Understanding Enthalpy: Confusion Cleared Up | Thermodynamics Course"

[Oliver321]

Hello everyone!
I have a course in thermodynamics this year, and there is a question about enthalpy that I cannot answer: given the definition of enthalpy H=U+PV and the integral form of the internal energy U=TS-PV we conclude that H=TS.
We normally say that enthalpy equals the heat exchanged in a isobaric processes. But where does the pressure appear in this equation? Related to this: the differential form of H is said to be dH=TdS+Vdp arising from dH=d(U+PV)=dU+d(PV). But if I do the same with the formula above dH=d(TS)=TdS+SdT I get a different result. How can this be?
And at least: We know that U=Q+W and U=TS-PV. Does this (in this case) mean, that Q=TS and W=-PV?

Thanks for every helping answer!

 

[anuttarasammyak]

the integral form of the internal energy U=TS-PV

$$dU=TdS-pdV=T(S,V)dS-p(S,V)dV$$
I am afraid integration of RHS does not coincide with TS-pV.

 

[Oliver321]

$$dU=TdS-pdV=T(S,V)dS-p(S,V)dV$$
I am afraid integration of RHS does not coincide with TS-pV.

Thanks for your awnser!
I know that the integration is in general not that trivial, but in this case it is. This is due to eulers homogeneous function theorem. See the Wikipedia article (https://en.wikipedia.org/wiki/Internal_energy) in section ‘internal energy of multi component systems‘.

 

[anuttarasammyak]

U=TS-PV holds in condition of constant temperature and constant pressure. In this condition
due to $dp=dT=0$,
$$dH=TdS+Vdp=TdS=TdS+SdT$$

 

[Oliver321]

U=TS-PV holds in condition of constant temperature and constant pressure. In this condition
due to $dp=dT=0$,
$$dH=TdS+Vdp=TdS=TdS+SdT$$

Thanks!
In my thermodynamics lecture notes, I can not find a indication that this only applies if T and P are constant. It is derived in a similar way like here: https://ps.uci.edu/~cyu/p115B/LectureNotes/Lecture6.pdf

Edit:
Could following be true: If H actually is TS than it follows dH=TdS+pdV=TdS+SdT and therefore TdS +pdV-TdS-SdT=0. This is true because of the Gibbs duhem relation. So it is true that H=U-pV=TS?

 

[anuttarasammyak]

$$dH=TdS+Vdp$$
$$d(TS)=TdS+SdT$$
So
$$d(H-TS)=Vdp-SdT\neq 0$$
in general.

 

[Oliver321]

$$dH=TdS+Vdp$$
$$d(TS)=TdS+SdT$$
So in general
$$d(H-TS)=Vdp-SdT\neq 0$$

But Vdp-SdT=0 because of the Gibbs Dulem relation (I am ignoring the chemical Potential)

 

[anuttarasammyak]

The Gibbs Dulem relation says
$$\sum_i N_i d\mu_i =Vdp-SdT$$
How do you ignore $\mu$ ?

 

[Oliver321]

The Gibbs Dulem relation says
$$\sum_i N_i d\mu_i =Vdp-SdT$$
How do you ignore $\mu$ ?

I have also ignored it in the definition of H=U-pV. U has the chemical potential in it. If I make the calculation with the chemical potential from the beginning on I get: d(H-TS)=Vdp-SdT-N$\mu$ which is zero.

 

[anuttarasammyak]

$$H-TS:=G$$
$$dG=Vdp-SdT+\mu dN=N d\mu + \mu dN =d (N\mu)$$
$$G(N,p,T)=N\mu(p,T)$$
$d\mu=0$ for process of p,T=constant. $dG=0$ for the process of N,p,T=const.

 

[Oliver321]

$$H-TS:=G$$
$$dG=Vdp-SdT+\mu dN=N d\mu + \mu dN =d (N\mu)$$
$$G(N,p,T)=N\mu(p,T)$$
$d\mu=0$ for process of p,T=constant, however, $\mu$ cannot be disregarded as far as material exist.

Yes, that’s absolutely right. I have a mistake in my calculation above. So here is what I think:

Let’s talk about only one sort particle:
H=U+pV=ST-pV+N$\mu$+pV=ST+N$\mu$
So if H is really ST+N$\mu$ than
d(H-ST-N$\mu$)=0

This is true because of Gibbs dulem:
d(H-ST-N$\mu$)=TdS+pdV+$\mu$dN-TdS-SdT-Nd$\mu$-$\mu$dN = pdV-TdS -Nd$\mu$ =0

So this is true. And also the referenced integral formula of U when I add $\mu$N.
Is this correct?

 

[anuttarasammyak]

Yea, $d(H-TS-N\mu)=d(H-TS-G)=d(U+PV-TS-G)=d(0)=0$.

