Question

assuming the density of air 1.295kg m-3 find the fall in barometric height in mm of hg at a height of 107 m above sea level . Take density of Mercury 13.6 ×1000 kg m3

Answer 1

Assuming no change (not really valid but ok for first approximation) in air density with increase in altitude the pressure difference of air can be given by formula which is a gross approximation for small height difference

Den_Hg*g*h_Hg= den_air*g*h_air

h_Hg = 107*1.295/13.6e3 in m

This will be approximately 10mm of Hg

In real life aviators use under constant temperature assumption the barometric formula below

P_h = P_o * exp(-mgh/KT)

where m is molar mass of air ~ 29 amu

And rest are standard thermodynamic constants. The above formula give P_h at 107m as 750.79 mmHg when P_0 is 760mm Hg

Which is closer to 9.2mm Hg