Atmospheric Pressure and Barometers

In short

The atmosphere is a vast layer of air — about 100 km thick — that presses down on everything at its base with a pressure of about 1.013 \times 10^5 Pa at sea level, called one atmosphere (1 atm). Torricelli's mercury barometer measures this by letting atmospheric pressure hold up a column of mercury in an evacuated glass tube; the column height works out to exactly h = P_{\text{atm}} / (\rho_{\text{Hg}} g) \approx 760 mm. The aneroid barometer achieves the same measurement without liquid, using the deformation of a sealed, partially evacuated metal chamber. Atmospheric pressure decreases with altitude roughly as P(y) = P_0 \exp(-y/H), where H \approx 8.4 km is the scale height — the reason mountaineers on Everest breathe from oxygen cylinders and your ears pop on a flight over the Himalayas.

Torricelli sealed one end of a metre-long glass tube, filled it with mercury, and inverted it into a dish of more mercury. The column did not fall all the way. It stopped at a height of about 76 centimetres — and the empty space above it was, for all practical purposes, a vacuum. What was holding the mercury up?

The answer was the atmosphere. The air pressing down on the mercury in the dish pushed mercury up the tube until the weight of the supported column matched the pressure of the surrounding air. This one experiment, done by Evangelista Torricelli in 1643, did two things at once: it produced the first barometer (a device for measuring atmospheric pressure) and it demonstrated that the atmosphere had weight — a physical reality that could be weighed on a scale.

This article is about that weighing. You will derive, from the hydrostatic formula you already know, why the column is exactly 760 mm; understand how a modern aneroid barometer achieves the same measurement without spilling mercury; build the barometric formula that tells you how pressure falls with altitude; and use all of it to answer practical questions — why pressure cookers cook faster, why altimeters in aircraft are just barometers with relabelled dials, and how a sudden drop in barometric pressure warns a Kerala fisherman of an approaching cyclone.

The weight of the atmosphere

Take a one-square-metre patch of ground at Marina Beach, Chennai (sea level, near enough). Above that patch, stretching up through the troposphere, through the stratosphere, and thinning out into space, is a column of air. It is mostly nitrogen and oxygen, and it is pulled down by gravity. Every molecule in that column has weight, and the total weight of the column presses down on your one square metre.

The result of that pressing is atmospheric pressure — force per unit area on the ground. By measurement, the total weight of air in a 1 m² column at sea level is about 1.013 \times 10^5 newtons, giving an atmospheric pressure of

P_{\text{atm}} = 1.013 \times 10^5 \text{ Pa}.

This is the standard atmosphere, abbreviated 1 atm. The "standard" is a conventional reference value defined once and for all; the actual atmospheric pressure at any given place and time varies by a few percent with weather and altitude, but sea-level pressures cluster near 1 atm.

How heavy is the air above you?

Pick one square metre. The atmospheric pressure there is 10^5 Pa, which means the column of air above it pushes down with 10^5 newtons — about 10 tonnes. A familiar square metre (a small table top) is supporting the weight of a small truck at every instant.

The reason you do not feel it is that the same pressure pushes up from below (through the table), from the sides (against every cell wall in your body), and at every surface in between. Your eardrum feels the difference when pressure is unequal on its two sides — at the bottom of a swimming pool, or in a lift descending quickly — not the absolute value. Evolution built your body in equilibrium with one atmosphere; the body internal cavities are at one atmosphere, and it is pressure changes that register as pain.

Why the column has a finite height

The atmosphere is not a solid block of fluid; it is a gas whose density falls off with altitude. The column does end — effectively at a height around 100 km, by convention — because there are only finitely many molecules of air per square metre above sea level. The total mass of atmosphere over your 1 m² patch, integrated over that whole height, comes out to about 10^4 kg — hence the 10-tonne weight and the 10^5 Pa pressure.

