# Gibbs free energy : is surface tension intensive and the surface area extensive?

As answered here Why Gibbs energy for nucleation theory?, one can write a surface Gibbs energy.

Also, in the context of Laplace Law we have the variation $$dU=(P_{ext}-P_{int})dV+\sigma dA$$ leading to : $$\Delta P=\frac{2\sigma}{R}$$

My question is that the conjugate variables are always one extensive and one intensive. Here $$A$$ is probably the extensive one and $$\sigma$$ the intensive one.

I understand that if you have $$A\sim N(surface)$$. So there is an extensivity on the surface. But is it enough thermodynamically speaking ? For instance, I assume we can't scale the variable like here https://web.archive.org/web/20070707224025/http://www.chem.arizona.edu/~salzmanr/480a/480ants/opensys/opensys.html : $$Gdx=-\mu_V V dx+\sigma A dx$$ since the variables surfacely and volumicaly extensive don't scale the same way.

I'm also wondering how to get the surface tension from energy itself, because the particles interact only with their neighbours and when you have a surface you have less neighbours. I assume that $$\sigma=\mu_V f$$ with $$f$$ a form factor. Could you give me a reference or an way to write this idea more formally please ? This is straightforward for the Ising Model where in 1D for instance the cost of an interface $$++++\mathbf{+-}---$$ is $$2J$$ since the interaction energy between the neigbouring spins is $$J \phi_i \phi_j$$. But how can one get an expression for liquids for instance ?