4.7 Homework Help regarding signs of work

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4.7 Homework Help regarding signs of work

Postby DHavo_1E » Tue Feb 11, 2020 2:46 pm


Could anyone explain how to determine the signs in the enthalpy equation: deltaH= deltaU + PdeltaV? In this question, work is found to be positive and when rearranging for deltaU, I would expect a negative internal energy. However, in the solution book, they wrote deltaU= deltaH + PdeltaV instead of deltaU=deltaH-PdeltaV. Do we account for the positive work after rearranging the equation then? Thank you!

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Re: 4.7 Homework Help regarding signs of work

Postby KaleenaJezycki_1I » Tue Feb 11, 2020 3:18 pm

Hi! This comes from the original equation deltaH= q+w

Step1: isolate q
So q=DeltaU-w

Step2: To get Delta H we are assuming that the reaction is occurring under constant pressure (qp)
Therefore DeltaH= DeltaU-w

Step3: BUT w=-PexDeltaV so the we have - -Pex which becomes positive to get:

DeltaH= DeltaU+ PexDeltaV

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