= Solution
Assume a well-mixed ideal gas with constant $g,c_p$, no latent heat, and sufficiently strong surface heating to maintain dry <convection> near the surface. An adiabatically displaced unit mass has specific volume $V=1/\rho$, so the stated relation and <hydrostatic equilibrium> give
$$
c_p\,dT=V\,dp=-g\,dz,
\qquad \boxed{\Gamma_d=-\frac{dT}{dz}=\frac g{c_p}\simeq9.8\,\mathrm{K\,km^{-1}}.}
$$
This is the dry neutral lapse rate; infrared absorption alone does not guarantee that a radiative profile is steep enough to trigger convection. Combining the same equations with the <ideal gas law> yields $T\propto p^{R_a/c_p}$ along that adiabat.
High above the main infrared-emitting region, optical depth to space becomes small. A transparent-to-sunlight atmosphere has no substantial direct solar heating there; a thin grey layer absorbs the upwelling planetary infrared flux $Q=(1-\alpha)S_0/4$ and emits from both faces. In the optically thin radiative-equilibrium limit,
$$
\sigma T_{\rm skin}^4\simeq Q/2,
\qquad \boxed{T_{\rm skin}\simeq\left[\frac{(1-\alpha)S_0}{8\sigma}\right]^{1/4}.}
$$
The factor four in $Q$ is projected-area versus whole-sphere averaging; the additional factor two counts emission upward and downward by the tenuous layer. For Earth-like $S_0\simeq1360\,\mathrm{W\,m^{-2}}$ and $\alpha\simeq0.30$, this is about $214\,\mathrm K$. Thus temperature approaches a height-independent skin value as $\tau\to0$, rather than continuing the dry adiabat to zero. This is an asymptotic grey-atmosphere approximation, not a proof that an entire finite stratosphere is exactly isothermal.
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