Inside a Drop of Mercury

Q. A drop of mercury has a radius of 3.00 mm at room temperature. Surface tension of mercury at that temperature (20 °C) is 4.65 × 10⁻¹ N/m. The atmospheric pressure is 1.01 × 10⁵ Pa. What is the pressure inside the drop, in pascals? expects 1.0131 × 10⁵ Pa

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

radius

r

3 mm

tension

S

0.465 N/m

atmosphere

P_{a}

1.01e+05 Pa

Outputs

pressure

P_{i}

The solution

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 Ruby

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

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

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

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