Why this matters: A soap bubble has TWO free surfaces (inner and outer). Each surface contributes 2S/R, giving a total $\Delta P$ = 4S/R. A liquid drop has only ONE free surface, so $\Delta P$ = 2S/R. The work done to increase bubble radius from r_{1} to r_{2} is W = 8πS(r_{2}^{2} − r_{1}^{2}) — the factor of 2 accounts for two surfaces.