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Question
A gas is 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 – 505J
W=−Pext ΔV=−2.5(4.50−2.50)
=−5 L atm=−5×101.325J=−506.625 J
ΔU=q+w
As, the container is insulated, thus q = 0
Hence, ΔU=w=−506.625J
W=−Pext ΔV=−2.5(4.50−2.50)
=−5 L atm=−5×101.325J=−506.625 J
ΔU=q+w
As, the container is insulated, thus q = 0
Hence, ΔU=w=−506.625J
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