Inside a Drop of Mercury

solved SurfaceTension

Q. A drop of mercury has a radius of 3.00 mm $r = 3.00$ mm at room temperature. Surface tension of mercury at that temperature (20 °C) is 4.65 × 10⁻¹ N/m $S = 4.65 \times 10^{−1}$ N/m. The atmospheric pressure is $P = P_{a}$ 1.01 × 10⁵ Pa $P_{a} = 1.01 \times 10^{5}$ Pa. What is the pressure inside the drop, in pascals?

Inputs

radius

r

3 mm

tension

S

0.465 N/m

atmosphere

P_{a}

1.01e+05 Pa

Outputs

pressure

P_{i}

How it's solved

Solved with SurfaceTension

Given

$r = 3.00$ mm
$S = 4.65 \times 10^{−1}$ N/m
$P_{a} = 1.01 \times 10^{5}$ Pa
$P = P_{a}$ from “The atmospheric pressure is”

Find $P_{i}$ .

Working

$r = 3.00$ mm $= 0.003$ m
The atmospheric pressure is $P = P_{a}$
$P = 101000$
$P = 101000$ Pa
excess pressure in a bubble in liquid(drop) $Δ P = \frac{2 \cdot S}{r}$
$Δ P = \frac{2 \cdot 0.465}{0.003}$
$Δ P = 310$ Pa
pressure inside a bubble(drop) $P_{i} = P + Δ P$
$P_{i} = 101000 + 310$
$P_{i} = 101300$ Pa
$P_{i} \approx 1.0131 \times 10^{5}$ Pato 5 significant figures

Conclusion

The air outside presses on the drop, and the mercury's curved skin squeezes it a little more; the smaller the drop, the harder that skin squeezes.

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 :inside_a_drop_of_mercury do4  kind :numerical5  describes "A drop of mercury has a radius of %{radius} at room temperature. Surface tension of mercury at that " \6            "temperature (20 °C) is %{tension}. The atmospheric pressure is %{atmosphere}. %{pressure}, in pascals?"7  given radius: "3.00 mm"8  given tension: "4.65e-1 N/m"9  given atmosphere: "1.01e5 Pa"10  asks pressure: "What is the pressure inside the drop"11end

solution.rb

1require_relative "../questions/inside_a_drop_of_mercury"2require_relative "../science/mechanical_properties_of_fluids/surface_tension"34solve :inside_a_drop_of_mercury, using: SurfaceTension do5  objects :drop67  take radius: drop.radius8  take tension: S9  take atmosphere: P_a1011  deduce_from "The atmospheric pressure is", drop.pressure => P_a1213  pressure = drop.pressure_inside14  answer pressure:, in: "Pa", figures: 51516  conclusion "The air outside presses on the drop, and the mercury's curved skin squeezes it a little " \17             "more; the smaller the drop, the harder that skin squeezes."18end

playground.rb

1require_relative "../solutions/inside_a_drop_of_mercury"23play_with :inside_a_drop_of_mercury do4  title "Inside a Drop of Mercury"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 drop	$P$	Pa	0 … ∞
thrust on drop	$F$	N	—
area of drop	$A$	m²	—
gauge pressure on ground	$P_{g,g}$	Pa	—
depth of ground	$h_{g}$	m	—
gauge pressure on drop	$P_{g}$	Pa	—
depth of drop	$h$	m	—
excess pressure of ground	$Δ P_{g}$	Pa	—
radius of ground	$r_{g}$	m	—
excess pressure of drop	$Δ P$	Pa	—
radius of drop	$r$	m	—
excess pressure in a soap bubble of ground	$Δ {P_s}_{g}$	Pa	—
excess pressure in a soap bubble of drop	$Δ P_s$	Pa	—
pressure inside of ground	$P_{i,g}$	Pa	—
pressure inside of drop	$P_{i}$	Pa	—
pull of the film on ground	$F_{S,g}$	N	—
slider length of ground	$l_{g}$	m	—
pull of the film on drop	$F_{S}$	N	—
slider length of drop	$l$	m	—
mass of drop	$m$	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(drop)

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(drop)

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

pressure adds the surface s(ground)

P = P_{a} + P_{g}

pressure adds the surface s(drop)

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(drop)

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(drop)

P_{i,g} = P_{g} + Δ P_{g}

pressure inside a bubble(ground)

P_{i} = P + Δ P

pressure inside a bubble(drop)

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(drop)

P = P_{a}

The atmospheric pressure is