# Derivation of chemical potential using Boltzmann entropy and number of microstates [closed]

If one has the Boltzmann equation for entropy $$S=k \ln(W)$$ where $$W=T^{C/k}V^{N}$$ is the number of microstates, and it is assumed that all the particles are indistinguishable

How would one derive an equation for chemical potential?

## 2 Answers

The first step is to prove the Fundamental Thermodynamic Relation from this definition of entropy. A simple derivation is given here. One arrives at the relation:

$$dE = T dS - p dV \tag{1}$$

This is then for a system consisting of a fixed number of particles. Then suppose that we have systems containing a different number of particles, then for each such system the number of particles is fixed, so the relation (1) applies to each system separately. We can then consider changing the number of particles while keeping S and V fixed and considering the difference between the internal energies of the different systems, so there is no problem with introducing the number of particles as an additional parameter. Eq. (1) can then be extended into:

$$dE = T dS - p dV +\mu dN \tag{2}$$

where, by definition, we have:

$$\mu \equiv \left(\frac{\partial E}{\partial N}\right)_{S, V}$$

We call $\mu$ the chemical potential.

This is then the starting point to from which one can derive other thermodynamic relations involving the chemical potential. E.g., we can write:

$$dF = -S dT - p dV +\mu dN \tag{3}$$

where $F= E - T S$ is the Helmholtz free energy. So, we have:

$$\mu = \left(\frac{\partial F}{\partial N}\right)_{T, V}$$

If you calculate te internal energy of your system, you can express $F$ in terms of $T$ and $V$ and use the above relation to evaluate the chemical potential.

Alternatively, you can use:

$$dS = \frac{1}{T}dE + \frac{p}{T} dV -\frac{\mu}{T} dN$$

So:

$$\mu = -T \left(\frac{\partial S}{\partial N}\right)_{E, V}$$

You then need to find $E$ as a function of $T$ for your system, so that you can keep that quantity constant when differentiating w.r.t. $N$.

Take a box, with a permeable partion evenly dividing its volume, so that energy and particles (for simplicity, particles of the same species) are free to move until equilibrium is established,

So intially the l.h.s of the box has $S_A$, $N_A$ and $U_A$ and, not surprisingly, the r.h.s has $S_B$, $N_B$ and $U_B$,

Assume that both the total particle number, $N$ and energy $U$, are fixed. This implies that the total entropy $S$ of this system is a function of $U_A$ and $N_A$.

At equilibrium, $S$ is maximised, so we can deduce that:

$\left(\frac {\partial S_{total}}{\partial U_A}\right)_{N_A, V_A} =0$, implying that the two systems are at equal temperature

and $\left(\frac {\partial S_{total}}{\partial N_A}\right)_{U_A, V_A} =0$.

So at equilibrium,

$\left(\frac {\partial S_{A}}{\partial N_A}\right) =\left(\frac {\partial S_{B}}{\partial N_B}\right)$, taking the partials at a fixed energy and volume.

By convention, place a factor of -1 on each side of the equality. Because thermal equilibrium is established, we can also place T in front of each derivative.

$\left(-T\frac {\partial S_{A}}{\partial N_A}\right) =\left(-T\frac {\partial S_{B}}{\partial N_B}\right)$

The chemical potential is then given by

$\left(-T\frac {\partial S}{\partial N}\right)_{U, N}= \mu$

At equilibrium $\mu_A=\mu_B$