Atmospheric pressures

The basic barometer can be used to measure air pressure changes with elevation, such as from the base of a mountain to the peak $($as air pressure is variable with weather, to do this properly, a team would have to take simultaneous measurements at the base and top of the mountain$).$ The change in air pressure can then be used to roughly estimate the height of the mountain. Consider Mount Kilimanjaro, in Tanzania, where the air pressure ranges from $89\mathrm{.}9$kPa near the base to $47\mathrm{.}2$kPa at the peak of the mountain. Please note that surveying and GPS are both practically superior methods here $($all of the following estimates are somewhat wrong$).$

$($a$)$ Assume an average density for air of $1\mathrm{.}2k\frac{g}{m^3}.$ How tall is the mountain by this estimate?

$5200m$

$5380m$

$($b$)$ Air is less dense at elevation, so instead calculate the air density at the average temperature and average pressure. How tall is the mountain by this estimate?

$($c$)$ The air temperature averages $12\mathrm{.}3C,$ but in reality the density of air varies with height. Estimate the height of the mountain based on integrating the ideal gas law $($see below$).$ Use the molecular weight for dry air. Note that $z$ is depth so that, for example, an elevation of $2km$ above sea level $(H=2000m)$ would be represented by $z=-2000m.$

$\{$:$[=\frac{MP}{RT}],[\frac{dP}{dz}=g]\}\frac{dP}{dz}=\frac{Mg}{RT}P=>{\displaystyle \underset{P(z=0)}{\overset{P(z=-H)}{}}}\frac{dP}{P}=\frac{Mg}{RT}{\displaystyle \underset{0}{\overset{-H}{}}}dz=-\frac{MgH}{RT}$

$5000m$

$($d$)$ Now assume that the mountain exists in the U$.$S$.$ Standard Atmosphere tabulated in the CRC Handbook of Chemistry and Physics $($see attached$).$ Estimate the elevation of the base and the elevation of the peak $-$ both above sea level $-$ by matching the given pressures to the table. Estimate the height of the mountain.