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When a sample of gas expands from 4.0 L to 12.0 L against a constant pressure of0.30 atm, the work involved is

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243.19 J

- 243.19 J

234.19 J

- 234.19 J

answer is B.

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Detailed Solution

$w = - p_{ext} Δ V = - (0.30 atm) (12.0 L - 4.0 L) = - 2.40 atmL (\frac{8.314 J}{0.082 atmL}) = - 243.3 J$

Thus the option B is correct.

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