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The work done during the expansion of a gas from a volume of 4 dm³ to 6 dm³ against a constant external pressure of 3 atm is (1 L atm = 101.32 J):

1. −6 J
2. −608 J
3. −304 J
4. −304 J

Correct answer: −608 J

Solution

The work done (W) is calculated using the formula W = −PΔV. Here, P = 3 atm, ΔV = (6 − 4) dm³ = 2 dm³ = 2 L. Substituting, W = −3 × 2 = −6 L atm. Converting to joules: −6 × 101.32 = −608 J.

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