A scuba diver at a depth of 41.0 m below the surface of the sea off the shores o

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)

Explanation / Answer

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

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