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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 ∆U of the gas in joules will be

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a

-500 J

b

-505 J

c

+505 J

d

1136.25 J

answer is B.

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Detailed Solution

w=-Pext ΔV=-2.5(4.50-2.50) =-5 L atm=-5×101.325 J=-506.625 J ΔU=q+w As, the container is insulated, thus q=0 Hence, ΔU=w=-506.625 J

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