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

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+304J

−304J

−6J

−cos⁡J

answer is D.

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Detailed Solution

W=−pΔV;W=−3×(6−4)W=−6×101.32(∴1Latm=101.32J)W=−608J

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