Answer:
V2 = 21.44cm^3
Explanation:
Given that: the initial volume of the bubble = 1.3 cm^3
Depth = h = 160m
Where P2 is the atmospheric pressure = Patm
P1 is the pressure at depth 'h'
Density of water = ρ = 10^3kg/m^3
Patm = 1.013×10^5 Pa.
Patm = 101300Pa
g = 9.81m/s^2
P1 = P2+ρgh
P1 = Patm +ρgh
P1 = 1.013×10^5+10^3×9.81×160.
P1 = 101300+1569600
P1 = 1670900 Pa
For an ideal gas law
PV =nRT
P1V1/P2V2 = 1
V2 = ( P1/P2)V1
V2 = (P1/Patm)V1
V2 = ( 1670900 /101300 Pa) × 1.3
V2 = 1670900/101300
V2 = 16.494×1.3
V2 = 21.44cm^3
The volume of the bubble can be determined using ideal gas law. The volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].
The formula of the pressure of the static fluid
P1 = P2+ρgh
Where,
P1 - pressure at depth 'h'
P2 - atmospheric pressure = [tex]\bold {1.013x10^5 }[/tex] = 1670900 Pa
h - Depth = 160m
ρ - Density of water = [tex]\bold {10^3\ kg/m^3}[/tex]
g- gravitational acceleration = [tex]\bold {9.81\ m/s^2}[/tex]
The initial volume of the bubble = [tex]\bold {1.3\ cm^3}[/tex]
[tex]\bold {P1 = 1.013x10^5+10^3\times 9.81\times 160}\\\\\bold {P1 = 101300+1569600}\\\\\bold {P1 = 1670900\ Pa}[/tex]
For an ideal gas,
PV =nRT
[tex]\bold {\dfrac {P_1V_1}{P_2V_2 }= 1}[/tex]
[tex]\bold {V2 = \dfrac { P_1}{P_2V_1}}[/tex]
So,
[tex]\bold {V2 = \dfrac {1670900 }{101300 }\times 1.3}\\\\\bold {V2 =21.44\ cm^3}[/tex]
Therefore, the volume of the bubble when it reaches surface is 21.44 [tex]\bold {cm^3}[/tex].
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