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Using an isothermal atmosphere with the stated scale height gives $p(30\mathrm{km})=10^5{e}^{-30/7}\mathrm{Pa}$ and $[\mathrm{M}]=p/(k_{B}T)=3.46\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}$ at $288\mathrm{K}$. The water atmospheric mixing ratio then gives $[{\mathrm{H}}_2\mathrm{O}]=6.92\times 10^{11}\mathrm{c}{\mathrm{m}}^{-3}$. Applying steady state to the short-lived electronically excited oxygen atom,Its HOx family source is $S=2k_3[\mathrm{O}({}^{1}D)][{\mathrm{H}}_2\mathrm{O}]\simeq 6.00\times 10^3\mathrm{c}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$. Balancing this with termination gives $2k_4[\mathrm{OH}][\mathrm{H}{\mathrm{O}}_2]=S$. Combining with the rapid partition, or solving both stationary radical equations including termination, givesConsequently the main catalytic odd-oxygen loss isThe Chapman cycle loss is $L_{\mathrm{Ch}}=2k_9[\mathrm{O}][{\mathrm{O}}_3]=5.40\times 10^5\mathrm{c}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$, so the given HO${}_x$ cycle is about 2.35 times larger. Both rates count two odd-oxygen units per completed O-plus-O${}_3$ destruction event. The omitted $v_4\simeq 3.00\times 10^3$ correction is only about $0.24\%$ of the main HO${}_x$ loss. The family abundance divided by its source is about $5.87\times 10^4\mathrm{s}$, explaining why its absolute stationary state is a stronger assumption than rapid OH/HO${}_2$ interconversion.

For the stated total number density, multiply the number atmospheric mixing ratio by $[\mathrm{M}]$. With NO${}_x$ defined as NO plus NO${}_2$,The water partial pressure is $30/1000=0.030$ of the total pressure. It is a 3% number fraction, not 30 ppm.

Before condensation, assume a closed parcel with fixed atmospheric mixing ratio, constant specific heat capacities, and negligible irreversible heating. The first law of thermodynamics for a dry adiabatic process isThe parcel remains in approximate pressure equilibrium with a hydrostatic environment. With the standard approximation that parcel and environmental densities are close, $dp/dz\simeq -\rho g$, henceThis is the dry adiabatic lapse rate, about $9.8\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ for air. More precisely the environmental hydrostatic balance gives $dT/dz=-g{\rho }_{\mathrm{env}}/(c_p{\rho }_{\mathrm{parcel}})$; the familiar constant dry adiabatic lapse rate uses their ratio close to one.

The dew-point depression is $6\mathrm{K}$, soKeeping $T_{D,s}=276.15\mathrm{K}$, $H\simeq 158.5\mathrm{km}$ and $c_p=1004\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ gives approximately $747\mathrm{m}$, consistent with the rounded estimate.

The real cloud base can differ because entrainment changes the parcel's atmospheric mixing ratio and temperature, surface humidity varies, and the environmental pressure profile need not match the assumed parcel profile. Finite cloud-droplet activation requires some atmospheric supersaturation, so the first visible cloud droplets need not occur exactly at the thermodynamic lifting condensation level. A parcel must also actually be lifted that far; environmental static stability can prevent this. The constant-latent heat and first-order Taylor expansion approximations produce smaller numerical differences.