A Soap Bubble's Excess Pressure

solved SurfaceTension

Q. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm $r = 5.00$ mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10⁻² N/m $S = 2.50 \times 10^{−2}$ N/m, in pascals?

Inputs

radius

r

5 mm

tension

S

0.025 N/m

Outputs

excess

Δ P_s

How it's solved

Solved with SurfaceTension

Given

$r = 5.00$ mm
$S = 2.50 \times 10^{−2}$ N/m

Find $Δ P_s$ .

Working

$r = 5.00$ mm $= 0.005$ m
excess pressure in a soap bubble with two surfaces(bubble) $Δ P_s = \frac{4 \cdot S}{r}$
$Δ P_s = \frac{4 \cdot 0.025}{0.005}$
$Δ P_s = 20$ Pa
$Δ P_s \approx 20.0$ Pato 3 significant figures

Conclusion

A soap bubble is a thin film with two surfaces, inner and outer, so it squeezes the air inside twice as hard as a single surface of the same radius would.

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 :a_soap_bubble_s_excess_pressure do4  kind :numerical5  describes "%{excess} of radius %{radius}, given that the surface tension of soap solution at the temperature " \6            "(20 °C) is %{tension}, in pascals?"7  given radius: "5.00 mm"8  given tension: "2.50e-2 N/m"9  asks excess: "What is the excess pressure inside a bubble of soap solution"10end

solution.rb

1require_relative "../questions/a_soap_bubble_s_excess_pressure"2require_relative "../science/mechanical_properties_of_fluids/surface_tension"34solve :a_soap_bubble_s_excess_pressure, using: SurfaceTension do5  objects :bubble67  take radius: bubble.radius8  take tension: S910  excess = bubble.excess_pressure_in_a_soap_bubble11  answer excess:, in: "Pa", figures: 31213  conclusion "A soap bubble is a thin film with two surfaces, inner and outer, so it squeezes the air " \14             "inside twice as hard as a single surface of the same radius would."15end

playground.rb

1require_relative "../solutions/a_soap_bubble_s_excess_pressure"23play_with :a_soap_bubble_s_excess_pressure do4  title "A Soap Bubble's Excess Pressure"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)