Blowing a Bubble Under Water

solved SurfaceTension

Q. The lower end of a capillary tube of diameter 2.00 mm $r = 1.00$ mm is dipped 8.00 cm $h = 8.00$ cm below the surface of water in a beaker. The surface tension of water at the temperature of the experiment is 7.30 × 10⁻² N/m $S = 7.30 \times 10^{−2}$ N/m. Taking 1 atmospheric pressure as 1.01 × 10⁵ Pa $P_{a} = 1.01 \times 10^{5}$ Pa, the density of water as 1000 kg/m³ $ρ = 1000$ kg/m³ and g as 9.80 m/s² $g = 9.80$ m/s², what is the pressure required in the tube in order to blow a hemispherical bubble at its end in water, in pascals?

Inputs

depth

h

8 cm

tension

S

0.073 N/m

atmosphere

P_{a}

1.01e+05 Pa

density

ρ

1000 kg/m³

gravity

g

9.8 m/s²

Outputs

pressure

P_{i}

How it's solved

Solved with SurfaceTension

Given

$h = 8.00$ cm
$S = 7.30 \times 10^{−2}$ N/m
$P_{a} = 1.01 \times 10^{5}$ Pa
$ρ = 1000$ kg/m³
$g = 9.80$ m/s²
$r = 1.00$ mm from “a capillary tube of diameter 2.00 mm”

Find $P_{i}$ .

Working

$h = 8.00$ cm $= 0.08$ m
$r = 1.00$ mm $= 0.001$ m
pressure of the fluid above(bubble) $P_{g} = ρ \cdot g \cdot h$
$P_{g} = 1000 \cdot 9.8 \cdot 0.08$
$P_{g} = 784$ Pa
pressure adds the surface s(bubble) $P = P_{a} + P_{g}$
$P = 101000 + 784$
$P = 101784$ Pa
excess pressure in a bubble in liquid(bubble) $Δ P = \frac{2 \cdot S}{r}$
$Δ P = \frac{2 \cdot 0.073}{0.001}$
$Δ P = 146$ Pa
pressure inside a bubble(bubble) $P_{i} = P + Δ P$
$P_{i} = 101784 + 146$
$P_{i} = 101900$ Pa
$P_{i} \approx 1.02 \times 10^{5}$ Pato 3 significant figures

Conclusion

The air in the tube has to match the water's pressure at that depth and then push a little more, to hold the bubble's curved skin open against its surface tension.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :blowing_a_bubble_under_water do4  kind :numerical5  describes "The lower end of a capillary tube of diameter 2.00 mm is dipped %{depth} below the surface of water in " \6            "a beaker. The surface tension of water at the temperature of the experiment is %{tension}. Taking " \7            "1 atmospheric pressure as %{atmosphere}, the density of water as %{density} and g as %{gravity}, " \8            "%{pressure}, in pascals?"9  given depth: "8.00 cm"10  given tension: "7.30e-2 N/m"11  given atmosphere: "1.01e5 Pa"12  given density: "1000 kg/m3"13  given gravity: "9.80 m/s2"14  asks pressure: "what is the pressure required in the tube in order to blow a hemispherical bubble at its end in water"15end

solution.rb

1require_relative "../questions/blowing_a_bubble_under_water"2require_relative "../science/mechanical_properties_of_fluids/surface_tension"34solve :blowing_a_bubble_under_water, using: SurfaceTension do5  objects :bubble67  take depth: bubble.depth8  take tension: S9  take atmosphere: P_a10  take density: rho11  take gravity: g1213  deduce_from "a capillary tube of diameter 2.00 mm", bubble.radius => "1.00 mm"1415  pressure = bubble.pressure_inside16  answer pressure:, in: "Pa", figures: 31718  conclusion "The air in the tube has to match the water's pressure at that depth and then push a little " \19             "more, to hold the bubble's curved skin open against its surface tension."20end

playground.rb

1require_relative "../solutions/blowing_a_bubble_under_water"23play_with :blowing_a_bubble_under_water do4  title "Blowing a Bubble Under Water"5end

