Torricelli's barometer used mercury. Pascal duplicated it using French wine of density $984\ \text{kg m}^{-3}$. Determine the height of the wine column for normal atmospheric pressure.

Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984kg m⁻³. Determine th

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984\ \text{kg m}^{-3}

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Concept understanding — Hydrostatic Pressure Balance

Hydrostatic Pressure Balance

Imagine you're standing at the bottom of a swimming pool. You feel pressure on your ears — and the deeper you go, the more intense that pressure becomes. Now think about a column of water above you: every kilogram of that water is being pulled down by gravity. That weight has to be supported by the water below it. The deeper you go, the more water is stacked above you, so the greater the weight pressing down.

That's the core intuition: pressure in a fluid at rest increases with depth because the fluid above has to be supported by the fluid below.

The Precise Statement

Hydrostatic pressure balance is the condition that holds for any fluid at rest in a uniform gravitational field. It says:

P_{\text{below}} = P_{\text{above}} + \rho g h

where:

$P_{\text{below}}$ is the pressure at a lower point,
$P_{\text{above}}$ is the pressure at a higher point,
$\rho$ is the density of the fluid (assumed constant),
$g$ is the acceleration due to gravity,
$h$ is the vertical depth between the two points.

Equivalently, the pressure gradient in the vertical direction is:

\frac{dP}{dz} = -\rho g

where $z$ increases upward. The minus sign tells you pressure decreases as you go up.

Why This Makes Sense

Take a thin horizontal slab of fluid of area $A$ , thickness $dz$ , at some depth. Its weight is $dW = \rho g A \, dz$ . For the slab to be in equilibrium (not accelerating), the net upward force from pressure must exactly balance this weight.

The upward force on the slab's bottom face is $P(z) A$ , and the downward force on its top face is $P(z+dz) A$ . The net upward force is:

P(z) A - P(z+dz) A = - \frac{dP}{dz} A \, dz

Setting this equal to the weight $\rho g A \, dz$ gives:

-\frac{dP}{dz} = \rho g

which is exactly the differential form above.

Note

This balance assumes the fluid is static — no flow, no acceleration. If the fluid moves, additional terms (like viscous forces or inertial effects) appear.

Key Implications

Pressure depends only on depth, not on the shape of the container. A tall thin tube and a wide shallow tank give the same pressure at the same depth — because only the vertical height of fluid above matters.
Pressure is the same at all points on the same horizontal level. If you move sideways at constant depth, $\rho g h$ doesn't change, so $P$ doesn't change.
Gases are compressible, so $\rho$ is not constant. For air, the density changes with pressure itself, leading to an exponential decrease — but the same principle applies locally.

A Common Mistake …

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A barometer measures atmospheric pressure by balancing it against a column of liquid; since wine is much less dense than mercury, it requires a proportionally taller column. For standard atmospheric pressure, the wine column reaches $10.5\ \text{m}$ .

Why liquids of different densities give different column heights

A barometer works on a beautifully simple principle: atmospheric pressure at the base supports a column of liquid in an evacuated tube. The pressure at the bottom must equal the weight per unit area of the liquid column above it.

For any liquid in hydrostatic equilibrium, the pressure exerted by a column of height $h$ is

P = \rho g h

where $\rho$ is the liquid's density and $g$ is gravitational acceleration. Since atmospheric pressure $P_{\text{atm}}$ is fixed (at a given location and time), a denser liquid like mercury needs a shorter column to produce the same pressure, while a lighter liquid like wine needs a much taller one.

Mercury's fame in barometry comes from its high density ( $13{,}600\ \text{kg m}^{-3}$ ), which keeps the instrument compact at about $76\ \text{cm}$ . Wine, being roughly $14$ times less dense, will require a column about $14$ times taller.

Step-by-step calculation

1. Identify the known quantities

Standard atmospheric pressure: $P_{\text{atm}} = 101{,}325\ \text{Pa}$ (or $1.01325 \times 10^5\ \text{Pa}$ )
Density of French wine: $\rho_{\text{wine}} = 984\ \text{kg m}^{-3}$
Gravitational acceleration: $g = 9.8\ \text{m s}^{-2}$ (or $9.81\ \text{m s}^{-2}$ for higher precision)

2. Apply the hydrostatic pressure formula

The atmospheric pressure must balance the pressure from the wine column:

P_{\text{atm}} = \rho_{\text{wine}} \cdot g \cdot h

3. Solve for the height $h$

Rearranging for $h$ :

h = \frac{P_{\text{atm}}}{\rho_{\text{wine}} \cdot g}

4. Substitute the numerical values

Using $g = 9.8\ \text{m s}^{-2}$ :

h = \frac{101{,}325}{984 \times 9.8} = \frac{101{,}325}{9{,}643.2} \approx 10.51\ \text{m}

If we use $g = 9.81\ \text{m s}^{-2}$ for slightly better accuracy: …

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Q.Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg m−3984\ \text{kg m}^{-3}984 kg m−3. Determine the height of the wine column for normal atmospheric pressure.

Why liquids of different densities give different column heights

Step-by-step calculation

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Q.Torricelli's barometer used mercury. Pascal duplicated it using French wine of density $984\ \text{kg m}^{-3}$ . Determine the height of the wine column for normal atmospheric pressure.

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