 Solution

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,ConsequentlyThe buoyancy minus bubble weight is $4\pi ({\rho }_w-{\rho }_a)g{a}^3/3$. Balancing it against the clean-bubble Stokes drag gives the instantaneous terminal velocitySet $z=0$ at the initial position. Combining this with the dissolution equation gives $da/dz=-D_a/(K{a}^3)$ and hence the dissolution-limited bubble rise lawThe expression applies until the bracket vanishes. The rise speed decreases continuously as the bubble shrinks. Equivalently, $z(t)=K(a_0^{2}t-D_a{t}^2)$, 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 $U_{\ast }$, the distinct result would be $a^2(z)=a_0^2-2D_{a}z/U_{\ast }$ and $z_{\mathrm{diss}}=U_{\ast }{a}_0^2/(2D_a)$. 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 $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ givesEquality means dissolution exactly on arrival at the surface; dissolution strictly before arrival needs a greater initial depth. This is a formal low-Reynolds number estimate. The same values predict $U(a_0)\simeq 0.033\mathrm{m}{\mathrm{s}}^{-1}$ and a radius-based Reynolds number ${\rho }_{w}U(a_0)a_0/{\eta }_w\simeq 3.3$; the diameter-based value is about $6.5$. Neither is $\ll 1$. 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.