Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The Gibbs-Duhem equation states

$$0~=~SdT-VdP+\sum(N_i d\mu_i),$$

where $\mu$ is the chemical potential. Does it have any mathematical (about intensive parameters) or physical meaning?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

The intensive variabes $${T=\left(\frac{∂U}{∂S}\right)_V} \ \ \ \ \text{and}\ \ \ {-P=\left(\frac{∂U}{∂V}\right)_S}$$ give the amount of change of the internal energy $U(S,V)$ w.r.t the extensive variables $S$ and $V$. One is heat and the other is work. There are however other extensive variables $N_i$ with slopes $$\mu_i=\left(\frac{∂U}{∂N_i}\right)_{S,V},$$ so $U$ should really read $U(S,V,\{N_i\})$.


The Gibbs free Energy $$G(T,P):=-H^{*S \rightarrow T}=-(-U^{*V \rightarrow P})^{*S \rightarrow T}$$ is connected to $U$ by Legendre transformations with respect to both of the slopes defined above. That (a priori purely mathematical) involutive transformation is a relation between tangents and points of two functions $f\longrightarrow f^*$.

The potential $G$ is said to be "natural" in $T$ and $P$. Similarly $F$ has applications in which $T$ and $V$ is of specific interest.

Considering another potential than the internal energy $U$ is particularly useful if the "slopes" are the parameters, which are held constant in the process you're interested in. Because then, of course, the new function doesn't change in them. For exmaple, if you know that $P=$const. in a process, then $\text d P=0$ and then the change of the enthalpy $H(S,P)$ is given by

$$\text d H=\left( \frac{∂H}{∂S} \right)_P\text d S+0=T\text d S=\delta Q. $$

I.e. the enthalpy change is just the heat flow. A typical such constant pressure situation where the enthalpy $H(S,P)$ as well as the free enthalpy $G(T,P)$ become relevant (the potentials, which depend on the intensive variable $P$ is when you consider chemical reactions in a free area/a laboratory room, becuase then the preassure is just 1 atm whatever happens. Similarly, when a constant external termperature $T$ is forced on your system, the change of the free energy $F(T,V)$ is

$$\text d F=0+\left( \frac{∂F}{∂V} \right)_T\text d V=P\text d V=\delta W, $$

and in this way the free energy relates to work the system can do (volume change in exchange for energy).

To come back from the motivation, in our case of $G$ the result of the Legendre transformation (expressed in terms of the initial $U$) turns out to be the fundamental relation

$$U=G-PV+TS.$$


To shed more light on the question you should have a look at the derivation of your statement, as well as the derivation of

$$G=\sum_i\mu_iN_i.$$

That is, the Gibs free energy is the sum of all the molar partial chemical energies: The two Ledgendre transformations stipped off the extensive variables $S$ and $V$ and what's left in the (extensive) energy $G$ can be expressed only using the expensive quanitities $N_i$.

What you get if you combine the two expressions for $G$ is the Gibbs duhem equation

$$\sum_i N_i d\mu_i=-SdT+VdP,$$

which can then be viewed as an expression restricting the slopes of the function $U(S,V,\{N_i\})$ you stated with. It thereby eliminates degrees of freedom.

share|improve this answer

One can interpret the Gibbs-Duhem relation as stating that there is necessarily one extensive degree of freedom in the system, so that the $k+2$ intensive parameters (with $k$ the number of components) only describe a $k+1$-dimensional "subspace". See the link below for a more detailed presentation:

http://blitiri.blogspot.com/2013/03/gibbs-duhem-and-euler-relations.html

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.