Problem 7.14 Enhanced - with Feedback Part A For every 10 m below the water surf
ID: 554263 • Letter: P
Question
Problem 7.14 Enhanced - with Feedback Part A For every 10 m below the water surface, the pressure exerted by the water adds 14.7 psi to atmospheric pressure, so 10 m below the surface would have a total pressure of 29.4 psi. If a swimmer has a lung volume of 6 L at sea level, what would the volume of her lungs be when she is at the bottom of a pool that is 5.0 m deep? Assume that the temperature and amount of the air in the lungs remain unchanged. Express your answer to two significant figures and include the appropriate units. You may want to reference ( pages 272-279) Section 7.2 while completing this problem. lue Units Submit My Answers Give Up Incorrect; Try Again; 2 attempts remaining Boyle's law states that pressure and volume are inversely proportional, so doubling the pressure would halve the volume rather than double it. Make certain to calculate the volume of the air at a depth of 5.0 m.Use Boyle's law for a gas sample at constant temperature: where P, and Vi are the initial pressure and volume, respectively, and Pr and Vr are the final pressure and volume of the gas sample, respectively, to calculate the final volume of the air in the swimmer's lungs. The key is to derive a relationship between the depth of the swimmer and the increase in pressure. Since the pressure doubles from the surface pressure of 14.7 psi, or 1 atm, to 29.4 psi, or 2 atm, at a depth of 10 m, the pressure can be calculated at any depth below the surface, assuming a linear dependence between the depth and the increase in pressure at 10 m. An increase in 1 atm per 10 m of depth (1 atm/i a conversion factor to calculate the increase in pressure at any depth below the surface. ContinueExplanation / Answer
From ideal gas equation equation
PV = nRT
At two locations
P1V1 / (n1T1) = P2V2 / (n2T2)
a linear relationship between depth and pressure..
Pressure at 5 m depth =
5m x (14.7 psi / 10m) = 7.35 psi
we know.
P1 = 14.7 psi
V1 = 6L
P2 = 7.35 psi + 14.7 psi = 22.05 psi
V2 = ?.
For constant temperature and moles
P1V1 / (n2T2) = P2V2 / (n2T2)
which becomes
P1V1 = P2V2
V2 = V1 x (P1 / P2)
V2 = 6L x (14.7 psi / 22.05 psi) = 4L