If the atmosphere had uniform density equal to the sea-level value (\rho_0 \approx 1.225 kg/m³), the column height would have to be

H_{\text{eff}} = \frac{P_{\text{atm}}}{\rho_0 g} = \frac{1.013 \times 10^5}{1.225 \times 9.8} \approx 8.4 \text{ km}.

This is the scale height of the atmosphere — the height of a fictitious uniform-density atmosphere that would produce the observed sea-level pressure. The real atmosphere extends much further, but because density falls off with altitude, most of the mass is concentrated below 10 km or so. The scale height sets the natural vertical length in every atmospheric calculation.

The mercury barometer

A glass tube, sealed at the top and evacuated above the mercury, stands inverted in a dish of mercury. Atmospheric pressure on the mercury in the dish pushes mercury up the tube. Equilibrium is reached when the weight of the supported mercury column matches the atmospheric pressure force.

Here is how Torricelli's reasoning works, translated into the hydrostatic formula from the previous chapter.

Setup. Take a glass tube longer than 800 mm, closed at one end. Fill it completely with mercury. Holding a finger over the open end, invert the tube and dip the open end into a dish of mercury. Remove your finger. Mercury falls from the tube until the column stabilises at a certain height above the dish. The space above the mercury in the tube is a vacuum (apart from a tiny bit of mercury vapour that can be neglected).

Step 1. Identify two points at the same height and apply the same-level rule from the previous chapter. Pick point A at the mercury surface inside the dish, and point B at the mercury inside the tube, at the same height as A. Because A and B are in a connected fluid at rest and at the same level, they have the same pressure:

P_A = P_B.

Why: two points at the same height in a connected static fluid are at the same pressure, regardless of the path between them — this is the key result of hydrostatics.

Step 2. Compute P_A. Point A is at the open mercury surface in the dish. The air directly above it pushes down with atmospheric pressure:

P_A = P_{\text{atm}}.

Step 3. Compute P_B. Point B is inside the mercury in the tube, at the dish-surface height. Above B is a column of mercury of height h (extending up to the top of the column), and above that is the vacuum at the top of the tube. Using the hydrostatic formula, the pressure at B is

P_B = P_{\text{vacuum}} + \rho_{\text{Hg}} g h = 0 + \rho_{\text{Hg}} g h = \rho_{\text{Hg}} g h.

Why: the pressure at any depth below a free surface is the surface pressure plus \rho g \times (\text{depth below that surface}). The "surface" above point B is the top of the mercury column, and the pressure there is essentially zero because a vacuum sits above it.

Step 4. Set P_A = P_B and solve for h.

P_{\text{atm}} = \rho_{\text{Hg}} g h \quad\Longrightarrow\quad h = \frac{P_{\text{atm}}}{\rho_{\text{Hg}} g}.

Step 5. Plug in numbers. Mercury has \rho_{\text{Hg}} = 13{,}600 kg/m³, g = 9.8 m/s², P_{\text{atm}} = 1.013 \times 10^5 Pa:

h = \frac{1.013 \times 10^5}{13{,}600 \times 9.8} = \frac{1.013 \times 10^5}{1.333 \times 10^5} \approx 0.760 \text{ m} = 760 \text{ mm}.

This is the entire Torricelli result. A mercury column at sea level, supported only by atmospheric pressure acting on the reservoir dish, stands 760 mm tall. The height depends on the mercury density, the local gravity, and the atmospheric pressure, and changes in any of those three shift the column. Of the three, atmospheric pressure varies the most over time — and that is what makes this device a meter for atmospheric pressure.

Why mercury specifically?

A barometer that used water instead of mercury would need a much taller tube. Setting P_{\text{atm}} = \rho_{\text{water}} g h with \rho_{\text{water}} = 1000 kg/m³ gives

h = \frac{1.013 \times 10^5}{1000 \times 9.8} \approx 10.34 \text{ m}.

