An ideal gas expands in volume from $4 {d m}^{3}$ to $6 {d m}^{3}$ against a constant external pressure of $3 a t m$ . The work done during the expansion is $[1 L a t m = 101.32 J]$

+ 340 J

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- 304 J

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- 6 J

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- 608 J

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Solution

The correct option is D $- 608 J$
$Work done (W) = - P Δ V$
$P = pressure = 3 a t m$
$Δ V = V_{f i n a l} - V_{i n i t i a l}$
$= 6 d m^{3} - 4 d m^{3}$
$= 2 d m^{3}$
$Δ V = 2 d m^{3} = 2 L$ [Since $1 d m^{3} = 1 L$ ]
Therefore,
$Work done = - 3 a t m \times 2 L$
$= - 6 L a t m$
$W = - 6 \times 101.32 J$
$= - 607.92 J$

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