The laws it drew on

quantity	written	units	within	from
mass	$m$	kg	0 … ∞	Pressure
gravity	$g$	m/s²	—	Pressure
thrust	$F$	N	—	Pressure
area	$A$	m²	—	Pressure
pressure	$P$	Pa	0 … ∞	Pressure
pressure at the surface	$P_{a}$	Pa	—	PressureAtDepth
density	$ρ$	kg/m³	0 … ∞	PressureAtDepth
depth	$h$	m	—	PressureAtDepth
gauge pressure	$P_{g}$	Pa	—	PressureAtDepth
surface tension	$S$	N/m	—	SurfaceTension
radius	$r$	m	—	SurfaceTension
excess pressure	$Δ P$	Pa	—	SurfaceTension
excess pressure in a soap bubble	$Δ P_s$	Pa	—	SurfaceTension
pressure inside	$P_{i}$	Pa	—	SurfaceTension
slider length	$l$	m	—	SurfaceTension
pull of the film	$F_{S}$	N	—	SurfaceTension
pressure of ground	$P_{g}$	Pa	0 … ∞
thrust on ground	$F_{g}$	N	—
area of ground	$A_{g}$	m²	—
pressure of bubble	$P$	Pa	0 … ∞
thrust on bubble	$F$	N	—
area of bubble	$A$	m²	—
gauge pressure on ground	$P_{g,g}$	Pa	—
depth of ground	$h_{g}$	m	—
gauge pressure on bubble	$P_{g}$	Pa	—
depth of bubble	$h$	m	—
excess pressure of ground	$Δ P_{g}$	Pa	—
radius of ground	$r_{g}$	m	—
excess pressure of bubble	$Δ P$	Pa	—
radius of bubble	$r$	m	—
excess pressure in a soap bubble of ground	$Δ {P_s}_{g}$	Pa	—
excess pressure in a soap bubble of bubble	$Δ P_s$	Pa	—
pressure inside of ground	$P_{i,g}$	Pa	—
pressure inside of bubble	$P_{i}$	Pa	—
pull of the film on ground	$F_{S,g}$	N	—
slider length of ground	$l_{g}$	m	—
pull of the film on bubble	$F_{S}$	N	—
slider length of bubble	$l$	m	—
mass of bubble	$m$	kg	0 … ∞

P_{s} = \frac{F_{s}}{A_{s}}

pressure is force over area

P_{g,p} = ρ \cdot g \cdot h_{p}

pressure of the fluid above

P_{p} = P_{a} + P_{g,p}

pressure adds the surface s

Δ P_{b} = \frac{2 \cdot S}{r_{b}}

excess pressure in a bubble in liquid

Δ {P_s}_{b} = \frac{4 \cdot S}{r_{b}}

excess pressure in a soap bubble with two surfaces

P_{i,b} = P_{b} + Δ P_{b}

pressure inside a bubble

F_{S,s} = 2 \cdot S \cdot l_{s}

a film pulls with both its faces

P_{g} = \frac{F_{g}}{A_{g}}

pressure is force over area(ground)

P = \frac{F}{A}

pressure is force over area(bubble)

P_{g,g} = ρ \cdot g \cdot h_{g}

pressure of the fluid above(ground)

P_{g} = ρ \cdot g \cdot h

pressure of the fluid above(bubble)

P_{g} = P_{a} + P_{g,g}

pressure adds the surface s(ground)

P = P_{a} + P_{g}

pressure adds the surface s(bubble)

Δ P_{g} = \frac{2 \cdot S}{r_{g}}

excess pressure in a bubble in liquid(ground)

Δ P = \frac{2 \cdot S}{r}

excess pressure in a bubble in liquid(bubble)

Δ {P_s}_{g} = \frac{4 \cdot S}{r_{g}}

excess pressure in a soap bubble with two surfaces(ground)

Δ P_s = \frac{4 \cdot S}{r}

excess pressure in a soap bubble with two surfaces(bubble)

P_{i,g} = P_{g} + Δ P_{g}

pressure inside a bubble(ground)

P_{i} = P + Δ P

pressure inside a bubble(bubble)

F_{S,g} = 2 \cdot S \cdot l_{g}

a film pulls with both its faces(ground)

F_{S} = 2 \cdot S \cdot l

a film pulls with both its faces(bubble)