I know that the $T dS$ equations are obtained from systems undergoing internally reversible processes, and I also understand where the first one comes from.

As for the second equation, it is obtained using both the enthalpy equation and the first $T dS$ equation.

What I am having a difficult time understanding is when the differential forms

$$dH = dU + \underline{d(PV)} = dU + \underline{pdV + Vdp.}$$

The underlined terms are what's confusing me. They are to equal each other. I do not know what they are trying to express and what the quantities mean.


2 Answers 2


You have

$$dU = TdS - pdV$$

Then add the term what's confusing you both sides and in the end you will notice why

$$d(U+pV) = TdS - pdV + pdV + Vdp = TdS + Vdp$$

First term on the left hand side is nothing but $dH = d(U+pV)$

  • $\begingroup$ Thank you. What I also find confusing is when expanding $d(PV)$, what does it mean? What are they trying to express? $\endgroup$
    – Qwin
    Apr 22, 2021 at 17:27
  • 1
    $\begingroup$ $pV$ is the work and you get the infinitesimal change in work from $pdV+Vdp$ where in an isobaric case this will lead to the first term and in case of isovolumetric this will lead to second term. $\endgroup$
    – Monopole
    Apr 22, 2021 at 17:41
  • $\begingroup$ +1. Ah, hence incompressible. $\endgroup$
    – Qwin
    Apr 22, 2021 at 17:49

Consider $P=x\;and\;V=y$, .
So, $PV=xy$
Now $P$ and $V$ can take any positive real values.
So $x\in\mathbb R^{+}$ and $y\in\mathbb R^{+}$.
Now take $g:\mathbb R^{+}\times\mathbb R^{+}\to\mathbb R$
s.t. $g(x,y)=xy$.
Now $dg=\frac{\partial g}{\partial x}\Big\rvert_y dx+\frac{\partial g}{\partial y}\Big\rvert_x dy \tag{1}$ We can see that
$\frac{\partial g}{\partial x}\Big\rvert_y =\frac{\partial xy}{\partial x}\Big\rvert_y =y$.
And, $\frac{\partial g}{\partial y}\Big\rvert_x =\frac{\partial xy}{\partial y}\Big\rvert_x = x$.

So, 1 becomes,

As $x=P$ and $y=V$. So, $g=PV$.
Thus $d(PV)=VdP+PdV$.

  • $\begingroup$ Thank you for your answer, although I only understood a little bit of that, I still did not understand it fully. Partial derivatives are still very new to me, if you could explain it in other words for me I would be grateful, and perhaps I could understand it at least a little better. $\endgroup$
    – Qwin
    Apr 22, 2021 at 17:41
  • 1
    $\begingroup$ $d(PV)$ is the change in $PV$ when $P$ and $V$ is changed to $dP$ and $dV$ respectively. $dP$ and $dV$ are very very small. So, $d(PV)=[(P+dP)(V+dV)] - PV= (PV+PdV+VdP+dPdV)-PV=PdV+VdP+dPdV=PdV+VdP$ ($dPdV$ can be neglected). So you can see that change in $PV$ is the the sum of product of $dV$ keeping $P$ fixed and product of $dP$ keeping $P$ fixed. Naively the form of $d(PV)$ you can relate to product rule of differentiation. $\endgroup$
    – Iti
    Apr 22, 2021 at 17:58
  • $\begingroup$ Thank you, I already marked an answer to the question, but this is helpful, as well. Why do they have to be very small? $\endgroup$
    – Qwin
    Apr 22, 2021 at 18:05
  • $\begingroup$ That's what the meaning of differentials.It is the very very small change in the quantity PV. $\endgroup$
    – Iti
    Apr 22, 2021 at 18:21
  • $\begingroup$ Haha, yes, you are right! $\endgroup$
    – Qwin
    Apr 22, 2021 at 18:24

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