6 March 2013

Gibbs-Duhem and Euler relations

In a previous post I discussed the derivation of Euler's equation in thermodynamics and proposed a geometrical illustration. Here, I will continue the discussion by including the Gibbs-Duhem relation. We will need the following ingredients:
The Euler equation:
\begin{equation}
U = TS -pV + \sum_i \mu_i N_i
\label{eq:Euler}
\end{equation}
First-order homogeneity:
\begin{equation}
U(\lambda S, \lambda V, \lambda N_i) = \lambda U(S, V, N_i)
\label{eq:homog}
\end{equation}
The fundamental relation:
\begin{equation}
\text{d}U = T\text{d}S -p\text{d}V + \sum_i \mu_i  \text{d}N_i
\label{eq:fund}
 \end{equation}
Taking the derivative of \eqref{eq:Euler} and subtracting \eqref{eq:fund} yields
The Gibbs-Duhem relation:
\begin{equation}
S\text{d}T -V\text{d}p + \sum_i N_i  \text{d}\mu_i =0
\label{eq:GD}
\end{equation}
 
To get some insight into the Gibbs-Duhem relation, let us start from the observation that the space dimension is \(k+2\), where \(k\) is the number of components. When working with the energy, we use the extensive variables \( S,V,N_i\), as shown in the Figure, but there are also \(k+2\) intensive parameters \( T, P, \mu _i\). As already discussed, the transformation that takes the system along the dashed line connecting the current point and the origin (the "diagonal") occurs at constant intensive parameters, so that all terms in \eqref{eq:GD} are identically zero. We used this path to prove the Euler relation, so we label it as such.

The intensive parameters do not "feel" a transformation along the Euler path. For them to change, the system must evolve in the "perpendicular" subspace1 (shown as a grey surface) of dimension \(k+1\). The Gibbs-Duhem relation \eqref{eq:GD} states that the variation of the intensive parameters is constrained to this subspace (that we duly label). An intuitive image is then that Euler is perpendicular to Gibbs-Duhem.

[Update 13/03/2013] A similar (but much simpler) situation occurs when mixing \(n\) different components. We can manipulate \(n\) parameters (the mass of each component) but can only set independently \(n-1\) intensive parameters (the mass concentrations) because the sum of all concentrations is always 1. The "missing" \(n\)-th variable is simply the total mass of the solution.



1. I use the quotes around the term perpendicular because \( (S,V,N_i)\) is not a metric space and we cannot define a scalar product (and hence an angle) on it. We can however define an affine transformation (scaling all parameters by the same amount \(\lambda\)) and say more properly that a change in the intensive parameters requires a non-affine transformation in the \( (S,V,N_i)\) space.

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