Answer:
1.82 cm
Explanation:
Utilize the equation [tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex] to calculate the change in volume and size of an air bubble.
P1 = pressure at 50m = [tex]P_{A}[/tex] + ρ*g*h (where [tex]P_{A}[/tex] = atmospheric pressure, ρ = density of water, g = acceleration due to gravity, h = height/depth)
P1 = 1.01 x 10⁵ Pa + (ρ x g x h)
= 1.01 x 10⁵ Pa + (1000 kg/m³ x 9.8 m/s² x 50 m )
= 1.01 x 10⁵ Pa + 4.9 x 10⁵ Pa
= 5.91 x 10⁵ Pa
V1 = [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex] [tex]r_{1}[/tex] = 10 cm = 1 x 10⁻² m
T1 = 10 °C = 10 + 273 = 283 K
P2 = [tex]P_{A}[/tex] = 1.01 x 10⁵ Pa because at the surface, pressure is equal to atmospheric pressure
V2 = [tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] [tex]r_{2}[/tex] = ??
T2 = 20 °C = 20 + 273 = 293 K
[tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex]
V₂ = P₁V₁T₂
P₂T₁
[tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] = P₁ x [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex] x T₂
P₂T₁
cancel out common terms
[tex]r_{2}[/tex]³ = 5.91 x 10⁵ Pa x (1 x 10⁻² m)³ x 293 k
1.01 x 10⁵ Pa x 283 k
[tex]r_{2}[/tex]³ = 757.9 x 10⁻⁹
[tex]r_{2}[/tex] = 9.1 x 10⁻³ m
[tex]r_{2}[/tex] = 0.91 cm
Therefore, bubbles diameter = 2r = 1.82 cm
Answer:
1.82 cm
Explanation:
The pressure done by a column of a liquid is called the hydrostatic pressure (Ph) and it can be calculated by:
Ph = Patm + ρgh
Where Patm is the atmospheric pressure under the column (101325 Pa), ρ is the density of the liquid (1000 kg/m³ for water), g is the gravity acceleration (9.8 m/s²), and h is the depth (50 m), so:
Ph = 101325 + 1000*9.8*50
Ph = 591325 Pa
Because the bubble is in equilibrium with the surroundings, its pressure is the same as the surroundings. Supposing a perfect sferic bubble, its volume is:
V = (4/3)*π*r³
Where r is the radius, which is half of the diameter, so r = 0.5 cm.
V = (4/3)*π*(0.5)³
V = 0.52 cm³
According to the ideal gas law, the multiplication of the pressure (P) by the volume (V) divided by the temperature (T) of a gas is constant, so if 1 is the state where the bubble is 50 m depth, and 2 the state at the surface:
P1*V1/T1 = P2*V2/T2
P1 = Ph = 591325 Pa
V1 = 0.52 cm³
T1 = 10°C + 273 = 283 K
P2 = 101325 Pa (atmosferic pressure)
T2 = 20°C + 273 = 293 K
591325*0.52/283 = 101325*V2/293
101325V2 = 318,354.3357
V2 = 3.14 cm³
V2 = (4/3)*π*r³
(4/3)*π*r³ = 3.14
r³ = 0.75
r = ∛0.75
r = 0.91 cm
The diameter is then 2*r = 1.82 cm.