A 10-metre-tall tube of water — four floors of a building — to measure one atmosphere. Mercury, being 13.6 times denser, shrinks the required column by a factor of 13.6, to a conveniently tabletop-sized 76 cm. Mercury is also dense enough that its surface vapour pressure is tiny and can be ignored, and it does not wet glass (so the meniscus is cleanly shaped — see Surface Tension and Capillarity).

Reading the barometer

The height of the column is read off a scale etched in millimetres alongside the tube. At sea level on a standard day, a well-made mercury barometer reads close to 760 mm. On a high-pressure day (cold, clear, sinking air) it might read 775 mm. On a low-pressure day preceding a storm, it might read 735 mm. Changes of 1–2 mm Hg over a few hours are typical weather indicators.

This is the origin of mmHg and torr as units of pressure. Both are defined so that "1 mmHg = 1 torr = the pressure of a 1-millimetre-tall column of mercury" — and if you multiply by \rho_{\text{Hg}} g you get 1 \text{ mmHg} = 133.3 Pa, the SI value.

The aneroid barometer

The mercury barometer is accurate and historically important, but it has obvious problems: mercury is toxic, the tube is fragile, and the thing cannot be carried up a mountain without spilling. The aneroid barometer (Greek a-neros, "no liquid") solves all three.

The aneroid capsule (evacuated, with a spring) flexes in response to changes in external pressure. A lever system amplifies the tiny deflection into a large pointer rotation on a calibrated dial.

The working part is a small sealed metal chamber — called a Vidi capsule or aneroid cell — evacuated to a low pressure and shaped like a thin corrugated disc. Because air inside the capsule is near vacuum, external atmospheric pressure tries to squash it flat; a stiff metal spring inside prevents total collapse. The equilibrium shape of the capsule is set by the balance between external atmospheric pressure and the spring's restoring force.

When the external pressure increases, the capsule is squeezed a little more, thinning by perhaps a few micrometres. When the external pressure decreases, the spring pushes the capsule back out. These micrometre deformations are amplified by a mechanical linkage — a system of levers and a fine chain wrapped around a pointer spindle — into a visible swing of a needle across a calibrated dial. The dial is marked directly in hectopascals (hPa), millibars (mbar), or inches of mercury, depending on the instrument.

Why the dial is sensitive

The aneroid barometer is sensitive because the lever linkage can multiply a micrometre-scale capsule deflection into a centimetre-scale pointer swing, and the pointer is read against a finely divided dial. A good pocket aneroid can resolve pressure changes of ~0.5 hPa (~0.4 mmHg), and a precision aircraft altimeter can do much better.

The trade-off is calibration. A mercury barometer is calibrated by the laws of physics — you measure the height of mercury and multiply by \rho_{\text{Hg}} g. An aneroid barometer has to be calibrated against a mercury barometer (or a deadweight pressure tester) because the spring constant of the capsule, the geometry of the linkage, and the exact vacuum level inside all affect the reading. Cheap aneroids drift over time and need periodic recalibration.

Altimeters — aneroids in disguise

An aircraft altimeter is nothing more than an aneroid barometer with the dial relabelled. Instead of reading "pressure in hPa," it reads "altitude in metres or feet," using the known relationship between altitude and pressure (the barometric formula, below). As you climb, the external pressure falls, the capsule expands a bit, and the pointer rotates to a higher altitude reading. Before landing, the pilot adjusts the altimeter's sub-dial to the local atmospheric pressure on the ground ("QNH" setting), which realigns the dial so that touchdown reads the airfield elevation.

Hikers, mountaineers, and even some smartphones (via built-in barometric sensors) use exactly the same idea.

How pressure varies with altitude

If the atmosphere had uniform density, pressure would fall linearly with altitude and hit zero at a clean boundary — the top of the atmosphere. Real gases do not behave that way; they compress under their own weight, so the density at the bottom is higher than the density at the top. Pressure falls off exponentially with altitude instead.

The barometric formula

Start from hydrostatic equilibrium (derived in the previous chapter), written with y measuring altitude upward:

\frac{dP}{dy} = -\rho g.

