Pressure and Its Measurement

Pressure is the force a gas exerts divided by the area it pushes on. Every gas shows this behavior. This includes the air in a bicycle tire, the atmosphere pressing on your skin, and the vapor above a liquid. The same handful of instruments and units describe all of it. This article builds the definition of pressure from force and area. It shows how barometers and manometers turn a column of liquid into a measurement. It also sorts out the zoo of units (atm, mmHg, torr, Pa, bar) that chemists must convert between constantly. Getting pressure right matters for an important reason. It is one of the three state variables — along with volume and temperature — that control how gases behave. The relations you will meet elsewhere are written in terms of pressure. These include Boyle's Law, Charles's Law, Avogadro's Law, and the Ideal Gas Law. So a shaky grasp of what "1 atm" or "760 torr" actually means will undermine everything built on top of it.

Key takeaways

Pressure is force per unit area: $P = F/A$ , with SI unit the pascal, $1\ \text{Pa} = 1\ \text{N/m}^2$ .
Atmospheric pressure arises from the weight of the air column above a point on Earth's surface.
A barometer measures atmospheric pressure; a manometer measures the pressure of a confined gas relative to the atmosphere.
Standard atmospheric pressure is defined as exactly $1\ \text{atm} = 760\ \text{mmHg} = 760\ \text{torr} = 101{,}325\ \text{Pa} = 1.01325\ \text{bar}$ .
Column height in a mercury device converts to pressure through $P = \rho g h$ , so denser fluids give shorter columns for the same pressure.
Key idea: pressure is a force-per-area quantity measured indirectly by balancing it against a column of liquid whose height you can read off a ruler.

The core insight

Gas pressure is the force per unit area exerted by molecular collisions on a surface; it is measured by balancing that force against a known column of liquid, and the resulting height converts into any of several equivalent units (atm, mmHg, torr, Pa, bar) via fixed conversion factors.

The physical picture

Picture a container of gas at the molecular level: billions of molecules moving randomly, each one occasionally striking a wall and bouncing off. Every collision delivers a tiny push. Averaged over the huge number of collisions per second, this produces a steady outward force on every unit of wall area — that steady force per area is the pressure of the gas. The same idea applies to the atmosphere itself. Air has mass, so the column of air stretching from sea level up to the edge of space has weight, and that weight pressing down on a unit area is atmospheric pressure. At sea level this column exerts a force equivalent to about $10^5$ newtons on every square meter — enough to support a column of mercury roughly 760 mm tall, or a column of water about 10.3 m tall, since mercury is far denser than water. Historically, chemists could not count molecular collisions directly, so they measured pressure indirectly by letting it push a liquid up or down a tube and reading the resulting height. This is the working principle behind two instruments: the barometer, which measures atmospheric pressure, and the manometer, which measures the pressure of a gas sample confined in a flask or tube. A simple mercury barometer is a glass tube, closed at one end, filled with mercury and inverted into an open dish of mercury. The mercury in the tube falls until the weight of the mercury column exactly balances the push of the atmosphere on the mercury surface in the dish; the space above the mercury in the tube is a near-vacuum (vapor pressure of mercury is negligible at room temperature). The height of the column, measured in millimeters, is a direct readout of atmospheric pressure — this is the origin of the unit "mmHg." A closed-end manometer works similarly but connects to a gas sample rather than to the open atmosphere; the height difference between the two mercury arms directly reads the gas pressure, since the closed arm above the mercury is vacuum. An open-end manometer connects one arm to the atmosphere, so the gas pressure equals atmospheric pressure plus or minus the observed height difference, depending on which side stands higher.

Figure

A mercury barometer: atmospheric pressure supports a 760 mm column at standard pressure

Figure

Open-end manometer: gas pressure exceeds atmospheric pressure by the height difference

The core relations

P = \frac{F}{A}

where: $P$ is pressure, $F$ is the force acting perpendicular to a surface, $A$ is the area over which that force is spread; SI unit is the pascal, $1\ \text{Pa} = 1\ \text{N/m}^2$ .

P = \rho g h

where: $\rho$ is the density of the manometer/barometer fluid, $g$ is the gravitational acceleration ( $9.81\ \text{m/s}^2$ ), $h$ is the height of the liquid column supported by the pressure being measured.

1\ \text{atm} = 760\ \text{mmHg} = 760\ \text{torr} = 101{,}325\ \text{Pa} = 1.01325\ \text{bar}

where: these are exact, defined equivalences (mmHg and torr are, to extremely high precision, interchangeable) used as conversion factors between pressure units.

