# Homework Help: Surfactant adsorption chemical equilibrium

1. Jun 16, 2014

### WannabeNewton

1. The problem statement, all variables and given/known data

A dilute solution of surfactants can be regarded as an ideal three-dimensional gas. As surfactant molecules can reduce their energy by contact with air, a fraction of them migrate to the surface where they can be treated as a two-dimensional ideal gas. Surfactants are similarly adsorbed by other porous media such as polymers and gels with an affinity for them.

(a) Consider an ideal gas of classical particles of mass $m$ in $d$ dimensions, moving in a uniform attractive potential $\epsilon_d$ in equilibrium with a thermal bath of temperature $T$. Show that the chemical potential with particle density $n_d$ is given by $\mu_d = -\epsilon_d + k_B T \ln (n_d \lambda^d)$ where $\lambda = \frac{h}{\sqrt{2\pi m k_B T}}$ is the thermal wavelength.

(b) If a surfactant lowers its energy by $\epsilon_0$ in moving from the solution it is in to the surface of the solution, calculate the concentration of floating surfactants at the surface as a function of the solution concentration $n = n_3$ at temperature $T$.

3. The attempt at a solution

I calculated (a) using the canonical ensemble so that $Z = \frac{V^N}{h^{3N}N!}e^{\beta \epsilon_d}\int \Pi _{i = 1}^N d^d p_i e^{-\beta \sum_j p_j^2/2m} = \frac{2\pi V^N}{h^{3N}N!}\frac{(2mE)^{(Nd -1)/2}}{(Nd/2 - 1)!}e^{\beta \epsilon_d} = \frac{V^N}{\lambda^{Nd}N!}\epsilon^{\beta \epsilon_d}$ where $V$ is the volume in $d$ dimensions that a single particle moves in. Then $F = -k_B T \ln Z = Nk_B T \ln n_d \lambda^d - N\epsilon_d - Nk_B T$ where $n_d = \frac{N}{V}$. Hence $\mu_d = \frac{\partial F}{\partial N}|_{T,V} = -\epsilon_d + k_B T \ln n_d \lambda^d$.

My problem is with (b). Let $\epsilon_0 \equiv \epsilon_3 - \epsilon_2$. Now in the canonical ensemble, the system doesn't exchange particles with the thermal bath but subsystems of the system can exchange particles with one another. In our case the surfactants in the solution will flow to the surface, driven by the difference in potential energy $\epsilon_0$ between the solution and surface. They should keep flowing until $dF_{\text{total}} = dF_2 - dF_3 = 0$ since the total Helmholtz free energy of the subsystems combined has to be minimized. But this implies $\frac{\partial F_2}{\partial N}|_{T,V} = \frac{\partial F_3}{\partial N}|_{T,V} \Rightarrow \mu_2 = \mu_3$ hence $n_2 = n_3 \lambda e^{-\beta\epsilon_0}$ which is certainly the wrong answer since this implies $n_2 \ll n_3$ for $\epsilon_0 \rightarrow \infty$ whereas it should be the other way around; in fact the correct answer should be $n_2 = n_3 \lambda e^{\beta\epsilon_0}$.

So where am I going wrong? Thanks in advance.

2. Jun 17, 2014

### Matterwave

[strike]How did you go from $F=Nk_BT\ln n_d\lambda^d-N\epsilon_d-Nk_BT$ to $\mu_d=k_BT\ln n_d\lambda^d-\epsilon_d$?

What happened to the last $-Nk_B T$ term?[/strike]

EDIT: REDACTED

Last edited: Jun 17, 2014
3. Jun 18, 2014

### bloby

Are $\mu_d = -\epsilon_d + k_B T \ln (n_d \lambda^d)$, $\epsilon_0 \equiv \epsilon_3 - \epsilon_2$ and $n_2 = n_3 \lambda e^{\beta\epsilon_0}$ given as the right answers?

4. Jun 18, 2014

### WannabeNewton

Yes blobly. This is problem 4.6 in Kardar if you're interested.

5. Jun 18, 2014

### WannabeNewton

Oops! I forgot that $\epsilon_0 < 0$ as per my sign convention so my answer agrees with the book's answer and the correct limit is actually $\epsilon_0 \rightarrow -\infty$ which gives $n_2 \gg n_3$ as desired.

And thank you to matterwave for setting my head straight on this :)

6. Jun 18, 2014

### Matterwave

This is one deeply confusing sign error, gotta admit.