 Solution

For dry ascent, $c_{p}dT=Vdp$ and hydrostatic pressure change is $dp/dz=-\rho g=-g/V$, so $\boxed{{\Gamma }_d=-dT/dz=g/c_p}$. During condensation $dq<0$, and $-Ldq>0$ offsets expansion cooling, making the saturated ascent less steep.

For a saturated parcel with $q=q_s(T,p)$, use $dq=q_{T}dT+q_{p}dp$ to obtainWith dilute vapor, $q_p\simeq -q_s/p$ and $q_T\simeq L{q}_s/(R_v{T}^2)$, this becomes ${\Gamma }_d[1+L{q}_s/(R_{a}T)]/[1+L^2{q}_s/(c_p{R}_v{T}^2)]$. In terrestrial conditions the denominator's latent-heating correction dominates. Warming increases saturation humidity strongly, so latent heat increasingly compensates expansion cooling and the moist lapse rate falls. This is a statement about saturated convective adjustment, not every instantaneous environmental temperature profile.