Question

Liquid water is nearly 1,000 times denser than air. Thus, for every 32.0 feet (9.75 m) a scuba diver descends below the water's surface, the pressure increases by 1.00 atm. Human lungs have a volume of approximately 3.50 L. If a scuba diver descends to a depth of 80.0 feet where the pressure is 3.50 atm (2.50 atm from the water and 1.00 atm from the air pressure), then by how much does the volume of a 3.50 L surface sample of air decrease

Answer 1

Answer:

 ΔV = -2.1 L

Explanation:

To solve this exercise we can use the ideal gas equation for two points

         PV = nRT

          P₁V₁ = P₂ V₂

where point 1 is on the surface and point 2 is at the desired depth,

          V₂ = $\frac{P_1}{P_2} \ V_1$

let's calculate

           V₂ = ( $\frac{1 atm}{2.5 atm}$ ) 3.5 L

            V₂ = 1.4 L

this is the new volume, the change in volume is

          ΔV = V₂ -V₁

          ΔV = 1.4-3.5

          ΔV = -2.1 L