Question

the ocean floor is, on average, 4267 m below sea level. What is the pressure in the atmosphere at this depth?

Answer 1

The pressure at the depth h in the ocean is given by (Stevin's law)

`p= p_0 + \rho g h`
where

`p_0 = 1.0 \cdot 10^5 Pa` is the atmospheric pressure
and
`\rho g h` is the pressure exerted by the column of water of height h=4267 m, with
`\rho = 1000 kg/m^3` being the water density and
`g=9.81 m/s^2`.
Substituting, we find

`p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa`
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so
`1 atm=1.0\cdot 10^5 Pa`, therefore we just need to divide by this number:

`p= (4.20 \cdot 10^7 Pa)/(1.0 \cdot 10^5 Pa/atm) =420 atm`