I came across an interesting question with some physical chemistry students today. Based on the following steps, we're uncertain whether the statement in the title is/could be true. Assuming $dN = 0$,

Enthalpy's natural variables are Entropy and Pressure:

$$dH(S,p) = TdS + Vdp \space (1)$$

Enthalpy can be expressed as a total derivative of Temperature and Pressure:

$$ dH(T,p) = \frac{\partial H}{ \partial T}\vert_p \space dT + \frac{\partial H}{\partial p}\vert_T \space dp \space (2)$$

Total Enthalpy is $$H = U + pV = TS - pV + \sum_i \mu_i N_i + pV = TS + \sum_i \mu_i N_i \space (3)$$

Thus, taking partial derivatives in (2), $$ dH(T,p) = SdT + 0 \space dp \space (4)$$

This means that, if $ dH(S,p) = dH(T,p)$,

$$ SdT = TdS + Vdp \space (5)$$ must be true. Can anyone see something wrong?

This seems to boil down to a math question: are two differentials of the same function (always) equal if expressed by different variables?

  • 2
    $\begingroup$ I don't agree with you assertion in $(3)$ that $U=TS-pV$. Where does that come from? $\endgroup$
    – user213305
    Oct 13, 2017 at 13:55
  • $\begingroup$ en.wikipedia.org/wiki/… -- maybe that is a bad assumption since T and P are variable? $\endgroup$
    – khaverim
    Oct 13, 2017 at 14:00
  • 2
    $\begingroup$ What that line $U=TS-pV+\sum_i \mu_i N_i$ actually tells you is that $G=\sum_i \mu_i N_i=U-TS+pV$ where $G$ is the Gibbs free energy. You've neglected the chemical potential $\mu$ of the different species in your system - which will have a hidden and complex relationship with $T$ and $p$. $\endgroup$
    – user213305
    Oct 13, 2017 at 14:07
  • $\begingroup$ True, thanks. Nonetheless, the focus of my question remains. The partial derivatives of H are still the same (no $\mu$ or $N$ dependence) $\endgroup$
    – khaverim
    Oct 13, 2017 at 14:10
  • $\begingroup$ en.wikipedia.org/wiki/Enthalpy#Other_expressions probably provides the relationship you want for the exact differential if $H(T,p)$ with a reference for the derivation - but as a rule $\frac{\partial \mu}{ \partial T}\ne 0$ hence leading your fourth equation to be incorrect. $\endgroup$
    – user213305
    Oct 13, 2017 at 14:25

1 Answer 1


The reasoning stated is partially correct, but the final relation you arrived to is incorrect. I will try to explain why and write the N dependence explicitly for completeness. The crucial thing is that when one writes an expression such as $$\left(\frac{\partial H}{\partial T} \right)_{P,N}$$ what one really means is "take the partial derivative of $H$ written as a function of $T$, $P$ and $N$ with respect to $T$". When you take the partial derivatives in equation (2) then, you should take them considering $H$ as a function of $T$, $P$ and $N$. You have to consider then the expression $H = H(T,P,N) = S(T,P,N)T + \mu(T,P) N$. If you differentiate that equation with respect to T and P the result is:

$$\left(\frac{\partial H}{\partial P} \right)_{T,N} = \left(\frac{\partial S}{\partial P} \right)_{T,N} T + \left(\frac{\partial \mu}{\partial P} \right)_{T} N~~~;~~~ \left(\frac{\partial H}{\partial T} \right)_{P,N} = \left(\frac{\partial S}{\partial T} \right)_{P,N} T + S + \left(\frac{\partial \mu}{\partial T} \right)_{P} N$$ If you replace those two relations in your equation (2) the result is:

$$ \mathrm{d}H = \left(\left(\frac{\partial S}{\partial P} \right)_{T,N} T+ \left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(\left(\frac{\partial S}{\partial T} \right)_{P,N} T + S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T.$$

You can indeed equate this with your equation (1), which is what you ask in your main question. This is the same thing one ordinarily does when expressing a scalar as a function of different sets of coordinates, for instance $f = f(x,y) = x^2 + y^2$ and $f = f(r,\theta) = r^2$ this means, equating both, that $x^2 + y^2 = r^2$, which is a relationship that must hold if you want both $(x,y)$ and $(r,\theta)$ to refer to $f$ (this last sentence may be a bit tautological, I hope what it means is clear, note that it is certainly not "always" true that $f(x,y)$ and $f(r,\theta)$ are equal, since they are two different functions, albeit expressed with the same letter, for instance, $f(x=1,y=0) = 1$ but $f(r=2,\theta= \pi) = 4$, there must be a specific relation between the coordinates for these to be equal). If you do this you obtain:

$$\left(\left(\frac{\partial S}{\partial P} \right)_{T,N} T+ \left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(\left(\frac{\partial S}{\partial T} \right)_{P,N} T + S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T = T~\mathrm{d}S + V~\mathrm{d}P.$$

Note that you would get the same expression even if you considered processes in which $\mu \mathrm{d}N$ wasn't cero, cause both terms would cancel out. If one remembers that $T\mathrm{d}S = T\left(\frac{\partial S}{\partial P} \right)_{T,N}\mathrm{d}P + T\left(\frac{\partial S}{\partial T} \right)_{P,N}\mathrm{d}T$ then this simplifies to:

$$\left(\left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T = V~\mathrm{d}P.$$

Dividing through by N:

$$\left(\frac{\partial \mu}{\partial P} \right)_{T} \mathrm{d}P + \;\left(\bar{S}+ \left(\frac{\partial \mu}{\partial T} \right)_{P} \right)~\mathrm{d}T = \bar{V}~\mathrm{d}P.$$

This is true if and only if:

$$\left(\frac{\partial \mu}{\partial P} \right)_{T} = \bar{V}~~;~~\left(\frac{\partial \mu}{\partial T} \right)_{P} = -\bar{S}.$$

These are correct relations and can also be deduced from the Gibbs-Duhem equation: $N\mathrm{d}\mu -V\mathrm{d}P + S\mathrm{d}T = 0$.


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