What is the normal pressure of the atmosphere at the summit of Mt. Everest, 8850m above sea level? (Assume g is constant and that the density of air, ρ, is proportional to the pressure, P, at any height. Use the values at sea level ρ0 = 1.29kg/m3 and P0 = 1.013 × 105N/m2.)

Answer 1

Well, to figure the proportional constant, you have to assume how thick is the atmosphere (ie, where ia density zero).

So assume a zero density at h=100km (the edge of space).
so density= m:(100km- h)+ b
and we quickly deduce that b=0 at h=100km; and
at sealevel
1.29=m*100km or m=1.29/1E5
so the density at h=8850
density=1.29E-5*8.850E3 kg/m^3

Use a similar argument to find pressure at 8850.

Answer 2

I just need the answer- what is it? Too many numbers for any normal human being to understand.