Pascal's Wine Barometer

Q. Toricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg/m³. Taking g as 9.8 m/s², what is the height of the wine column for normal atmospheric pressure of 1 atm, in metres? expects 10.5 m

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

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Q. Toricelli's barometer $P_{a} = 0$ Pa used mercury. Pascal duplicated it using French wine of density 984 kg/m³ $ρ = 984$ kg/m³. Taking g as 9.8 m/s² $g = 9.8$ m/s², what is the height of the wine column for normal atmospheric pressure of 1 atm $P = 1$ atm, in metres?

The playground

Inputs

density

ρ

984 kg/m³

gravity

g

9.8 m/s²

atmosphere

P

1 atm

Outputs

height

h

The solution

Solved with PressureAtDepth

Given

$ρ = 984$ kg/m³
$g = 9.8$ m/s²
$P = 1$ atm
$P_{a} = 0$ Pa from “Toricelli's barometer”

Find $h$ .

Working

$P = 1$ atm $= 101325$ Pa
pressure adds the surface s(column foot) $P = P_{a} + P_{g}$
$101325 = 0 + P_{g}$
$P_{g} = 101325$ Pa
pressure of the fluid above(column foot) $P_{g} = ρ \cdot g \cdot h$
$101325 = 984 \cdot 9.8 \cdot h$
$h = 10.51 m$
$h \approx 10.5$ mto 3 significant figures

Conclusion

Above the column is empty space, so the wine alone must press as hard as the atmosphere; wine is far lighter than mercury, so its column stands far taller.

The Ruby

1require_relative "../physics"23question :pascal_s_wine_barometer do4  kind :numerical5  describes "Toricelli's barometer used mercury. Pascal duplicated it using French wine of density %{density}. " \6            "Taking g as %{gravity}, %{height} for normal atmospheric pressure of %{atmosphere}, in metres?"7  given density: "984 kg/m3"8  given gravity: "9.8 m/s2"9  given atmosphere: "1 atm"10  asks height: "what is the height of the wine column"11end

1require_relative "../questions/pascal_s_wine_barometer"2require_relative "../science/mechanical_properties_of_fluids/pressure_at_depth"34solve :pascal_s_wine_barometer, using: PressureAtDepth do5  objects :column_foot67  take density: rho8  take gravity: g9  take atmosphere: column_foot.pressure1011  deduce_from "Toricelli's barometer", P_a => "0 Pa"1213  height = column_foot.depth14  answer height:, in: "m", figures: 31516  conclusion "Above the column is empty space, so the wine alone must press as hard as the atmosphere; " \17             "wine is far lighter than mercury, so its column stands far taller."18end

1require_relative "../solutions/pascal_s_wine_barometer"23play_with :pascal_s_wine_barometer do4  title "Pascal's Wine Barometer"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
pressure of ground	$P_{g}$	Pa	0 … ∞
thrust on ground	$F_{g}$	N	—
area of ground	$A_{g}$	m²	—
pressure of column foot	$P$	Pa	0 … ∞
thrust on column foot	$F$	N	—
area of column foot	$A$	m²	—
gauge pressure on ground	$P_{g,g}$	Pa	—
depth of ground	$h_{g}$	m	—
gauge pressure on column foot	$P_{g}$	Pa	—
depth of column foot	$h$	m	—
mass of column foot	$m$	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

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

pressure is force over area(ground)

P = \frac{F}{A}

pressure is force over area(column foot)

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(column foot)

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

pressure adds the surface s(ground)

P = P_{a} + P_{g}

pressure adds the surface s(column foot)