= Saturated vapor column on a dry adiabat
{title2=$W\sim\frac{c_{pA}T_s e_s(T_s)}{gL}$}
Prescribe a <dry adiabat> for a background <ideal gas> A,
$$
T(p)=T_s(p/p_s)^x,\qquad x=R_A/c_{pA}>0,
$$
where $R_A$ is its <specific gas constant>, $c_{pA}$ its constant specific <heat capacity> at fixed <pressure>, and $T_s,p_s$ the surface <temperature> and <pressure>. A dilute saturated vapour B has <specific gas constant> $R_B$, constant <latent heat> $L$, and <saturation vapor pressure> $e_s(T)$. Put $\epsilon=R_A/R_B$ and $\beta=L/(R_BT_s)$.
The ideal-vapour <Clausius-Clapeyron relation> integrates to
$$
e_s(T)=e_s(T_s)\exp\left[-\frac L{R_B}\left(\frac1T-\frac1{T_s}\right)\right].
$$
Its dilute vapour mass fraction is approximately $\epsilon e_s/p$. Under constant-$g$ <hydrostatic balance>, the vapour <mass path> is therefore
$$
W=\frac\epsilon g\int_0^{p_s}\frac{e_s(T(p))}{p}\,dp
=\frac{\epsilon e_s(T_s)e^\beta}{gx}E_1(\beta),\qquad
E_1(\beta)=\int_\beta^\infty\frac{e^{-y}}y\,dy.
$$
For the second equality, set $\zeta=\log(p_s/p)$ and then $y=\beta e^{x\zeta}$. Thus the <exponential integral> follows from the combination of adiabatic cooling and rapidly falling <saturation vapor pressure>.
For $\beta\gg1$, <integration by parts> gives $e^\beta E_1(\beta)\sim\beta^{-1}(1-\beta^{-1}+2\beta^{-2}-\cdots)$, hence
$$
W\sim\frac{c_{pA}T_s e_s(T_s)}{gL}\left[1-\frac1\beta+O(\beta^{-2})\right].
$$
The column is weighted toward its warm lower part. Diluteness must hold throughout the column. This calculation prescribes the <dry adiabat>; if <condensation> heating changes that profile, a consistent <moist adiabat> is needed instead.
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