# Why is the change in enthalpy computation for a change of state done like this?

Suppose that a system whose initial state is Pressure = $$P_1$$, Volume = $$V_1$$; and it is taken through a process after which final pressure and volume are $$P_2$$ and $$V_2$$ respectively. Now change in enthalpy is defined as
$$\Delta H= \Delta U +P\Delta V+V\Delta P$$
The thing which I am confused with is which pressure and which volume do we hold constant for their respective parts in the calculation? Till now I had dealt only with problems involving the $$P\Delta V$$ part, and it was calculated as $$P_2(V_2-V_1)$$, same as irreversible work. But when I came across a problem in which Pressure also varied with volume, the second part needed to be taken as $$V_1(P_2-P_1)$$.Why do we take $$P_2$$ as constant in one case and $$V_1$$ in the other?
One reason which I thought of was that in an irreversible process, The pressure is first changed by a finite amount and then the volume varies. Thus the pressure throughout the process was $$P_2$$. This can be a rationale for the $$P\Delta V$$ part, but I am not sure if it is reasonable to say that the volume at start of the process was $$V_1$$, which resulted in this method of calculation.
So.. what is the reason?

• The expression you should use for a finite change is $\Delta H = \Delta U + \Delta (pV)$ Nov 11, 2020 at 13:49
• $d(pV)=p.dV + V.dp + dp.dV$, $\Delta(pV) = p.\Delta V + V . \Delta p + \Delta p . \Delta V$. But dp.dV can be neglected with infinitely small error, so it is never explicitly mentioned nor used. $\Delta p . \Delta V$ cannot be neglected. Draw it as geometrical scenario of 1 rectangle consisting of 4 rectangles to see it. Nov 11, 2020 at 14:33
• @Pournik See my answer. Nov 11, 2020 at 14:59
• @ChetMiller thanks, I have seen it. Nov 11, 2020 at 15:16

## 1 Answer

You're confused because the equation is incorrect. It should read $$\Delta H=\Delta U+\Delta (PV)$$If you insist on writing it the way you did, then P should be $$\frac{P_1+P_2}{2}$$ and V should be $$\frac{V_1+V_2}{2}$$. That will give the same result as the first equation I wrote.

• Nice analysis. Not obvious on first thought, but if one imagines the geometry of 1 rectangle + 2 trapezoids, instead of 1+2+1 rectangles, it becomes obvious. Nov 11, 2020 at 17:09
• Actually, the process path doesn’t have to be straight lines. The equations I gave are totally algebraic. Nov 11, 2020 at 17:56
• It is a state change, independent on path. You just divide rectangle deltap.deltaV and each triangle part add to p.deltaV or V deltap. In geometric picture it is very obvious. (Not really drawn it, just imaginating it ) Nov 11, 2020 at 17:59
• THANKS. I thought you were actually saying that they had to be straight lines. Your assessment is correct, of course. Nov 11, 2020 at 22:20