6
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ I don't agree with you assertion in $(3)$ that $U=TS-pV$. Where does that come from? $\endgroup$ – user213305 Oct 13 '17 at 13:55
  • $\begingroup$ en.wikipedia.org/wiki/… -- maybe that is a bad assumption since T and P are variable? $\endgroup$ – khaverim Oct 13 '17 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 '17 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 '17 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 '17 at 14:25
4
$\begingroup$

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: $N\mathrm{d}\mu -V\mathrm{d}P + S\mathrm{d}T = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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