Blowing a Bubble Under Water

Q. The lower end of a capillary tube of diameter 2.00 mm is dipped 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. Taking 1 atmospheric pressure as 1.01 × 10⁵ Pa, the density of water as 1000 kg/m³ and g as 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? expects 1.02 × 10⁵ Pa

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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?

The playground

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}

The solution

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 Ruby

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

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

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

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)