= Solution
A thin soap bubble has two interfaces, so the <Young–Laplace equation> gives $\Delta p=4\gamma/R=8\gamma/d$. Hence
$$
\boxed{d=\frac{8\gamma}{\Delta p}=\frac{8(0.025)}{10}=0.020\,\mathrm m=2.0\,\mathrm{cm}.}
$$
At equal <surface tension>, the smaller bubble has the higher internal <pressure>. Air diffuses through their separating film from the smaller to the larger bubble, so the smaller shrinks and the larger grows. The pressure difference increases as the smaller shrinks. This is gas transfer driven by curvature, not necessarily immediate merging of the interfaces; coalescence additionally requires film rupture.
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