How Do You Calculate ΔH and Determine Volume Changes in Thermodynamics?

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deviljay
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Hi All!
I ran across a homework problem ans am confused by the question.
here it is:
For a certain process at constant pressure, ∆U = − 225 kJ and +52 kJ of expansion work is
done. What is ∆H for this process and does the volume increase or decrease?

I know that ∆U=q+w and therefore q=-277 kJ, also I know that change in enthalpy = ∆U-P∆V.
However I am all confused on the relationship between these variables.

I appreciate any feedback. Thank you for your time.
 
on Phys.org
delta U = delta H - work ( done on the system or by the system )
delta U , internal energy can be gained or lost
delta H , is negative if heat is evolved and positive if heat is absorbed.
work = PdeltaV and is negative if done by the system , as in expansion
and is positive if done on the system.
So the double negative gave you the + 52 kJ
 
deviljay said:
Hi All!
I ran across a homework problem ans am confused by the question.
here it is:
For a certain process at constant pressure, ∆U = − 225 kJ and +52 kJ of expansion work is
done. What is ∆H for this process and does the volume increase or decrease?

I know that ∆U=q+w and therefore q=-277 kJ, also I know that change in enthalpy = ∆U-P∆V.
However I am all confused on the relationship between these variables.

I appreciate any feedback. Thank you for your time.

correction to my previous reply: there is no double negative giving a + 52 kJ
delta U = delta H - work,PV so in the above problem it should be
-225 kJ = -173 kJ - 52 kJ
 
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