Question

A scuba diver at a depth of 41.0 m below the surface of the sea off the shores of Panama City, where the temperature is 5.0⁰C, releases an air bubble with volume 19.0 cm³. The bubble rises to the surface where the temperature is 23.0⁰C. What is the volume of the bubble immediately before it breaks the surface? The specific gravity for seawater is 1.025.

Answer 1

T1 =5 ⁰C = 5+273 = 278 K

T2 = 23 ⁰C =23+273 = 296 K

V1 =19 cm^3

h =41 m , g =9.8 m/s^2

Specific gravity for sea water is =1.025

density of water =1000 kg/m^3

Then density of sea water is d=1.025x1000 kg/m^3

d =1025 kg/m^3

Pressure at surface P2 =1 atm = 1.01x10^5 Pa

Pressure at bottom P1 =P2+dgh = (1.01x10⁵) +(1025*9.8*41) =5.13 x10⁵ Pa

From the relation between P,V and T

(P1V1)/T1 =(P2V2)/T2

V2 = P1V1T2/P2T1 = [5.13x10⁵ x 19cm³ x 296]/[ 1.01 x10⁵ x278]

V2 = 102.75cm³