Why: the minus sign is because pressure decreases as you go up. (In the earlier chapter y was depth, so dP/dy was positive; here y is altitude, so the sign flips.)

For a gas, density depends on pressure via the ideal gas law: PV = nRT gives

\rho = \frac{m}{V} = \frac{M P}{R T},

where M is the molar mass of air (~0.029 kg/mol), R = 8.314 J/(mol·K) is the universal gas constant, and T is the temperature. Substituting:

\frac{dP}{dy} = -\frac{Mg}{RT} P.

Why: this is now a differential equation relating P to its own derivative, with the coefficient -Mg/(RT) that sets the characteristic length. The solution is an exponential.

Assuming T is constant (the isothermal atmosphere approximation), this integrates to

\boxed{\; P(y) = P_0 \, e^{-y/H}, \qquad H = \frac{RT}{Mg}. \;}

The length H is the scale height of the atmosphere — the altitude at which pressure falls to 1/e (about 37%) of its sea-level value.

Plugging in numbers

Use T = 288 K (15 °C, roughly sea-level average), M = 0.029 kg/mol, R = 8.314 J/(mol·K), g = 9.8 m/s²:

H = \frac{8.314 \times 288}{0.029 \times 9.8} = \frac{2394}{0.284} \approx 8400 \text{ m} = 8.4 \text{ km}.

This matches the scale height we found earlier from P_{\text{atm}}/(\rho_0 g) — that was no accident.

So at an altitude of 8.4 km, atmospheric pressure is down to P_0 / e \approx 0.37 \, P_0 \approx 0.37 atm. At 16.8 km (two scale heights), P_0 / e^2 \approx 0.14 \, P_0. At Mount Everest's summit (8.85 km), the formula predicts about 0.35 \, P_0 \approx 0.35 atm, which is close to the measured value (actual ~0.33 atm; the real atmosphere is cooler at altitude, which refines the formula).

Atmospheric pressure falls exponentially with altitude. The scale height $H \approx 8.4$ km is the altitude where pressure has decayed to $1/e$ of sea-level pressure. Indian landmarks on the curve: most of the country is essentially at sea level; Shimla and Leh sit noticeably higher; Everest's summit is at a third of sea-level pressure.

The isothermal approximation — where it breaks

The barometric formula assumes constant temperature throughout the column. The real troposphere cools at about 6.5 °C per kilometre of altitude (the standard lapse rate). Including this correction gives a slightly sharper pressure fall-off — at Everest's summit, the isothermal model predicts 0.35 atm, but the real value is about 0.33 atm, because the air there is much colder than at sea level.

For JEE problems, use the isothermal formula P = P_0 e^{-y/H} with H \approx 8 km. It is right to within a few percent in the lower atmosphere.

Worked examples

Example 1: Using a mercury barometer to measure pressure at Shimla

A mercury barometer in Shimla (altitude ≈ 2200 m, temperature 15 °C) reads a column height of 583 mm. The mercury density is 1.36 \times 10^4 kg/m³, and g = 9.8 m/s². (a) Find the atmospheric pressure at Shimla. (b) Compare with the isothermal prediction P(y) = P_0 e^{-y/H} using P_0 = 1.013 \times 10^5 Pa, H = 8.4 km.

A mercury barometer at Shimla's altitude reads a shorter column than at sea level, because the overlying atmosphere is thinner.

Step 1. Use the barometer equation P = \rho_{\text{Hg}} g h.

Convert h to SI: h = 583 mm = 0.583 m.

P = (1.36 \times 10^4)(9.8)(0.583) = 7.77 \times 10^4 \text{ Pa}.

Why: a mercury barometer converts a column height directly into pressure via the hydrostatic formula, because the vacuum above the column means the entire pressure at the dish-level point comes from the weight of the column.

Step 2. Convert to atmospheres for easy comparison.

