . a t m = 101.3 J$ )

- 4.7 k c a l

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- 2.7 k c a l

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- 6.7 k c a l

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- 8.7 k c a l

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Solution

The correct option is B $- 2.7 k c a l$
At STP, $160 g o f O_{2}$ or $5$ mole of $O_{2}$ will occupy $22.4 \times 5 = 112 L$ volume.
Now, if the volume is doubled, it means $V_{2}$ = 224 L
So, ( $V_{2} - V_{1}$ ) $= 224 - 112 = 112 L$
At STP = P = 1 atm
Now,putting values
$w = - P \times (V_{2} - V_{1}) = - 1 \times 112 L a t m = - 112 L a t m$
since,
$1 L a t m = 101.3 J$
so,
$- 112 L a t m = - 112 \times 101.3$ J =
$w = - 11345.6$ J

Since, 1 cal = 4.18 J
$w = \frac{11345.6}{4.18}$
$= - 2711.7 c a l$ $= - 2.7 k c a l$

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