# How to go from $H=TS+\mu N$ to $dH=TdS+VdP$?

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?

• hint: en.wikipedia.org/wiki/Homogeneous_function (set $k=1$) Commented Mar 26, 2022 at 23:59
• Why are you not doing $dU=TdS+SdT-PdV-VdP+\mu dN+Nd\mu$? Commented Mar 27, 2022 at 0:22
• @BioPhysicist Because the dependent variables of $U$ is defined (or postulated) to be $\{S,V,N\}$ before the differentiation happened? Commented Mar 27, 2022 at 0:25
• @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? Commented Mar 27, 2022 at 0:28

(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.)

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.

• 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? Commented Mar 27, 2022 at 0:12
• 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). Commented Mar 27, 2022 at 0:16
• 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? Commented Mar 27, 2022 at 0:30
• 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....) Commented Mar 27, 2022 at 0:35