A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the surface (where the pressure is 1.00 atm). The temperature at the bottom is 4.00 ∘C, and the temperature at the surface is 23.0 ∘C. What is the ratio of the volume of the bubble as it reaches the surface (Vs) to its volume at the bottom (Vb) ? If Jaques were to hold his breath the air in his lungs would be kept at a constant temperature. Would it be safe for Jacques to hold his breath while ascending from the bottom of the lake to the surface?

A) the ratio of volumes of the bubble is Vs/Vb = 3.74

B) would not be safe since Vs/Vb= = 3.5 and there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

Explanation:

assuming the gas of the bubble behaves as ideal gas

P * V = n * R * T

where P = absolute pressure, V = volume occupied by the gas, n = number of moles, R = ideal gas constant, T = absolute temperature

if we assume that the mass of the bubble remains constant (that is, it does not capture other bubbles during ascension of disaggregate into smaller ones and there is no mass transfer into the bubble due to diffusion)

inicial state) Pb * Vb = n * R * Tb

final state) Ps * Vs = n * R * Ts

dividing both equations

(Ps/Pb) (Vs/Vb) = Ts/Tb

therefore

Vs/Vb = (Ts/Tb) (Pb/Ps)

since Tb = 4°C = 277 K and Ts = 23°C = 296 K

Vs/Vb = (Ts/Tb) (Pb/Ps) = (296K/277K) * (3.5 atm/1 atm) = 3.74

B) if the T remains constant Ts=Tb and thus

Vs/Vb = (Ts/Tb) (Pb/Ps) = 1 * (Pb/Ps) = 3.5 atm/1 atm = 3.5

it would not be safe since there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.