Answer:
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.
