= Solution
For a spherical liquid droplet, the <Kelvin equation> is
$$
\ln\frac{p(r)}{p(\infty)}=\frac{2\gamma V_m}{rRT}.
$$
Here $V_m=0.018/957=1.881\times10^{-5}\,\mathrm{m^3\,mol^{-1}}$, so
$$
\boxed{r=\frac{2(0.0585)(1.881\times10^{-5})}{8.314(375)\ln2}=1.02\times10^{-9}\,\mathrm m.}
$$
The formal answer is \b[about $1.0\,\mathrm{nm}$]. At molecular dimensions, bulk <surface tension>, incompressible-liquid and continuum curvature assumptions are only approximate. There is one liquid-vapour interface here, unlike a soap bubble's two interfaces.
Back to article page