P = \frac{7.77 \times 10^4}{1.013 \times 10^5} \approx 0.767 \text{ atm}.

Step 3. Predict the same pressure from the barometric formula.

P(2200 \text{ m}) = 1.013 \times 10^5 \times \exp\left(-\frac{2200}{8400}\right) = 1.013 \times 10^5 \times e^{-0.2619}.

Using e^{-0.2619} \approx 0.7696:

P \approx 1.013 \times 10^5 \times 0.770 \approx 7.80 \times 10^4 \text{ Pa} \approx 0.770 \text{ atm}.

Why: the exponent -y/H = -2200/8400 = -0.262 describes how many scale-heights above sea level you are. Being 0.26 scale-heights up reduces pressure by a factor of e^{-0.26} \approx 0.77, matching the measured reading closely.

Step 4. Compare.

Measured: P = 7.77 \times 10^4 Pa. Predicted: P = 7.80 \times 10^4 Pa. Agreement within 0.4%.

Result: The atmospheric pressure at Shimla is about 7.77 \times 10^4 Pa (~77% of sea-level pressure, or ~583 mmHg), in good agreement with the isothermal barometric formula.

What this shows: A barometer is a direct-reading pressure meter — the column height multiplied by \rho_{\text{Hg}} g gives the pressure without any calibration curve. The isothermal barometric formula matches such readings to within a few percent in the lower atmosphere, which is why meteorology, aviation, and altimetry rely on it.

Example 2: Why a pressure cooker cooks dal faster in Leh than in a regular pot in Chennai

In Chennai (sea level, P_{\text{atm}} = 1.013 \times 10^5 Pa), you cook dal in an open pot. The water boils at 100 °C. In Leh (altitude 3500 m), atmospheric pressure is lower and water boils at a lower temperature. A pressure cooker in Leh raises the internal pressure by 1 atmosphere above the local atmospheric pressure. Find (a) the atmospheric pressure at Leh, (b) the absolute pressure inside the cooker in Leh, and (c) comment on whether the cooker compensates for the altitude.

The safety valve on an Indian pressure cooker lifts when the internal pressure exceeds the external by a fixed amount (typically 1 atm = 100 kPa). So the gauge pressure is fixed regardless of altitude, but the absolute pressure depends on the local atmosphere.

Step 1. Atmospheric pressure at Leh.

Use the isothermal barometric formula with y = 3500 m, H = 8400 m:

P_{\text{atm, Leh}} = 1.013 \times 10^5 \times \exp\left(-\frac{3500}{8400}\right) = 1.013 \times 10^5 \times e^{-0.417}.

With e^{-0.417} \approx 0.659:

P_{\text{atm, Leh}} \approx 1.013 \times 10^5 \times 0.659 \approx 6.67 \times 10^4 \text{ Pa} \approx 0.66 \text{ atm}.

Step 2. Absolute pressure inside the cooker.

The cooker's valve is rated at a gauge pressure of 1 atm (typical for Indian kitchen pressure cookers). The absolute pressure inside is

P_{\text{in}} = P_{\text{gauge}} + P_{\text{atm, Leh}} = 1.013 \times 10^5 + 6.67 \times 10^4 = 1.68 \times 10^5 \text{ Pa} \approx 1.66 \text{ atm}.

Why: the valve holds the internal pressure at a fixed excess over the outside. In Leh the outside is lower, so the absolute inside pressure is lower than it would be in Chennai (where the same cooker would run at 2 atm absolute), but both are well above 1 atm — so water inside boils well above 100 °C in either location.

Step 3. Does the cooker compensate?

The absolute pressure inside the Leh cooker (1.66 atm) is higher than the sea-level atmospheric pressure in Chennai (1 atm). Water inside the Leh cooker boils at the temperature where its saturation vapour pressure equals 1.66 atm, which (from steam tables) is about 115 °C. By contrast, water in an open pot in Leh boils at about 88 °C (because atmospheric pressure is only 0.66 atm, and water's vapour pressure reaches 0.66 atm at ~88 °C).

