# Surface Tension - Lung Alveoli

So, the way I understand this is as follows :

The alveoli (pretend they're bubbles) have diameters of the order of microns implying a massive pressure required to inflate them by the Young-Laplace equation.

$p_{in}-p_{out}=\frac{2\gamma}{r}$

However, the presence of pulmonary surfactant molecules (lets just pretend they're like detergents molecules in washing liquid) can effectively reduce the surface tension at the unexpanded alveoli and hence allow easy inflation.

Now this bit I don't understand :

As the alveoli expand the distance between the individual surfactant molecules on the alveoli increases and hence the surface tension rises again therefore decreasing the rate of expansion.

What is the mathematical connection between surface tension and separation between surfactant molecules ? How can I rationalise the statement in bold ?

Suppose the surface tension of a pure interface (without surfactants) is $\gamma_0$. Now adding the surfactants lowers the surface tension to $\gamma(a) = \gamma_0 - \Pi(a)$, where $a$ is the area per surfactant (or the inverse of the surface density; one can use either, but $a$ is more customary). The term $\Pi(a)$ is called the surface pressure precisely because it acts like a two-dimensional pressure. Let me elaborate.
As to how to model $\Pi(a)$, the simplest models might be that of ideal gas, or van der Waals equations of state. Some more elaborate, but still quite elementary methods would be some lattice models (a la Langmuir adsorption model), and at the more advanced end you have density functional theory and fundamental measure theory and the stuff that branches off of them. The literature on the subject is huge.
• I can see how this connects to my question and I really appreciate the detail. With specific reference to my question in bold could you say that increasing the surface area of the alveoli during inspiration (assuming constant no. of surfactant molecules) $a$ will increase and consequently the surface pressure will decrease in magnitude meaning that the rate of expansion slows ? – Ari Ben Canaan Jun 3 '14 at 14:20
• @AriBenCanaan Rate of expansion slows as compared to what (before inhalation I guess the system should be in equilibrium)? We could guess that $\frac{da}{dt} = k(\gamma_\text{eq}-\gamma(a))$, where $k$ is a compressibility constant which does in general depend on $a$, and $\gamma_\text{eq}$ is the equilibrium tension defined by the Young-Laplace equation. Now obviously as the system gets closer and closer to the equilibrium tension, the forces get smaller and smaller and the rate of expansion therefore also becomes slower (the system approaches equilibrium roughly with an exponential decay). – alarge Jun 3 '14 at 15:47