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

 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 =

let's calculate

           V₂ = ( ) 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

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

the answer is a)

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