# Deriving a relation between entropy, volume and particles through the chemical potential

I'm a teaching assistant on a first year course on Introduction to Thermodynamics based on the book "Molecular Driving Forces" by Dill & Bromberg.

There's a question which asks the students to derive the following relation between S, V and N:

$S(V,N) = Nkln(V)$

A simple solution that is hinted is to use the differential form of the fundamental entropy equation and realizing the following identities (assuming an ideal gas):

$\frac{P}{T} = (\frac{\partial S}{\partial V})_{U,N}=\frac{Nk}{V}$

$ds = Nk\int \frac{dV}{V} \rightarrow \Delta S = Nkln(\frac{V_2}{V_1})$

Now, I've meditated about and have been asked by several students the following question: could we arrive to the same (or equivalent) expression through a chemical potential route? After all, we need an expression that relates S both to N and V. Unfortunately, I have not been able to give a satisfactory answer.

Here's what I've tried, arriving to some ugly looking equation that intuitively looks incorrect to me:

Use the definition: $\frac{\mu_j}{T}=-(\frac{\partial S}{\partial N_j})_{U,V,N_{j\neq i}}$, which also stems from the fundamental differential entropy equation.

Now I make the assumption that the chemical potential is equal to the density of molecules in the ideal gas. This is what I'm most dubious about, although I couldn't think of another way in which to have a relation to it. If that holds, I can then write the following identities:

$\mu = \frac{n}{V}$ then $P=\mu kT$

$\frac{-P}{T^2k} = \frac{\mu}{T} = (\frac{\partial S}{\partial N_j})_{U,V,N_{j\neq i}}$

Now because $P = \frac{nkT}{V}$,

$\frac{-P}{T^2k} = \frac{-nkT}{VT^2k}=\frac{-n}{VT}=(\frac{\partial S}{\partial N_j})_{U,V,N_{j\neq i}}$

Which I think can be solved:

$\int dS = \int \frac{-n}{VT}dn \rightarrow \Delta S = -\frac{\Delta n^2}{2VT}$

This result doesn't really make much sense to me (a negative entropy (?)). I'm a bit confused, because the first 'route' seems so simple but this second attempt looks fairly involved.

So, what I wrong in my logic? Apologies beforehand if I'm doing something egregiously wrong, it's been awhile since I've had to solve these 'pen and paper' problems.

• I'm not sure how to help, but I'm sure that $\int n \, dn \neq \Delta n$. – ralk912 Mar 17 '18 at 3:30
• Also, I think that using this relationship between chemical potential and entropy won't lead to your result, since this relationship holds $V$ constant, and you want to get to entropy as a function of a variable $V$ and $n$. But, that I might have wrong. – ralk912 Mar 17 '18 at 3:48
• @ralk912 d'oh! Thanks for the heads up on the integral, will update the correct result now. Also, as far as I'm aware, isn't that what's happening on the first correct solution to the problem? Where $N$ is being treated as a constant, although $S$ depends on both $V$ and $n$. – j_eiros Mar 18 '18 at 19:58
• Well, $N$ is treated constant because it's for that particular case (i.e. this is valid for gas expansions/compressions). If both $N$ and $V$ were to vary, then the first derivation wouldn't be valid. You're trying to get to an expression where $N$ is constant and $V$ varies from one where $V$ is constant and $N$ varies. – ralk912 Mar 18 '18 at 21:19
• Also, from a dimensional analysis, you're assumption says that $\mu$ has $\pu{mol dm-3}$ units, while the definition implies the correct $\pu{J mol-1}$ units for chemical potential. So I think that assumption is not valid. – ralk912 Mar 18 '18 at 21:26