Result: The cooker in Leh operates at an absolute pressure of about 1.68 \times 10^5 Pa (1.66 atm). This pushes the boiling point of water inside the cooker to ~115 °C — a higher cooking temperature than even an open pot in Chennai. So the cooker more than compensates for the altitude.

What this shows: A pressure cooker is precisely an altitude-correction device. At high altitude, an open pot cannot reach 100 °C (dal takes forever), but a cooker raises the internal absolute pressure above atmospheric by a fixed amount regardless of altitude, pushing the boiling point to 110–120 °C everywhere. This is why Indian cookers are sold the same way in Chennai and Ladakh — the engineering works identically on both sides of the country.

Common confusions

"Atmospheric pressure pushes only downward." No — atmospheric pressure is scalar, acting equally in all directions on any surface at that altitude. It pushes up on the bottom of your hand just as hard as it pushes down on the top. The net force on a free body sitting on a flat table is zero — the equal-and-opposite pushes from all sides cancel. This is the same wedge-argument point as in Pressure in Fluids.
"The column in a mercury barometer depends on tube width." It does not. Take a barometer tube 1 cm wide or 3 cm wide — both settle to 760 mm at sea level. The hydrostatic formula P = \rho g h has no area dependence. A wider tube holds more total mercury, but the pressure at the dish level is the same.
"A barometer measures the vacuum, not pressure." A barometer measures atmospheric pressure by using a vacuum. The vacuum is the reference — the thing with pressure zero — and the mercury column rises until the weight of the supported mercury just matches atmospheric pressure. Without the vacuum at the top, the reading would be contaminated by whatever gas was there.
"If you drop the barometer at sea level, the reading becomes zero while it falls." Interesting question. During free fall, the apparent gravitational acceleration on the mercury is zero (the barometer is in free fall together with its contents), so \rho g h evaluates to zero — the mercury would rise up the tube and refill the vacuum as long as the fall lasts. In free fall, a mercury barometer reads zero regardless of the actual atmospheric pressure.
"Atmospheric pressure and wind pressure are the same." They are not. Atmospheric pressure is the static hydrostatic pressure of the atmosphere on everything — about 10^5 Pa at sea level. Wind pressure (dynamic pressure) is \tfrac{1}{2} \rho v^2, the kinetic-energy-per-volume of moving air, and it is typically only a few hundred pascals even in a storm. The two are related by Bernoulli's equation, not identical.
"A pressure cooker is just a heavier pot." No — a pressure cooker uses a sealed lid with a safety valve to trap steam. As water heats, the steam would normally escape; trapped, it raises the internal absolute pressure. The valve ensures the pressure does not keep rising indefinitely (it opens at a set value). The physics is thermodynamic: higher pressure raises the boiling point of water, which lets you cook at >100 °C.

If you have the barometer derivation, the aneroid mechanism, and the basic exponential pressure fall-off, you have everything a JEE problem can throw at you. What follows is a more careful look at the barometric formula (including the temperature-varying atmosphere), a derivation of the scale height from statistical mechanics, and a connection to why aeroplane cabins are pressurised.

The barometric formula with a varying temperature

The standard troposphere cools with altitude at a rate of about L = 6.5 K/km. Using T(y) = T_0 - L y and the gas law, the hydrostatic equation becomes

\frac{dP}{dy} = -\frac{Mg}{R(T_0 - L y)} P.

This is separable. Integrating both sides:

\int \frac{dP}{P} = -\frac{Mg}{R} \int \frac{dy}{T_0 - L y}.

The right-hand integral evaluates to (1/L) \ln(T_0 - L y) with a sign, and a bit of algebra gives the standard atmosphere formula:

P(y) = P_0 \left(1 - \frac{L y}{T_0}\right)^{Mg/(RL)}.

