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The work done during the expansion of a gas from a volume of 4dm3 to 6dm3 against a constant external pressure of 3atm

is (1Latm=101.32J)

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a
+304J
b
−304J
c
−6J
d
−cos⁡J

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detailed solution

Correct option is D

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

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