= Solution
Choose $x$ vertically upward. Let $G(x,t)$ be the gas-occupied part of a pipe cross-section, of area $\alpha A$. Define the phase mean <mass density>, <mass>-weighted <velocity>, second moment of <velocity>, and phase mean <pressure> by
$$
\bar\rho_g=\frac{1}{\alpha A}\int_G\rho_g\,dA,\quad
\bar v=\frac{\int_G\rho_g v_x\,dA}{\alpha A\bar\rho_g},\quad
\overline{v^2}=\frac{\int_G\rho_g v_x^2\,dA}{\alpha A\bar\rho_g},\quad
\bar p_g=\frac{1}{\alpha A}\int_Gp_g\,dA.
$$
The <void fraction> $\alpha$ is a volume fraction, not a <mass> fraction. The gas <momentum> in a slice of length $dx$ is $A\alpha\bar\rho_g\bar v\,dx$; its advected <momentum flux> is $A\alpha\bar\rho_g\overline{v^2}$. The two end <pressures> give the net force $-A\partial_x(\alpha\bar p_g)\,dx$. Include every other vertical force on the gas, including interfacial forces, in $M_g\,dx$. <Newton's second law> for this fixed slice, followed by division by $A\,dx$, therefore gives
$$
\boxed{\partial_t(\alpha\bar\rho_g\bar v)+\partial_x(\alpha\bar\rho_g\overline{v^2})=M_g/A-\partial_x(\alpha\bar p_g).}
$$
This is a <cross-section averaged phase momentum balance>. In particular, <mass> weighting makes the <momentum> and its flux exact averages; one must not replace $\overline{v^2}$ by $\bar v^2$ without an additional closure. With liquid mean <velocity> $\bar u$, constant liquid <mass density> $\rho_\ell$, and liquid second moment $\overline{u^2}$, the same calculation gives
$$
\boxed{\partial_t[(1-\alpha)\rho_\ell\bar u]+\partial_x[(1-\alpha)\rho_\ell\overline{u^2}]=M_\ell/A-\partial_x[(1-\alpha)\bar p_\ell].}
$$
With the common <pressure> $p=\bar p_g=\bar p_\ell$, the specified <pressure> contributions to $M_g,M_\ell$ cancel the phase-fraction <derivatives> from the <product rule>. The remaining <pressure> forces per unit mixture volume are $-\alpha p_x$ and $-(1-\alpha)p_x$.
Assume that the bubbles at a given height have the stated radius $a$. Their number per unit mixture volume is $N=3\alpha/(4\pi a^3)$. The <drag force> on a rising bubble opposes its relative <velocity>, so multiplication of the single-bubble law by $N$ gives the signed magnitude
$$
D_v=\frac{3C_D\rho_\ell\alpha}{4a}\bar v|\bar v|.
$$
<Gravity> acts downward; the <drag force> on the liquid is the opposite of that on the gas. With no wall force,
$$
\boxed{M'_g/A=-\alpha\bar\rho_g g-D_v,\qquad M'_\ell/A=-(1-\alpha)\rho_\ell g+D_v.}
$$
A bubble retains its <mass> in the absence of transfer between phases. Consequently $a^3\bar\rho_g=a_0^3\rho_0$ and
$$
\boxed{a=a_0(\rho_0/\bar\rho_g)^{1/3}.}
$$
For clarity suppress bars on $p,\rho_g,v$ in the steady calculation. On a macroscopic variation length $\ell$, gas <inertia> per unit volume scales like $\alpha\rho_gv^2/\ell$, whereas the quadratic <drag force> scales like $\alpha\rho_\ell v^2/a$. Their ratio is $O[(\rho_g/\rho_\ell)(a/\ell)]$. Thus $\rho_g=O(\rho_0)\ll\rho_\ell$, together with the scale separation $a\ll\ell$ inherent in this averaging model, justifies the leading-order neglect of gas <inertia>. Small reference density alone does not justify it in an arbitrarily short or strongly compressed region. Neglect gas weight at the same order, with the <liquid fraction> not vanishing. For a stationary liquid, summing the two <momentum> equations gives
$$
p_x=-g[\alpha\rho_g+(1-\alpha)\rho_\ell]\sim-\rho_\ell g(1-\alpha).
$$
The leading gas balance is $0=-\alpha p_x-D_v$. Substitution of the bubble-radius law yields
$$
\boxed{p_x=-\rho_\ell g(1-\alpha)=-\frac{3C_D\rho_\ell}{4a_0}(\rho_g/\rho_0)^{1/3}v|v|.}
$$
The gas <continuity equation>, with constant pipe area and no <mass> exchange, is $\partial_t(\alpha\rho_g)+\partial_x(\alpha\rho_gv)=0$. Hence the steady gas <mass flux> $J=\alpha\rho_gv$ is constant. For $0<\alpha<1$ the <pressure> equation selects $v>0$, and
$$
v^2=\frac{4a_0g}{3C_D}(1-\alpha)(\rho_0/\rho_g)^{1/3},\qquad
J^2=\frac{4a_0g}{3C_D}\rho_0^{1/3}\alpha^2(1-\alpha)\rho_g^{5/3}.
$$
Evaluate the latter constant at a reference point $\rho_g=\rho_0$, $\alpha=\alpha_0$ and divide. The <terminal-drift density law for compressible bubbles> is
$$
\boxed{\frac{\rho_g}{\rho_0}=\left[\frac{\alpha_0^2(1-\alpha_0)}{\alpha^2(1-\alpha)}\right]^{3/5}.}
$$
The exponent $3/5$ comes from combining <mass> conservation with the density dependence of bubble radius; no gas equation of state was needed.
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