By direct calculation of differentiation
$$dU=TdS-pdV+\mu dN$$
$$d(pV)=pdV+Vdp$$
$$d(-TS)=-TdS-SdT$$
$$d(-G)=-\mu dN-Nd\mu$$
Sum of RHS =$-SdT+Vdp-Nd\mu$=0, Gibbs Duhem relation.

 

[Oliver321]

Yea, $d(H-TS-N\mu)=d(H-TS-G)=d(U+PV-TS-G)=d(0)=0$.

Thank you!

 

[Chestermiller]

If U=TS-PV, then G=0

 

[vanhees71]

But on the other hand there are Euler's relations, following from the extensivity of $U$ and its "natural independent thermodynamical quantities", $S$ and $V$ (discussing the most simple case of a gas with a fixed number of particles). So we have
$$U(\lambda S,\lambda V)=\lambda U(S,V).$$
Take the drivative of this equation wrt. to $\lambda$ and set $\lambda=1$ you get
$$S \partial_S U + V \partial_V U=U.$$
On the other hand
$$\mathrm{d}U=\mathrm{d} S T - \mathrm{d} V P,$$
from which indeed
$$T=\partial_S U, \quad P=-\partial_V U.$$
From that
$$U=ST-PV.$$
The complete Legendre transform which eliminates all extensive variables must lead to 0, and indeed
$$G=U-ST+PV=0.$$
The free enthalpy (Gibbs potential) is only non-trivial if you have more extensive variables, e.g., the particle number. Then the internal energy is
$$U=U(S,V,N)$$
and
$$\mathrm{d} U=T \mathrm{d} S - P \mathrm{d} V + \mu \mathrm{d}N,$$
where we have an additional intensive variable is the chemical potential $\mu$. Then the Euler relation reads
$$U=ST-PV+\mu N,$$
and the Gibbs potential is
$$G=\mu N \neq 0.$$
Now to the original posting #1. One has to be aware which are the "natural independent variables" for the used thermodynamic potential. Then no problems occur:

With the same technique you also prove the relation for the enthalpy
$$H=U+P V.$$
From
$$\mathrm{d} H=\mathrm{d} U + \mathrm{d} P V+V \mathrm{d} P = \mathrm{d} S T + \mathrm{d} P V$$
you see that the natural independent variables for $H$ are $S$ and $P$ and
$$\partial_S H=T, \quad \partial_P H=V.$$
Now from the extensivity of $S$ and the intensivity of $P$ you have
$$H(\lambda S,P)=\lambda H(S,P).$$
Taking the derivative of this equation wrt. $\lambda$ and then setting $\lambda=1$ you get
$$H(S,P)=S T(S,P).$$
So you have
$$\mathrm{d} H=T \mathrm{d} S + S \mathrm{d} T, \qquad (1)$$
but to find the original relation you need to consider $T$ as a function of $S$ and $P$. Since $T$ is intensive we have
$$T(\lambda S,P)=T(S,P)$$
Taking the derivative wrt. $\lambda$ and then setting $\lambda=1$ yields
$$\partial_S T(S,P)=0.$$
From this you get
$$\mathrm{d} H=T \mathrm{d} S + \mathrm{d} P S \partial_P T(S,P). \qquad (2)$$
But now
$$T(S,P)=\partial_S H(S,P) \; \Rightarrow \; \partial_P T(S,P)=\partial_P \partial_S H(S,P) = \partial_S \partial_P H(S,P)=\partial_S V(S,P). \qquad (3)$$
Since now $V$ is extensive we have
$$V(\lambda S,P)=\lambda V(S,P).$$
Once more the trick with the $\lambda$ derivative yields
$$V(S,P)=S \partial_S V(S,P).$$
Now using (3) in (2) and then this relation we get
$$\mathrm{d} H = T \mathrm{d} S + \mathrm{d} P S \partial_S V=T \mathrm{d} S+ \mathrm{d} P V.$$
So the important point is to keep in mind which are the "natural independent variables" for a used potential. Then you get no contradictions in the formalism with the Legendre transformations and the Euler relations.

That's also, why it's important to keep in mind which variables are held fixed in the Euler relations. E.g., you have
$$T=\partial_S U(S,V)$$
and
$$T=\partial_S H(S,P),$$
but you get $T$ as a function of different other thermodynamical variables using the different potentials.

To emphasize which variables are to be held constant in the partial derivatives in Maxwell relations one often writes the above relations
$$T = \left (\frac{\partial U}{\partial S} \right)_V=\left (\frac{\partial H}{\partial{S} \right)_{P},$$
i.e., you indicate with a subscript which other independent variable has to be kept fixed.

All this can of course extended if you consider systems with more independent variables, as if there is one or several particle numbers additional independent variables.