∆U and ∆H


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ng1D
Posts: 43
Joined: Wed Sep 18, 2019 12:17 am

∆U and ∆H

Postby ng1D » Sun Mar 15, 2020 7:55 am

Why does when q = qP then ∆U = ∆H - P∆V?

Gabriella Bates 2L
Posts: 113
Joined: Thu Jul 11, 2019 12:15 am

Re: ∆U and ∆H

Postby Gabriella Bates 2L » Sun Mar 15, 2020 7:58 am

By definition, qp = delta H. Therefore, under constant pressure, qp = delta H and w = P(delta V). We can then substitute these into the equation delta U = q + w, yielding delta U = delta H + P(delta V).


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