Hey Dejan,

Good question, I'll just briefly lay out the derivation here and try to explain it in words:

We'll need a couple equations to prove this:

1. w = -PdV that is work is a change in volume at constant pressure.

2. dE = w + q

3. dH = dE + d(PV)

For the above equation 2, d(PV) refers to either a change in volume or a change in pressure, which is not equal to work, because work is defined as a change in volume at constant pressure. So we artifically restrict the system to constant pressure.

At constant pressure, d(PV) becomes PdV, and we rewrite 3. as below:

4. dH = dE + PdV at constant pressure

With this information, we can substitute 1 into 4, giving:

5. dH = dE - w at constant pressure

We can also substitute 2. into 5, yielding:

5. dH = w + q - w at constant pressure

This simplifies to:

6. dH = q at constant pressure

So yes, there is PV work being done, but that does not affect your definition of dH as it is the change in dE without work considerations at constant pressure. The work done is not considered because of the definition of dH intentionally excludes it.

I hope this helps!

Katt