The work during expansion of a gas from a volume of 4L to 6L against a constant external pressure of 3 atm is (1 atm –L = 101.32J)

- 6J

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- 608 J

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+ 304 J

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- 304 J

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Solution

The correct option is B - 608 J
$W = - P_{e x t} Δ V$
$= - 3 \times 2$
$= - 6 a t m - L$
$= - 6 \times 101.32 J o u l e s$
$= 608 J o u l e s$

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