P_{\text{gas}} = P_{\text{atm}} \pm \Delta h \quad \text{(open-end manometer, in mmHg)}

where: the sign is $+$ if the gas-side arm's mercury stands lower than the open arm (gas pressure exceeds atmospheric), and $-$ if the gas-side arm stands higher (gas pressure is below atmospheric); $\Delta h$ is the height difference between the two mercury surfaces.

Derivation ADeriving

P = \rho g h

for a liquid column

Goal. Show that the pressure exerted by, or needed to support, a column of liquid of height $h$ and density $\rho$ is $P = \rho g h$ , and use it to justify why mercury (not water) is the practical barometer fluid.

Step 1. Consider a column of liquid of cross-sectional area $A$ and height $h$ , sitting on top of an imaginary horizontal surface at its base. Its volume is

V = A h

Step 2. Its mass follows from density,

m = \rho V = \rho A h

Step 3. Its weight — the force it exerts on the surface beneath it due to gravity — is

F = m g = \rho A h g

Step 4. Pressure is force per unit area, so divide by $A$ :

P = \frac{F}{A} = \frac{\rho A h g}{A} = \rho g h

The area cancels completely, which is the key physical insight: the pressure supported by a column depends only on the fluid's density and the column's height, never on the tube's width. A barometer with a wide bore and one with a narrow bore, filled with the same liquid, show the same height for the same atmospheric pressure.

Step 5 — why mercury. To support standard atmospheric pressure ( $101{,}325\ \text{Pa}$ ) with water ( $\rho = 1000\ \text{kg/m}^3$ ), solve for $h$ :

h_{\text{water}} = \frac{P}{\rho g} = \frac{101{,}325}{1000 \times 9.81} \approx 10.33\ \text{m}

A water barometer would need to be over ten meters tall. With mercury ( $\rho = 13{,}534\ \text{kg/m}^3$ at typical lab conditions):

h_{\text{Hg}} = \frac{101{,}325}{13{,}534 \times 9.81} \approx 0.763\ \text{m} = 763\ \text{mm}

This matches the familiar 760 mm figure closely (the small difference comes from using standard $g$ and a slightly idealized mercury density); mercury's very high density is precisely why it was chosen — it keeps the instrument bench-sized rather than building-sized.

Worked examples

Example 1 — Reading an open-end manometer. A flask of gas is connected to an open-end mercury manometer. The mercury in the arm open to the atmosphere stands 34.5 mm higher than the mercury in the arm connected to the flask. Atmospheric pressure that day, read from a nearby barometer, is 748.2 mmHg. Find the gas pressure in mmHg, then convert to atm and to kPa. Because the flask-side mercury is pushed down (lower) relative to the open side, the gas is pushing harder than the atmosphere, so its pressure exceeds atmospheric by the height difference:

P_{\text{gas}} = P_{\text{atm}} + \Delta h = 748.2\ \text{mmHg} + 34.5\ \text{mmHg} = 782.7\ \text{mmHg}

Convert to atm using $1\ \text{atm} = 760\ \text{mmHg}$ :

P_{\text{gas}} = 782.7\ \text{mmHg} \times \frac{1\ \text{atm}}{760\ \text{mmHg}} = 1.0299\ \text{atm}

Convert to kilopascals using $1\ \text{atm} = 101.325\ \text{kPa}$ :

P_{\text{gas}} = 1.0299\ \text{atm} \times \frac{101.325\ \text{kPa}}{1\ \text{atm}} = 104.4\ \text{kPa}

As a limiting-case check: if the two mercury levels had been equal ( $\Delta h = 0$ ), the formula gives $P_{\text{gas}} = P_{\text{atm}}$ exactly, which is physically required — no height difference means no net pressure difference between the gas and the atmosphere.

Example 2 — Converting a bizarre pressure reading into every common unit. A chemistry lab in an older textbook reports a gas pressure of 0.8892 bar. Express this in atm, mmHg (torr), Pa, and psi is not requested — stick to the four target units. Start from the bar and convert to Pa, using $1\ \text{bar} = 100{,}000\ \text{Pa}$ exactly:

P = 0.8892\ \text{bar} \times \frac{100{,}000\ \text{Pa}}{1\ \text{bar}} = 88{,}920\ \text{Pa}

Convert Pa to atm, using $1\ \text{atm} = 101{,}325\ \text{Pa}$ :

P = \frac{88{,}920\ \text{Pa}}{101{,}325\ \text{Pa/atm}} = 0.8776\ \text{atm}

Convert atm to mmHg (equivalently torr), using $1\ \text{atm} = 760\ \text{mmHg}$ :

P = 0.8776\ \text{atm} \times 760\ \text{mmHg/atm} = 667.0\ \text{mmHg} = 667.0\ \text{torr}

A quick sanity check on internal consistency: since $1\ \text{bar}$ is slightly less than $1\ \text{atm}$ ( $1.01325\ \text{bar} = 1\ \text{atm}$ ), any pressure below 1 bar should convert to something below 0.987 atm — and $0.8776\ \text{atm}$ is indeed comfortably in that range, confirming no unit-flip error occurred.