Plugging in L = 0.0065 K/m, T_0 = 288 K, M = 0.029 kg/mol, g = 9.8, R = 8.314, the exponent Mg/(RL) comes out to about 5.26. This is the formula civil aviation altimeters use below 11 km. Above 11 km, the temperature in the real atmosphere becomes roughly constant (the stratosphere), and an isothermal formula takes over.

For JEE, stick with the isothermal version — it is easier and good to a few percent. The varying-temperature formula is for engineering-grade altimetry.

The scale height from the equipartition theorem

The scale height H = RT/(Mg) = k_B T / (m g), where m = M/N_A is the mass of one air molecule, has a clean physical interpretation in statistical mechanics. Consider one air molecule in the atmosphere. The potential energy to lift it to altitude y is m g y. By the Boltzmann distribution, the probability of finding a molecule at altitude y (at temperature T) scales as \exp(-m g y / k_B T).

P(y) \propto \exp\left(-\frac{m g y}{k_B T}\right) = \exp\left(-\frac{y}{H}\right), \quad H = \frac{k_B T}{m g}.

So the scale height is the altitude at which the gravitational potential energy of a molecule equals the typical thermal energy k_B T. Below that altitude, thermal agitation dominates and molecules are well-mixed; above it, the exponential suppression takes over. This is the microscopic view of the barometric formula: the atmosphere is a dilute thermal gas, and its pressure profile is just the Boltzmann distribution of one molecule in a gravitational field.

Aircraft cabin pressurisation

A commercial aircraft cruises at ~11 km altitude, where the atmospheric pressure is about 0.22 atm (from P_0 e^{-11/8.4} \approx 0.27; the real number is slightly lower because of the real temperature profile). At that pressure, the partial pressure of oxygen is too low for sustained consciousness — humans black out within minutes. Cabins are therefore pressurised. The standard pressurisation brings the cabin "altitude" to about 2400 m (Shimla-equivalent), which corresponds to an internal absolute pressure of about 0.76 atm.

The pressurisation system works by bleeding compressed air off the turbine compressor stages, cooling it, and routing it into the cabin; an outflow valve regulates the cabin-outside pressure differential. Structurally, the fuselage must withstand this differential: at cruise, P_{\text{cabin}} - P_{\text{outside}} \approx 0.76 - 0.22 = 0.54 atm \approx 5.5 \times 10^4 Pa. Over the area of a fuselage window (say 0.1 m²), this is about 5500 N of net outward force — which is why the windows are rounded (to distribute stress) and triple-layered.

The barometric formula is, in other words, a working engineering equation. It is not just an exam problem.

Why the atmosphere does not escape to space

The scale height answer to this question is the Jeans escape criterion. A gas escapes Earth's gravity if its typical molecular speed exceeds the escape velocity v_{\text{esc}} = \sqrt{2 g R_E} \approx 11 km/s. The typical thermal speed of a nitrogen molecule at 300 K is v_{\text{th}} = \sqrt{3 k_B T / m} \approx 0.5 km/s — much less than v_{\text{esc}}. Only a tiny fraction of the Maxwell distribution reaches escape speed, and that fraction determines the atmospheric loss rate.

For hydrogen and helium, v_{\text{th}} is much larger (because m is smaller), and Earth does lose these gases over geological time. That is why the atmosphere is overwhelmingly N₂ and O₂, with only trace hydrogen and helium — the lighter gases have mostly leaked away.

Where this leads next

Pressure in Fluids — the hydrostatic formula P = P_0 + \rho g h on which the barometer argument was built.
Pascal's Law and Hydraulic Machines — how the same "pressure is transmitted equally" idea powers hydraulic brakes and lifts.
Archimedes' Principle and Buoyancy — the upward force on a submerged body, derived from the pressure difference across its top and bottom surfaces.
Kinetic Theory of Gases — the microscopic picture where atmospheric pressure is the cumulative impact of gas molecules against a surface.
Thermodynamics of Phase Change — why the boiling point of water depends on pressure, closing the loop on the pressure-cooker argument.