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?

  • $\begingroup$ The expression you should use for a finite change is $\Delta H = \Delta U + \Delta (pV)$ $\endgroup$
    – Buck Thorn
    Commented Nov 11, 2020 at 13:49
  • $\begingroup$ $d(pV)=p.dV + V.dp + dp.dV$, $\Delta(pV) = p.\Delta V + V . \Delta p + \Delta p . \Delta V$. 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. $\endgroup$
    – Poutnik
    Commented Nov 11, 2020 at 14:33
  • $\begingroup$ @Pournik See my answer. $\endgroup$ Commented Nov 11, 2020 at 14:59

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Poutnik
    Commented Nov 11, 2020 at 17:09
  • $\begingroup$ Actually, the process path doesn’t have to be straight lines. The equations I gave are totally algebraic. $\endgroup$ Commented Nov 11, 2020 at 17:56
  • $\begingroup$ 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 ) $\endgroup$
    – Poutnik
    Commented Nov 11, 2020 at 17:59
  • $\begingroup$ THANKS. I thought you were actually saying that they had to be straight lines. Your assessment is correct, of course. $\endgroup$ Commented Nov 11, 2020 at 22:20

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