A diver named Jacques observes a bubble of air rising from the bottom of a lake

Question

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 degree C, and the temperature at the surface is 23.0 degree C. What is the ratio of the volume of the bubble as it reaches the surface (V_s) to its volume at the bottom (V_b)? If Jaques were to hold his breathe the air in his lungs would be kept at a constant temperature. Would it be safe for Jacques to hold his breathe while ascending from the bottom of the lake to the surface?

Explanation / Answer

given that

Ps = 1 atm

Pb = 3.50 atm

Ts = 23 degree C = 23+273 = 296 K

Tb = 4 deg C = 4+273 = 277 K

air in bubble will act as an ideal gas

so, from the ideal gas equation

P*V = n*R*T

P*V/T = n*R          (n*R is a constant value)

so , Ps*Vs / Ts = Pb*Vb / Tb

Vs / Vb = Pb*Ts / Ps*Tb

Vs / Vb = 3.50*296 / 1*277

Vs / Vb = 3.74

(b)

volume at surface is more than volume at bottom

if jacques holds his breath while ascending , the volume of air in his lungs will expand and lungs would have to stretch outside to balance the increased volume which would be unsafe for him.

so the answer is no.

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