Question

1. At the surface, the average total lung capacity of human lungs is about 6 liters of air (~0.21 ft³). Free divers typically fill their lungs with air and hold their breath while they dive. Assuming that Boyle’s Law accurately reflects changes in the air volume in a free divers’ lungs, calculate the change in volume that would occur in a free diver’s lungs at the following depths below the surface of the ocean: 10m, 20m, 35m, 45m. Finally, the world record for constant weight free diving depth (where divers must hold their breath, swim to depth and swim to the surface without dropping weight) is 111 m. Calculate the volume of air in a free diver’s lungs at that depth (111m). Remember that you must include the weight of the atmosphere in your calculations!

Answer 1

let

Po = P1atm = 1.0134*10^5 pa
Vo = 6 L (or) 0.21 ft^3

we know, density of ocean water, rho = 1030 kg/m^3

at depth h1 = 10 m,
P1 = Po + rho*g*h1

= 1.013*10^5 + 1030*9.8*10

= 2.0224*10^5 pa

use, P1*V1 = Po*Vo

V1 = Vo*(Po/P1)

= 6*(1.0134/2.0224)

= 3.01 L <<<<<<<<---------------Answer

at depth h2 = 20 m,
P2 = Po + rho*g*h2

= 1.013*10^5 + 1030*9.8*20

= 3.0318*10^5 pa

use, P2*V2 = Po*Vo

V2 = Vo*(Po/P2)

= 6*(1.0134/3.0318)

= 2.006 L <<<<<<<<---------------Answer

at depth h3 = 35 m,
P3 = Po + rho*g*h3

= 1.013*10^5 + 1030*9.8*35

= 4.5459*10^5 pa

use, P3*V3 = Po*Vo

V3 = Vo*(Po/P3)

= 6*(1.0134/4.5459)

= 1.338 L <<<<<<<<---------------Answer

at depth h4 = 45 m,
P4 = Po + rho*g*h4

= 1.013*10^5 + 1030*9.8*45

= 5.5553*10^5 pa

use, P4*V4 = Po*Vo

V4 = Vo*(Po/P4)

= 6*(1.0134/5.5553)

= 1.094 L <<<<<<<<---------------Answer

at depth h5 = 111 m,
P5 = Po + rho*g*h5

= 1.013*10^5 + 1030*9.8*111

= 12.21734*10^5 pa

use, P5*V5 = Po*Vo

V5 = Vo*(Po/P5)

= 6*(1.0134/12.21734)

= 0.4977 L <<<<<<<<---------------Answer