1
$\begingroup$

I know for a closed system by using Legendre transformation:

$$d[H(S,P,N)]=d(U+PV)=TdS-PdV+PdV+VdP=TdS+VdP$$

But by direct differentiation:

$$d[H(S,P,N)]=d(TS+\mu N)=TdS+\mu dN=TdS$$

The two equations above don't match with each other. And same problem appears when directly differentiating thermal potential $F$ and $G$. So what is it that I am doing wrong here?

$\endgroup$
4
  • $\begingroup$ hint: en.wikipedia.org/wiki/Homogeneous_function (set $k=1$) $\endgroup$
    – hyportnex
    Commented Mar 26, 2022 at 23:59
  • $\begingroup$ Why are you not doing $dU=TdS+SdT-PdV-VdP+\mu dN+Nd\mu$? $\endgroup$ Commented Mar 27, 2022 at 0:22
  • $\begingroup$ @BioPhysicist Because the dependent variables of $U$ is defined (or postulated) to be $\{S,V,N\}$ before the differentiation happened? $\endgroup$
    – P'bD_KU7B2
    Commented Mar 27, 2022 at 0:25
  • $\begingroup$ @hyportnex,$$\begin{align} H(S,P,N) &= TS-μN = (S,P,N)\cdot \nabla H(S,P,N) \\ &= S\left(\frac{\partial H}{\partial S}\right)_{P,N} + P\left(\frac{\partial H}{\partial P}\right)_{S,N} + N\left(\frac{\partial H}{\partial N}\right)_{S,P}\\&=ST+0-Nμ \end{align}$$ I still can't pull the $VdP$ term out of it though, is there anything I am missing? $\endgroup$
    – P'bD_KU7B2
    Commented Mar 27, 2022 at 0:28

2 Answers 2

3
$\begingroup$

(1) Starting from $H=U+PV$, we have

$$dH=d(U+PV)=dU+PdV+VdP.$$

In conjunction with the fundamental relation $dU=TdS-PdV+\mu dN$, we obtain

$$dH=TdS+VdP+\mu dN.$$

(2) Starting from $H=TS+\mu N$, we have

$$dH=d(TS+\mu N)=TdS+SdT+\mu dN+Nd\mu.$$

In conjunction with the Gibbs–Duhem relation $SdT-VdP+Nd\mu=0$, we obtain

$$dH=TdS+VdP+\mu dN.$$

Same result. (For each, drop the final term for a closed system.)

$\endgroup$
3
$\begingroup$

Since $H = TS+\mu N$, recall that the energy is $E = TS-PV + \mu N$, then $H = E + PV$. If we take the total differential of H we find, \begin{align} dH& = dE + PdV + VdP\\ & = TdS - PdV + PdV + VdP\\ & = TdS + VdP, \end{align} where we have used $dE = TdS - PdV + \mu dN$ assuming $Nd\mu = 0$ in the second line (which is common).

Hope this helps, let me know if you have any questions.

$\endgroup$
4
  • $\begingroup$ Thanks, but this is not what I am confused about. What happened is when I tried to directly differentiate $H$, so $dH=d(TS+μN)=TdS$ for a closed system, and the $VdP$ term is missing. Why would direct differentiation gives the wrong answer in this case? $\endgroup$
    – P'bD_KU7B2
    Commented Mar 27, 2022 at 0:12
  • 1
    $\begingroup$ If you go that route, you need to take the total differential of $H$, for what you put this would be $d(TS + \mu N) = TdS + SdT + \mu dN + Nd\mu$. If this is troublesome, recall that really this is a time dependent system, and divide out by a $dt$. But from here, you would have to use the differential for energy, $dE$ to restore the $VdP$ term. The way I did it is just quicker, that's all. But in short, I think you are taking the differential wrongly. To expand a little more, use the product rule on the $TS$ and $\mu N$... (the derivative is a linear operator in a sense). $\endgroup$
    – MathZilla
    Commented Mar 27, 2022 at 0:16
  • $\begingroup$ I thought $H$ and $U$ are pre-defined as $H(S,P,N)$ and $U(S,V,N)$, thats why I didn't use the product rule. But as you are saying here, if I choose to directly differentiate $dH$, I shouldn't define the independent variable $\{S,P,N\}$ for $H$ at the first place? $\endgroup$
    – P'bD_KU7B2
    Commented Mar 27, 2022 at 0:30
  • 1
    $\begingroup$ Recall that enthalpy is defined as a function of $H(S,P,N)$, which are the macrostates of the system, and in particular, we took our differentials with respect to a reversible process. If you are a little lost, take the simplest system you can think of, say a function $f(x,y)$. I have no idea what it is equal to, but I can use some calculus to get: $df(x,y) = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$. And, it may turn out to be, that this is really, $df(x,y) = ydx + xdy$ (you can see where the Maxwell relations come from now too....) $\endgroup$
    – MathZilla
    Commented Mar 27, 2022 at 0:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.