B2 - variation of temperature with height

considering an air parcel at height $z$ moving up by $d z$
considering an adiabatic ( $δ Q = 0$ ) and reversible process, ie. isotropic
the the first law of thermodynamics:

d U = δ W + δ Q

d U = - p d V

d (p V) = p d V + d p V = n R d T

d U = n C_{V, m} d T

where, $C_{V, m}$ is the molar heat constant at constant volume

also, $C_{p, m} = C_{V, m} + R$ , which is the molar heat constant at constant pressure
the specific heat capacity at constant volume, $C_{V} = C_{V, m} / m_{r}$

\begin{matrix} d U = - n R d T + V d p \\ n C_{V, m} d T = n R d T + \frac{n R T}{p} d p \\ (C_{V, m} + R) \frac{d T}{d z} = \frac{R T}{p} \frac{d p}{d z} \\ \frac{d T}{d z} = - \frac{R T ρ g}{p C_{p, m}} = - \frac{R T ρ g}{ρ R^{'} T C_{p, m}} \\ = - \frac{R}{R^{'}} \frac{g}{C_{p, m}} = - \frac{m_{r} g}{C_{p, m}} \\ ∴ \frac{d T}{d z} = - \frac{g}{C_{p}} \end{matrix}

Γ = - \frac{d T}{d z} = \frac{g}{C_{p}}

for dry air: $Γ_{a i r} \approx 10$ K km $^{- 1}$ , which is called Dry Adiabatic Lapse Rate (DALR)
DALR is the rate of decrease of temperature with height for a dry air parcel that does not exchange hear with its surroundings nor gains heat from condensation of water
for moist air: $C_{p}$ for $H_{2} 0 = 1900$ J kg $^{- 1}$ K $^{- 1}$
in a damp/moist atmosphere, fraction of $H_{2} O_{(g)}$ by volume $χ_{m} < 4 %$ , and average is $1 %$ , by mass $χ \approx 0.6 %$ , so $C_{p}$ for moist air $\approx C_{p}$ for dry air, and $Γ_{m o i s t} \approx Γ_{d r y} \approx 10$ K km $^{- 1}$
for saturated air: liquid water condensed, releasing heat, $= L d x$ , where $L$ is the latent heat of vaporisation, and $d x$ includes other factors such as water content, temperature, pressure, etc
note that the heat released is still internal to the parcel, hence it is still adiabatic
similarly to above:

\begin{matrix} c_{p, m} \frac{d T}{d z} = \frac{R T}{p} d p + L d x \\ \frac{d T}{d z} = - \frac{g}{C_{p}} + \frac{L}{C_{p}} \frac{d x}{d z} \\ - {(\frac{d T}{d z})}_{s a t u r a t e d} = - {(\frac{d T}{d z})}_{d r y} - \frac{L}{C_{p}} \frac{d x}{d z} \\ ∴ Γ_{s a t u r a t e d} = Γ_{d r y} - 4 {K km}^{- 1} = 6 {K km}^{- 1} \end{matrix}

this is the SALR
consider a $3000$ m tall mountain, and a cloud spanning from $1000$ m to $3000$ m on one side
the temperature at ground level is $20 °$ C
consider a moist air parcel at ground level that is rising up the cloudy side due to wind
at $1000$ m, the parcel's temperature would drop to $20 - 10 = 10 °$ C (use $Γ_{m o i s t}$ )
after that, as the cloud begins (saturated air), the temperature at the mountain top would be $10 - 6 \times 2 = - 2 °$ C
at $3000$ m, the air parcel will be dry as it will have deposited all of its water
now, on the other side (no clouds), as the parcel descends, it will gain $3 \times 10 = 30 °$ C, reaching $28 °$ C at ground level
the two sides are called windward and leeward sides