In some books, the chemical potential is defined in term of the Gibbs energy as $$ \mu_i \equiv \left( \frac{\partial G}{\partial n_i} \right)_{P,T,n_j} $$ where $n_j$ indicates all mole numbers except the ith species mole number.

In other books, it is defined in term of the Helmholtz energy as

$$ \mu_i \equiv \left( \frac{\partial A}{\partial n_i} \right)_{V,T,n_j} $$

Looking up at the Physics Stack Exchange, I could also find:

$$ \mu_i \equiv \left( \frac{\partial U}{\partial n_i} \right)_{S,V,n_j} $$

And I suppose (by analogy) that one may also define it as

$$ \mu_i \equiv \left( \frac{\partial H}{\partial n_i} \right)_{S,P,n_j} $$

Are all these definitions equivalent? Why?

Or they give rise to different quantities and one should be consistent when using one definition or the other?

I have tried to work out one definition from the other using thermodynamics identities but I could not make it out.

  • They are just simplifications depending on what is taken to be constant. – Gowtham Jan 2 '15 at 18:06
up vote 2 down vote accepted

Let's start from the second TD law: $$ \tag 1 dU(S,V) = TdS - pdV, $$ In case of variable number of particles it is modified by the adding $\mu dn$ term to the right side: $$ dU(S, V, n) = TdS - pdV + \mu d n. $$ The other TD potentials are related with $U$ by Legendre transformation with variables $T, S, p, V$, but not $\mu$ or $n$. So it's obviously that the relation $$ \mu = \left( \frac{\partial Y (n, X)}{\partial n}\right)_{X = const} $$ is true for all of four TD potentials $Y$.

  • 1
    Thanks! I had to read other sources to fully understand the answer, but your mention to Legendre transformation showed me the way. Reviewing "ܶNonnatural derivatives" was also important. The paper arxiv.org/abs/0806.1147 by Zia, Redish and McKay was very useful for understanding the Legendre transformation. – toliveira Jan 3 '15 at 21:26

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