[G O D I V A]

I just started work on this enthalpy and I'm a little confused.

It says:

dH = dq + dw + pdV + Vdp

If the system is in mechanical equilibrium with its surroundings at a pressure p, and does only expansion work we can write dw = -pdV and obtain *

dH = dq + Vdp

Now we impose the condition that the heating occurs at constant pressure by writing dp = 0 **. Then

dH = dq.

* Is this saying that p = p_ex ≠ constant?

** and now you are saying that p = p_ex = constant?

[renge ishyo]

In both conditions the pressure is treated as constant. If you examine the partial derivatives you will see this more clearly.

dH = dq + dw + d(PV)

dH = dq + dw + P(dV)_P + V(dP)_V

You can see that the condition for the P(dV) term in the partial derivative to begin with is for the pressure to be kept constant momentarily for each infinitesmal change in V. The term that deals with changes in pressure dP is then cancelled by the next condition. It is just mathematical formalism. The main idea is that under conditions of constant pressure (such as doing a reaction that is open to the atmosphere at atmospheric pressure) where only pressure volume work is allowed, dH is equal to dq. My favorite derivation of dH arises from the first law and may prove useful to you:

1st law:

dU = dq + dw

Assume pV work only at constant P such that dw = -PdV; then:

dU = dq - PdV

dU + PdV = dq

Define dH = dU + PdV, then:

dH = dq

Sloppy in some respects, but I still remember it and that says something