A gas is allowed to expand in a well insulated container against constant external pressure of 2.5 atm from an initial volume of 2.50L to a final volume of 4.50L. The change in internal energy △U of the gas in joules will be:
A
+505J
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B
+1136.25J
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C
−500J
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D
−505J
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Solution
The correct option is D−505J The change in the internal energy is given by the following expression. ΔU=q+W Since container is well insulated, q=0. Hence, ΔU=W=−PΔV ΔU=W=−2.5atm×(4.50L−2.50L) ΔU=W=−5.0Latm Convert the unit to J ΔU=W=−5.0Latm×101.33J1Latm=−505J