A bubble forms on the ocean floor far away from the coast. Assume the density of...

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Question

A bubble forms on the ocean floor far away from the coast. Assume the density of the salt water is 1.05 g/cm3, and that the o

Answer 1

Answer #1

(a) pressure P at 0.6 km deep inside sea water that has density 1.05 g/cm³

P = pressure at surface + pressure due to water column = 1.013 $\times$ 10⁵ + ( $\rho$ g h )

P = 1.013 $\times$ 10⁵ + ( 1050 $\times$ 9.8 $\times$ 600 ) = 62.753 $\times$ 10⁵ Pa

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(b) for ideal gas, we have, P V = n R T

where V is volume = (4/3) $\pi$ r³ , where r is radius

n is number of moles, R is gas constant, T is temperature in Kelvin

n = ( P V ) / ( R T ) = ( 62.753 $\times$ 10⁵ $\times$ (4/3) $\pi$ $\times$ (0.75)³ $\times$ 10^-6 ) / (8.31 $\times$ 276.5)

n = 4.826 $\times$ 10^-3 mole

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(c) Radius at surface

we have , P V = n R T

V = ( n R T ) / P = ( 4.826 $\times$ 10^-3 $\times$ 8.31 $\times$ 296.5 ) / ( 1.013 $\times$ 10⁵ ) = 1.1738 $\times$ 10^-4 m³

Radius :- (4/3) $\pi$ r³ = 1.1738 $\times$ 10^-4

radius r = 3.037 cm

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(d) velocity of sound v is given by,

$v = \sqrt{\frac{\gamma R T}{M}}$

where $\gamma$ is adiabatic constant of air, R is gas constant and M is molar mass of air

for air $\gamma$ =1.4 and M = 29 g

velocity of sound = [ ( 1.4 $\times$ 8.31 $\times$ 296.5 ) / ( 29 $\times$ 10^-3 ) ]^1/2 $\approx$ 345 m/s

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(e) diameter of bubble = 2 $\times$ 3.037 = 6.074 cm

wavelength = 2d = 12.148 cm

velocity v = frequency $\times$ wavelength

frequency = velocity / wavelength = 345 / ( 12.148 $\times$ 10^-2 ) $\approx$ 2840 Hz

Answer 2