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.

(Hint: Remember that both pressure and temperature change. Also, don't forget that the diver is diving in SEA water.)

(Hint: The answer is not 96.22728847 cc, or 96.227 cc or 90.64 cm³)

Answer 1

here from the given data,

T1 =5 0C = 5+273 = 278 K

T2 = 23 0C =23+273 = 296 K

V1 =19 cm^3

h is height = 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 of ideal gas pV = nRT (

V2 = P1V1T2/P2T1

V2 = 5.13e 5*19 * 296)/(1.01e 5* 278)

V2 = 102.75 cm^3