If a scuba diver fills his lungs to full capacity of 5.4 LL when 8.0 mm below the surface, to what volume would his lungs expand if he quickly rose to the surface? Assume he dives in the sea, thus the water is salt.

9.62 L

Explanation:

Considering the temperature to be constant, then:

PV = constant.

At 8.0 mm below the surface, pressure would be:

P₁ = ρ g h + P₀

where, ρ is the density, g is the gravitational acceleration, h is the depth and P₀ is the atmospheric pressure.

ρ = 1020 kg/m³ (density of sea water)

P₁ = 1020 kg/m³ * 9.8 m/s² * 8.0 * 10⁻³ m + 1.01 * 10⁵ Pa

P₁=79.968 + 1.01 * 10⁵ Pa = 1.8*10⁵ Pa

P₁V₁ = P₂V₂

At the surface, P₂ = P₀

V₂ = (1.8*10⁵ Pa) (5.4L) / (1.01 * 10⁵ Pa)

V₂ = 9.62 L