= Solution
{c}
Retain the clean-bubble approximation $\eta_a\ll\eta_w$ used in parts (b) and (c), with constant <mass densities> and <diffusion coefficient>. For a spherical bubble,
$$
M=\frac{4\pi}{3}\rho_a a^3,\qquad
4\pi\rho_a a^2\dot a=-4\pi aD_a\rho_a.
$$
Consequently
$$
\boxed{\dot a=-\frac{D_a}{a},\qquad a^2(t)=a_0^2-2D_at,\qquad
t_{\mathrm{diss}}=\frac{a_0^2}{2D_a}.}
$$
The <buoyancy> minus bubble weight is $4\pi(\rho_w-\rho_a)ga^3/3$. Balancing it against the <clean-bubble Stokes drag> gives the instantaneous <terminal velocity>
$$
\dot z=U(a)=\frac{(\rho_w-\rho_a)g}{3\eta_w}a^2\equiv Ka^2.
$$
Set $z=0$ at the initial position. Combining this with the dissolution equation gives $da/dz=-D_a/(Ka^3)$ and hence the <dissolution-limited bubble rise> law
$$
\boxed{a(z)=\left[a_0^4-\frac{12\eta_wD_a}{(\rho_w-\rho_a)g}z\right]^{1/4},\qquad
z_{\mathrm{diss}}=\frac{(\rho_w-\rho_a)ga_0^4}{12\eta_wD_a}.}
$$
The expression applies until the bracket vanishes. The rise speed decreases continuously as the bubble shrinks. Equivalently, $z(t)=K(a_0^2t-D_at^2)$, providing a direct check on the elimination of time.
The phrase “constant terminal velocity” must here mean rapid adjustment to a force balance at each radius. A literally time-independent speed is incompatible with that balance because $U\propto a^2$. If instead an external mechanism maintained a fixed $U_*$, the distinct result would be $a^2(z)=a_0^2-2D_az/U_*$ and $z_{\mathrm{diss}}=U_*a_0^2/(2D_a)$. In particular, holding the initial terminal speed fixed doubles the predicted rise distance; it is not the freely rising shrinking-bubble model.
Using the supplied values and $g=9.81\,\mathrm{m\,s^{-2}}$ gives
$$
t_{\mathrm{diss}}\simeq50\,\mathrm s,\qquad
\boxed{d\gtrsim z_{\mathrm{diss}}\simeq0.82\,\mathrm m.}
$$
Equality means dissolution exactly on arrival at the surface; dissolution strictly before arrival needs a greater initial depth. This is a \b[formal low-<Reynolds number> estimate]. The same values predict $U(a_0)\simeq0.033\,\mathrm{m\,s^{-1}}$ and a radius-based <Reynolds number> $\rho_wU(a_0)a_0/\eta_w\simeq3.3$; the diameter-based value is about $6.5$. Neither is $\ll1$. Thus the numerical input does not consistently satisfy the <Stokes flow> assumption initially, and the $0.82\,\mathrm m$ result should not be presented as an accurate oceanic prediction. Within the stated ideal model, bubble compression, contaminants and changes in dissolved-gas concentration are also excluded.
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