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Question
A gas allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be
A gas allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be
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Solution
The correct option is B −505 J
W=−PextΔV=−2.5(4.50−2.50)
W=−5 L atm=−5×101.325 L. bar=−506.625 J
ΔU=q+W
As, the container is insulted, thus q=0
Hence, ΔU=W=−506.625J≈−505 J
W=−PextΔV=−2.5(4.50−2.50)
W=−5 L atm=−5×101.325 L. bar=−506.625 J
ΔU=q+W
As, the container is insulted, thus q=0
Hence, ΔU=W=−506.625J≈−505 J
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