## [tex]\Delta U = \Delta H[/tex]

$\Delta U=q+w$

David Sarkissian 1K
Posts: 29
Joined: Fri Sep 28, 2018 12:21 am
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### [tex]\Delta U = \Delta H[/tex]

I'm just a little confused when you're suppose to disregard that $\Delta U = \Delta H$ and instead use the equation $\Delta H = \Delta U + \Delta n_{gas}RT$ instead? What are the conditions that need to be met?

Ashley Zhu 1A
Posts: 69
Joined: Fri Sep 28, 2018 12:16 am

### Re: [tex]\Delta U = \Delta H[/tex]

The two equations you have there are essentially the same equations derived from delta(U) = delta(H) + w where w is work. Since w = -Pdelta(V) where P is pressure and delta V is the change in volume, we can rewrite the equation as delta(U) = delta(H) - Pdelta(V). By the ideal gas law, PV = nRT, -PdeltaV = -delta(n)RT so we can rewrite the equation as delta(U) = delta(H) - delta(n)RT, which if you rearrange it, is delta(H) = delta(U) + delta(n)RT (the second equation you have there). Thus from this, we can see that the main difference between the two equations you listed is that one includes the change in internal energy of a system as a result of doing work. If no work is done (no change in volume of the system), then w = 0 and delta(U) = delta(H). It's not so much as disregarding delta(U) = delta(H) as knowing when you have done work.

lukezhang2C
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

### Re: [tex]\Delta U = \Delta H[/tex]

So delta U = delta H is only considered a very specific case of the original equation, which is delta U = q + w = q - P delta V. You can only use delta U = delta H when q = delta H and you are under constant pressure, otherwise you cannot use the other equation.