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:
$p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm$