# Equilibrium condition for Kelvin's Vapor Pressure-curvature equation

This is regarding the Correction due to 'Curvature of the drop' made by Lord Kelvin to the Clausius-Clapeyron Equation.

So, the idea is that, during large scale nucleation of water vapour into cloud droplets, the equilibrium vapour pressure over a droplet of radius r, is given by $$e_s^{\ curved}(r) = e_s(\infty)*e^{\frac{2\sigma}{\rho_w R_vTr}}$$ where, T = temperature, $$R_v$$ is gas constant for water vapour, $$\rho_w$$ is density of drop, $$\sigma$$ is Surface tension, and $$e_s(\infty)$$ is the equilibrium vapour pressure over a plain surface at temp T.

In order to derive this equation, we used the Gibbs Free Energy, and calculated that for this drop formation process, $$\Delta G = - R_v T ln(\frac{e}{e_s}) \rho_w \frac{4}{3} \pi r^3 + \sigma 4 \pi r^2$$

Now, when $$e>e_s$$, then the graph of $$\Delta G$$ vs r, initially increases up to a critical radius R and then decreases as shown:

$\Delta G$ vs r, S=e/e_s" />

Then, to find the equilibrium condition, we said that The drop is in equilibrium with the surrounding when the radius = R, since the slope of $$\Delta G$$ wrt r is 0 at r = R

However, I feel that the equilibrium condition arises for the radius r, for which, $$\Delta G$$ = 0. This interpretation does affect the final formula much. It just replaces the "2" with a "3".

Q: Which one of these interpretations is correct and why? Why is the other interpretation incorrect?

Kelvin's Equation

A sample Derivation - A similar derivation is seen in the book "Atmospheric Science - by Wallace and Hobbs"