Question

What pressure, in atmospheres, is exerted on the body of a diver if he is 38 ft below the surface of the water when atmospheric pressure at the surface is 750 mmHg ? Assume that the density of the water is 1.00g/cm3=1.00×103kg/m3. The gravitational constant is 9.81m/s2, and 1Pa=1kg/m−s2.
Express your answer using two significant figures.

Answer 1

The pressure of diver = atmospheric pressure + water pressure

atmospheric pressure = 750 mmHg (as given) = 750 / 760 atm = 0.987 atm

Water pressure is

P = hρg

where

h = height of water = 38 ft

1 ft = 0.3048

38 ft = 11.58 m

ρ = density  = 1000 Kg / m³

g = gravitational constant  = 9.81 m/s2

P = 11.58 X 1000 X 9.81 = 113599.8 Kg / m s^2 Or N /m^2

1 N / m^2 = 1 pa = 9.869 X 10^-6 atm

P = 113599.8 Pa = 1.12 atm

Total pressure = 1.12 + 0.987 atm = 2.107 atm = 2.1 atm (two significant figures)