= Solution
Take $z=0$ at the base and assume gas leaves freely at the liquid surface, without being retained as foam. There is no new bubble production after filling. Neglecting the liquid velocity, gas-volume conservation is a <scalar conservation law>
$$
\phi_t+\partial_zj(\phi)=0,\qquad j(\phi)=V_s\phi(1-\phi).
$$
An impermeable base has no incoming bubbles. Thus a clear region develops below the uniform bubbly region. The clearing discontinuity satisfies the <Rankine-Hugoniot condition>, with states $0$ and $\phi_0$:
$$
s=\frac{j(\phi_0)-j(0)}{\phi_0-0}=V_s(1-\phi_0).
$$
For the dilute regime $0<\phi_0<1/2$, the characteristic speeds obey $j'(0)>s>j'(\phi_0)$, so this is the admissible compressive <shock>. Before it reaches the surface,
$$
\phi(z,t)=\begin{cases}0,&0<z<st,\\ \phi_0,&st<z<H.\end{cases}
$$
The remaining bubbles move upwards at $s$ and escape at the top. Averaging the <void fraction> over the entire cylindrical volume gives
$$
\boxed{\Phi(t)=\phi_0\max\left(1-\frac{V_s(1-\phi_0)t}{H},0\right),\qquad
t_{\rm clear}=\frac{H}{V_s(1-\phi_0)}.}
$$
This <bubble clearing in a cylindrical vessel> describes bubbles in the liquid, not a hypothetical retained gas layer above it. A closed surface or long-lived foam would require a different upper boundary condition.
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