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Topic: Enthalpy transition  (Read 3889 times)

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Offline Disturbia

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Enthalpy transition
« on: November 14, 2007, 12:32:23 AM »
Hi

I'm having a hard time trying to understand what does a high value of thee difference between delta H and delta U means ? (U = internal energy)
We have that
H = U + pV

which can be written as H - U = pV
So let's say i want to calculate delta(H) - delta(U) for the transition graphite-diamond under uncommon condition like 500 kbar of pressure. I find that delta(H) - delta(U) gives around -100 kJ/mol wchi is pretty high but why ?

Offline Yggdrasil

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Re: Enthalpy transition
« Reply #1 on: November 16, 2007, 07:15:37 PM »
You are correct that H-U =pV.  This means that ΔH - ΔU = Δ(pV).  For a phase change at constant pressure this means that:

ΔH - ΔU = pΔV

So, the volume of your solid changes as you transition from diamond to graphite.  Under normal conditions, ΔV is small, so the difference in ΔH and ΔU is also small.  However, for large pressure, this difference is no longer small (as you can see from the equation above).

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