Limiting cases & common traps

Situation	What happens	Why
$\Delta h \to 0$ in an open manometer	$P_{\text{gas}} \to P_{\text{atm}}$	No height difference means the two sides are in balance
Wider or narrower barometer tube	$h$ unchanged for the same $P_{\text{atm}}$	Area cancels in $P=\rho g h$
Switching fluid from Hg to water	$h$ increases by factor $\approx 13.6$	$h \propto 1/\rho$ , and $\rho_{\text{Hg}}/\rho_{\text{water}} \approx 13.6$
Confusing mmHg with mm of the sample liquid	Wrong pressure computed	$\rho g h$ only equals a pressure in mmHg if the column is mercury; other liquids need their own $\rho$
Forgetting the sign convention in open manometers	Adding when should subtract, or vice versa	Always ask which side's mercury is lower — that side has the greater gas-facing pressure pushing it down

The chemist behind it

The barometer's invention is credited to Evangelista Torricelli in 1643, a student of Galileo, who filled a glass tube with mercury and inverted it into a dish, noticing that the mercury always settled at roughly the same height regardless of the tube's shape or tilt. Torricelli correctly reasoned that the space above the mercury was a vacuum and that the surrounding air's weight — not some mysterious "nature abhors a vacuum" force — was what held the column up. This was a radical claim in its time, since it implied the atmosphere itself had measurable weight and pushed on everything at Earth's surface. Blaise Pascal later carried barometers up a mountain and showed the mercury column fell with altitude, confirming that atmospheric pressure decreases as the air column above thins out — direct experimental proof of Torricelli's reasoning, and the origin of the pascal as a unit name.

Origins · 1 minEvangelista Torricelli and how this was discovered

Evangelista Torricelli

1608–1647 · Devised the mercury barometer and gave the first correct physical explanation of atmospheric pressure, replacing vague appeals to a vacuum-abhorring nature with a quantitative account of air's weight.

Ask the tutor: How would the height reading on a mercury barometer change if it were carried to the top of a tall mountain, and why?

Why is mercury used in barometers instead of water?

Mercury's density (about $13{,}534\ \text{kg/m}^3$ ) is roughly 13.6 times that of water. Since column height for a given pressure scales as $h = P/(\rho g)$ , a denser fluid needs a much shorter column to balance the same atmospheric pressure — about 760 mm for mercury versus over 10 meters for water. This keeps the instrument a practical benchtop size.

Is "torr" exactly the same as "mmHg"?

For essentially all chemistry purposes, yes. The torr was defined so that $1\ \text{torr} = 1/760\ \text{atm}$ , matching the mmHg scale to within a few parts in $10^7$ — a difference far smaller than any measurement uncertainty in a general chemistry laboratory. You can treat them as interchangeable.

Why does the top of a barometer tube read as "vacuum" rather than trapped air?

When the tube is filled completely with mercury and then inverted into the dish without letting any air bubble in, the only thing that can occupy the space left as mercury falls is mercury vapor, whose pressure at room temperature is negligible (a tiny fraction of a millimeter of mercury). This near-perfect vacuum is what lets the height of the column read atmospheric pressure directly, with no correction needed for gas trapped above it.

How does pressure connect to the gas laws?

Pressure is one of the four quantities — along with volume, temperature, and amount of gas — that describe the state of a gas sample. Boyle's Law relates pressure and volume at constant temperature, Charles's Law relates volume and temperature at constant pressure, and Avogadro's Law relates volume and moles at constant temperature and pressure. All three combine into the single Ideal Gas Law, $PV = nRT$ , where consistent pressure units (matched to the gas constant $R$ you are using) are essential for a correct numeric answer.

What is the difference between an open-end and closed-end manometer?

A closed-end manometer has one arm sealed with a vacuum above the mercury, so the height difference between the two arms directly equals the gas pressure with no reference to atmospheric pressure at all. An open-end manometer has one arm open to the atmosphere, so you must add or subtract the height difference from the current atmospheric pressure (read separately from a barometer) to get the gas's actual pressure.