Why does the enthalpy equation include work (PV term) twice?

JeweliaHeart

Hello. I am a thermodynamics novice trying to gain a better understanding of state functions, particularly enthalpy.

I understand that enthalpy is defined as

"A measure of the total energy of a thermodynamic system, including internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure."

The equation:

ΔH=ΔU(internal energy) + ΔPV

confuses me b/c


ΔU= q(heat added) - w(work done by system on environment)

so

ΔH really means:

ΔH=q - w + ΔPV


There are two terms of work (w and ΔPV) and b/c of the opposite sign, they cancel out, leaving only q. This means ΔH= q which is at odds with the accepted definition of enthalpy. Where did I mess up?

DrDu

That not at odds with the usual definition. Mostly enthalpy is determined by measuring heat.
However ##\Delta(PV)=P\Delta V=w## only if p is constant and volume work is the only kind of work the system is performing.

JeweliaHeart

Okay,
so you are saying that enthalpy is only equivalent to heat if pressure is held constant?

Meaning,

ΔH=q - PΔV + PΔV= q (only when pressure is constant and only PV work is being exerted)

Otherwise, when pressure is not constant the equation should like this, perhaps?:

ΔH=q - w + PΔV

And the work defined by the 'w' above includes all forms of work, whether PV or mechanical, etc?

If so, that makes a little more sense. It's just that all the example problems I've encountered with ΔU only use PV work and no other form.

DrDu

In general, ##d(PV)=PdV+VdP##. The second term will not vanish when P is not constant while the first term gives the volume work done in an infinitesimal step.
Hence ## \Delta H=q-w+\int PdV +\int V dP## in general.
If there is no work done other than volume work, this reduces to
## \Delta H=q+\int V dP##.
An example of non-volume work is e.g. the work done when stirring a viscous fluid.