Calculating Pressure and Volume Increase on Rising from Depth d

[dimpledur]

Homework Statement

A diver rises quickly to the surface from a depth d. If she did not exhale the gas from her lungs before rising, by what factor would her lungs expand? Assume the temperature to be constant and the pressure in the lungs to match the pressure outside the diver’s body. The density of seawater is ρ. Express your answer in terms of d, ρ, g, and Patm (atmospheric pressure).

Homework Equations

P1V1=P2V2

P2=P1 +pgd

The Attempt at a Solution

P2 (Patm) is a distance d below P1. SO,

P1V1=P2V2
Patm = P1 +pgd
P1 = Patm - pgd
(Patm - pgd)V1=PatmV2
V1 = V2Patm / (1-pgd)

is that right?

 

[dimpledur]

No one can tell?

 

[nlightner]

Yes, that is correct. To further explain, the first equation you listed is Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is constant. The second equation is the hydrostatic equation, which relates pressure to depth in a fluid. In this case, the pressure inside the lungs (P1) is equal to the atmospheric pressure (Patm) minus the pressure due to the weight of the water (pgd) at a depth d. The last equation you wrote is just rearranging the first equation to solve for V1. So, when the diver rises to the surface, the pressure outside and inside her lungs will be equal (P2 = Patm), and using the last equation, we can see that her lung volume will increase by a factor of (1-pgd)/Patm. This means that the lung volume will increase by a larger factor for deeper depths (larger d) and for denser seawater (larger ρ), but will decrease if the atmospheric pressure (Patm) increases.