= Solution
{c}
Let $e_v$ be the parcel's <water vapor> <partial pressure>. Fixed <mole fraction> before <condensation> means $e_v/p$ is constant. Its <dew point> $T_D$ is defined by $E(T_D)=e_v$, where $E$ is the <saturation pressure>. The <Clausius-Clapeyron relation> gives
$$
\frac{L_v}{R_vT_D^2}\frac{dT_D}{dz}=\frac{d\ln e_v}{dz}=\frac{d\ln p}{dz}
\simeq-\frac g{R_d T}.
$$
The last approximation uses <hydrostatic balance>, the <ideal gas law>, and an environmental <temperature> close to the parcel's. Rearrangement yields
$$
\frac{d\ln T_D}{dz}=-\frac{gR_v}{L_vR_d}\frac{T_D}{T}.
$$
For a modest initial <dew-point depression>, $T_D/T\simeq1$ in absolute <temperature>, so
$$
\boxed{T_D(z)\simeq T_{D,s}e^{-z/H},\qquad H=\frac{L_vR_d}{gR_v}.}
$$
With $L_v\simeq2.5\times10^6\,\mathrm{J\,kg^{-1}}$, $R_d\simeq287\,\mathrm{J\,kg^{-1}\,K^{-1}}$, and $R_v\simeq461.5\,\mathrm{J\,kg^{-1}\,K^{-1}}$, $H\simeq1.58\times10^5\,\mathrm m$. This <dew point> scale is much longer than the atmospheric <pressure scale height>.
The <exponential function> is an approximation, not the exact consequence of a dry <adiabatic process>. Retaining $T=T_s-\Gamma_d z$ gives $p/p_s=(T/T_s)^{c_p/R_d}$ and the useful comparison
$$
\frac1{T_D(z)}=\frac1{T_{D,s}}-\frac{R_v}{L_v}\frac{c_p}{R_d}\ln\!\frac{T(z)}{T_s}.
$$
It also shows explicitly that expansion lowers the <dew point> even before any <water vapor> condenses.
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