(\[Mech-Eq\]) we have $$\label{dN/dt-simple}
N = \frac{4 \pi}{3 k T} R^3 \left[ \Pi + \frac{2 \sigma}{R}\right]\; .$$ Differentiating Eq. (\[dN/dt-simple\]), using Eq. (\[n\_g\]), we obtain $$\label{dN/dt}
\frac{dN}{dt} = 4 \pi n_g R^2 \frac{dR}{dt} \left[ 1 + \frac{R_\sigma}{R} \right]\; ,$$ where $$\label{R_s}
R_\sigma \equiv \frac{4}{3} \frac{\sigma}{\Pi}$$ is the characteristic size of the bubble.

Material balance between the dissolved gas and the gas in the growing bubble gives us the following equality $$\label{Balance}
\frac{dN}{dt} = 4 \pi R^2 j_D\; .$$ Substituting expression for diffusion flux density Eq. (\[Flux\]) and the rate of change of the number of molecules Eq. (\[dN/dt\]) into material balance equation Eq. (\[Balance\]) we have $$\label{Dyn-Eq-1}
n_g \frac{dR}{dt} \left[ 1 + \frac{R_\sigma}{R} \right] = D \frac{n_0 - n_R}{R}\; .$$ Or, exploiting Eqs. (\[s\]) and (\[n\_g\]), we equivalently have $$\label{Dyn-Eq-2}
R \dot{R} \left[ 1 + \frac{R_\sigma}{R} \right] = D s \frac{n_0 - n_R}{n_\infty}\; .$$

Using Eqs. (\[zeta\]), (\[Mech-Eq\]) and, correspondingly, (\[Chem-Eq-Henry\]) and (\[Chem-Eq\]), the fraction in the r. h. s. of Eq. (\[Dyn-Eq-2\]) can be expressed as $$\label{Conc-Frac-1-Henry}
\frac{n_0 - n_R}{n_\infty} = \zeta - \frac{2 \sigma}{\Pi R}$$ for Henry’s law, and $$\label{Conc-Frac-1}
\frac{n_0 - n_R}{n_\infty} = \zeta + 1 - \sqrt{1 + \frac{2 \sigma}{\Pi R}}$$ for Sievert’s law. After exploiting 