Atmospheric Pressure & Measurement

Pressure at the ground is the weight of everything stacked above each square metre of it. Standard sea-level pressure, p₀ = 101 325 Pa, therefore means p₀/g = 1.03 × 10⁴ kg of air on every square metre; spread over the whole Earth, that puts the atmosphere's mass near 5.3 × 10¹⁸ kg. You are not crushed because pressure pushes on every face of a surface and the fluids in your body push outward at nearly the same pressure: a thin sheet with air on both sides feels no net force at all. A force appears only across a difference. Pump the air out of a sealed box with a 0.30 m × 0.40 m lid and the lid carries 101 325 Pa × 0.12 m² = 1.2 × 10⁴ N, the weight of a 1.2-tonne car. Every instrument in this lesson works the same way: it registers a pressure difference, and the reading means nothing until you know the pressure on its other side.

Fill a long glass tube with mercury, invert it into a dish, and the column sinks until it stands about 0.76 m above the dish's surface, leaving an almost empty space at the top; mercury's vapour pressure at room temperature is a fraction of a pascal. Apply the equal-level rule: the open surface of the dish and the point inside the tube at the same height share one pressure. Outside, that is pₐₜₘ; inside, it is the near-zero pressure at the top plus the weight of the column, ρgh. So pₐₜₘ = ρgh, and the height alone measures absolute pressure. Only vertical height enters, so tilting or widening the tube leaves h unchanged. The liquid matters: mercury, 13.6 times as dense as water, gives a compact 760 mm, while a water barometer would stand 10.3 m tall, and about 0.24 m less in practice, because water vapour at 2.3 kPa fills the space above and pushes back.

Join one arm of a U-tube to a vessel and leave the other open. At the level of the liquid surface in the vessel's arm the pressure is p; at the same level in the open arm it is pₐₜₘ plus the weight of the liquid standing above that level. So p = pₐₜₘ + ρgΔh, with Δh positive when the open arm stands higher; the gas in the tubes adds nothing measurable, since air is over 800 times less dense than water. The instrument has shown only ρgΔh, the gauge pressure. If the vessel's arm stands higher, Δh is negative and the vessel holds a partial vacuum. Sensitivity follows from Δh = Δp/(ρg): a 2.0 kPa difference moves mercury 15 mm but water 204 mm, so water or light oil suits small differences and mercury suits large ones. Tilt the reading arm at θ to the horizontal and the liquid travels Δh/sin θ along it, 5.8 times as far at 10°.

Most dials are built like manometers and read zero when open to the air. A tyre gauge showing 230 kPa means 230 kPa above the atmosphere, 331 kPa absolute. A blood pressure of 120 mm Hg is 16.0 kPa above the pressure around the arm. A vacuum gauge at −60 kPa, on a day when the barometer reads 101.3 kPa, leaves 41 kPa absolute in the vessel, and no gauge reading can fall below −pₐₜₘ because absolute pressure cannot fall below zero. The physics decides which one to use. The net force on a wall with air on its far side needs gauge pressure; anything that depends on how many molecules are present, such as the gas laws, boiling points or a gas's density, needs absolute. And pₐₜₘ itself moves: weather shifts it by a few kPa, and a rigid bottle sealed at sea level reads 21 kPa on a gauge at an 80 kPa mountain site, although nothing inside it has changed.

The hydrostatic relation dp/dz = −ρg holds in air as it does in water; what fails is treating ρ as a constant. Do it anyway, with sea-level density ρ₀ = 1.23 kg m⁻³, and p = p₀ − ρ₀gz: pressure falls 12 Pa for every metre climbed and reaches zero at H = p₀/(ρ₀g) = 8.4 km. Locally this is a very good model. Across a 100 m tower the drop is 1.2 kPa, just 1.2% of p₀, and at about 36 Pa per 3 m storey a sensor that resolves a few pascals can tell which floor it is on. Upward it fails, and the failure shows without any better theory: the model puts the top of the atmosphere below the 8849 m summit of Everest and predicts about −5 kPa there, yet climbers breathe on the summit, where barometers read roughly a third of an atmosphere. A constant ρ cannot know that the air above is thinner than the air below.

For an ideal gas of molar mass M at temperature T, density follows pressure: ρ = pM/(RT). Put that into dp/dz = −ρg and the equation becomes dp/dz = −p/H, with H = RT/(Mg). Separate and integrate up from the ground, ∫dp/p = −∫dz/H, to get ln(p/p₀) = −z/H, so p = p₀e^(−z/H). At 288 K with M = 0.0290 kg mol⁻¹, H = 8.42 km, the same length as before, since ρ₀ = p₀M/(RT) makes p₀/(ρ₀g) and RT/(Mg) identical. The straight line is the exponential's tangent at the ground: they agree within 1% up to 1.1 km, are 8% apart at 3 km, and at Everest's height give 35.4 kPa against −5.2 kPa. This model has a limit too. Real air cools by roughly 6.5 K per kilometre through the lowest 11 km, so its pressure falls a little faster; the standard atmosphere gives 31 kPa at 8849 m.