Atmospheric pressure

The Earth's atmosphere is not weightless. Air is matter — mostly nitrogen and oxygen molecules — and gravity pulls on it just as it pulls on a liquid. Because the atmosphere behaves like a very tall, very light "fluid ocean" resting on the Earth's surface, every point on the surface (and every object in the air) has the entire remaining column of air above it pressing down. That downward weight, spread over area, is atmospheric pressure. It is not felt as a directional push because air molecules dart around and collide with a surface from every angle, so the pressure acts equally in all directions — sideways and upward on the walls of a container just as much as downward on its lid.

Why a barometer measures it. You cannot weigh the atmosphere on a scale, so instead a barometer balances atmospheric pressure against something you can measure: the pressure of a liquid column, $p = h\rho g$. In the classic mercury barometer, a glass tube longer than about 76 cm, closed at one end, is filled completely with mercury and inverted into an open dish of mercury with the closed end pointing up. The mercury inside the tube immediately starts to fall — but only until the weight of the mercury column left standing in the tube exactly balances the atmospheric pressure pushing down on the mercury surface in the open dish below. At that point everything stops moving: a stable, self-balancing scale.

where: $p_{\text{atm}}$ is atmospheric pressure, $h$ is the height of the mercury column left standing in the tube, $\rho_{\text{Hg}}$ is the density of mercury, and $g$ is the acceleration due to gravity. At sea level, on an ordinary day, this balance settles with $h \approx 76\ \text{cm}$ — which is why "76 cm of mercury" is itself used as a everyday way of stating one atmosphere of pressure, without ever converting to pascals.

The empty space at the top matters. The gap that opens up above the mercury column, at the closed end of the tube, is not filled with air — it is a near-perfect vacuum, called the Torricellian vacuum, after Evangelista Torricelli, who first built this device. Because nothing pushes down from inside that space, the entire support for the mercury column comes from the atmosphere pressing on the open dish outside. If even a little air leaked into the top of the tube, it would push down on the mercury too, and the reading would fall below the true atmospheric pressure — a real barometer must keep that space sealed and empty.

Altitude changes the reading. Carry the barometer up a hill and the mercury column shortens. This has nothing to do with the barometer itself — it is because there is genuinely less air left above you as you climb, so the air column pressing down is shorter and lighter, and it can only balance a shorter mercury column in return. This same falling-with-altitude behaviour is why aircraft cabins are pressurised and why unaided breathing gets harder on a high mountain, and it is also monitored on the ground: a sudden drop in atmospheric pressure at a fixed location, tracked over hours, is one of the classic early signs used in weather forecasting.

A mercury barometer at sea level reads a column height of $76\ \text{cm}$ (density of mercury $= 13{,}600\ \text{kg/m}^3$, density of water $= 1000\ \text{kg/m}^3$). What height of water column would be needed to balance the same atmospheric pressure?

Step 1. Set the two liquid-column pressures equal, since both balance the same atmosphere: $h_{\text{Hg}}\,\rho_{\text{Hg}}\,g = h_{\text{water}}\,\rho_{\text{water}}\,g$, and cancel $g$ to get $h_{\text{Hg}}\,\rho_{\text{Hg}} = h_{\text{water}}\,\rho_{\text{water}}$.

Step 2. Substitute the numbers: $0.76 \times 13600 = h_{\text{water}} \times 1000$, so $h_{\text{water}} = \dfrac{0.76 \times 13600}{1000} = \dfrac{10336}{1000}$.

Step 3. Compute the height: $h_{\text{water}} = 10.336\ \text{m}$. A water barometer would need a column just over 10 metres tall to show the same atmospheric pressure that mercury shows in a mere 76 cm.

Competitive stretch

If a student instead used an oil of density $800\ \text{kg/m}^3$, the same ratio method gives $h_{\text{oil}} = h_{\text{Hg}} \times \dfrac{13600}{800} = 0.76 \times 17 = 12.92\ \text{m}$ — even taller than the water column, since oil is lighter than water. The pattern holds generally: the lighter the liquid, the taller the required tube, which is precisely why every practical barometer uses a dense liquid like mercury rather than water or oil.

• Confusing the mercury column's height with atmospheric pressure itself. The height is only a convenient stand-in measure of the pressure, made possible because mercury's density never changes; it is not the pressure.
• Believing the whole tube above the dish is a vacuum. Only the small gap above the mercury column — the Torricellian vacuum — is empty; the rest of the tube down to the dish is completely filled with mercury.
• Treating atmospheric pressure as constant everywhere on Earth. It falls steadily with altitude and shifts slightly from day to day with the weather at any one place.
• Turning this topic into a pascal-conversion numerical. ICSE keeps atmospheric pressure qualitative; the only calculation expected is comparing column heights of different liquids that balance the same pressure, as in the worked example above.

Ask the tutor: Curious why the barometer tube must be filled completely before inverting, or want to check your grasp of the Torricellian vacuum? Ask the AI tutor to walk through it step by step.