The Atmosphere: Pressure, Temperature and Density with Height - Physics for Pilot Assessments

The atmosphere is the largest and most familiar application of everything in this module, and it is where the gas laws stop being laboratory exercises. By the end of this lesson you should be able to explain why air pressure falls with height and why the fall is not a straight line, say what a lapse rate is and quote the standard one, work out how density follows from pressure and temperature together, and define the International Standard Atmosphere, QNH and density altitude in terms of the physics behind them. This lesson stops at the physics and the definitions: which altimeter setting is used at which stage of a flight, and how performance figures are read off a chart, are operational matters for a flying course rather than a physics one.

Why pressure falls with height

You are at the bottom of an ocean of air, and the reason it presses on you is the same reason water presses on a diver: the weight of everything stacked above. Climb, and there is less air above you to be supported, so the pressure is less. That is the whole of the qualitative explanation, and it is worth noticing that it makes no reference to the air being thin or cold or moving — only to how much of it is overhead. For a liquid, the matter and fluids module turned that argument into a formula, and it is worth having in front of you here because the comparison with air is the point of the next section.

Pressure due to the fluid above — for a fluid of constant density

p = \rho g h

$p$: the pressure due to the fluid above, in pascals (Pa)
$\rho$: the density of the fluid, in kilograms per cubic metre (kg/m³)
$g$: gravitational field strength, about 10 newtons per kilogram (N/kg) on Earth
$h$: the depth below the surface, in metres (m)

Read the conditions attached to that formula carefully, because they are what fails in air: it assumes a single fixed density all the way down.

Why the fall is not linear

A liquid is very nearly incompressible, so its density is the same near the surface as it is at the bottom, and the formula above then says that pressure grows in a straight line with depth — go twice as deep and the pressure due to the liquid doubles. Air is a gas, and from the states-of-matter material a gas has a great deal of empty space between its particles and can therefore be squeezed. So the air near the ground is compressed by the weight of everything above it and is dense, while the air high up has little above it and is thin. There is no single density to put into the formula, and there is no sharp surface to measure a depth from either.

What follows from that is a fall which is steep at the bottom and gentler higher up. Climbing the first thousand feet takes you above a thousand feet of dense air and removes a lot of weight from your shoulders; climbing from 30,000 to 31,000 feet takes you above a thousand feet of thin air and removes far less. Each successive slice of altitude therefore removes a smaller amount of pressure than the one below it — but a roughly constant fraction of what is left, because the amount of air in a slice is itself proportional to how much is pressing on it. A constant fraction removed per unit of height is what produces the characteristic curve, and the practical form of it is that the pressure roughly halves for every 18,000 feet of altitude.

The dashed line is what a fluid of fixed density would do, and the curve is what air actually does. They agree near the ground, which is why the rule of thumb of about 27 feet per hectopascal works at low level and nowhere else.

That gives you two independent ways to describe the same fall, both worth carrying. Near sea level the gradient is about one hectopascal for every 27 feet of altitude, a figure introduced with the aviation units in the mechanics module; the hectopascal, also worth restating, is a hundred pascals, and standard sea-level pressure of 1013 hectopascals is the same pressure as 1013 millibars or 29.92 inches of mercury. Higher up that gradient stretches out, so at 18,000 feet it takes roughly 50 feet to lose a hectopascal, and higher still more again. The halving rule is the same statement in a form that survives to altitude: 1013 hectopascals at sea level, about 500 at 18,000 feet, about 250 at 36,000 feet.

Temperature falls with height, and the lapse rate

Temperature also falls as you climb, through most of the lower atmosphere, and there are two reasons worth giving. The first is where the heating comes from: sunlight passes through the air with relatively little absorption, is absorbed by the ground, and warms the air from below — so the atmosphere is heated at its base, and the further you are from that source the cooler it is. The second is the gas behaviour from the previous lesson. Air that rises finds itself surrounded by lower pressure, so it expands; expanding means pushing the surrounding air outwards, which is doing work on it, and that work is paid for out of the rising air's own internal energy. Its particles therefore end up slower than they were, and slower particles mean a lower temperature. No heat was removed from the air at all — it cooled because it expanded.

The rate at which temperature falls with height is called the lapse rate, and it is quoted in degrees per thousand feet. Its real value varies constantly — with the season, the time of day, the weather and the height band — so any single figure has to be an agreed convention rather than a measurement. The standard figure is 1.98 °C per thousand feet, universally worked with as 2 °C per thousand feet, starting from 15 °C at sea level and continuing up to 36,090 feet, where the temperature reaches −56.5 °C. Above that height, in the standard model, the temperature is taken as constant; the boundary itself is called the tropopause.

Standard temperature at altitude

T = 15 - 2h

$T$: the standard temperature at that altitude, in degrees Celsius (°C)
$15$: the standard sea-level temperature, in degrees Celsius (°C)
$2$: the standard lapse rate, in degrees Celsius per thousand feet (°C per 1000 ft) — strictly 1.98
$h$: the altitude, in thousands of feet — so 20,000 ft is h = 20

That relationship is only valid up to the tropopause, since above it the standard temperature stops falling, and it describes a convention rather than the sky outside the window.

Density follows from pressure and temperature together

Density is where the two previous sections meet, and the link comes straight out of the combined gas law. For a fixed mass of gas, pressure times volume divided by temperature is a constant. Density is mass divided by volume, so volume is mass divided by density — and substituting that in puts the density into the relationship: pressure times mass, divided by density times temperature, is constant. The mass is fixed, so it can be absorbed into the constant, leaving the statement that pressure divided by the product of density and temperature is fixed. Rearranged for the density, that says the density is proportional to the pressure and inversely proportional to the absolute temperature.

Air density from pressure and temperature

\rho \propto \frac{p}{T}

$\rho$: the density of the air, in kilograms per cubic metre (kg/m³)
$\propto$: "is proportional to" — halve the right-hand side and the density halves
$p$: the pressure, in pascals (Pa) or any consistent unit
$T$: the absolute temperature, in kelvin (K)

Two consequences follow, and between them they account for almost everything that matters about air density. Lower the pressure at a fixed temperature and the density falls in proportion, because the same particles are spread through more space. Raise the temperature at a fixed pressure and the density falls, because the particles have to spread further apart to keep striking the walls just as hard — which is Charles's law from the previous lesson, seen from the density end. On a climb the two effects pull against each other: the pressure falls, which thins the air, while the temperature falls, which would thicken it. The pressure effect wins comfortably, so density does fall with height — but more slowly than pressure does, which is why the density has only halved by about 22,000 feet while the pressure has halved by 18,000.

The International Standard Atmosphere

The real atmosphere is different every day, at every place and at every hour, and that is a problem for anyone who has to calibrate an instrument or compare one set of measurements with another. The solution is an agreed fiction: a single set of values that everybody uses as a reference, called the International Standard Atmosphere, or ISA. It fixes sea-level pressure at 1013.25 hectopascals, sea-level temperature at 15 °C, sea-level density at 1.225 kilograms per cubic metre, and the lapse rate at 1.98 °C per thousand feet up to 36,090 feet, above which the temperature is held at −56.5 °C.

The important thing about that list is what kind of thing it is. It is not a description of today's weather, not an average of measurements you could go and check, and not a prediction. It is a convention, adopted so that an altimeter built in one country reads the same as one built in another, and so that a performance figure quoted for an aircraft means something definite. The real atmosphere agrees with it only by coincidence, and the standard way of describing a real day is as a departure from it — a day 10 degrees warmer than the standard value for that altitude is described as ISA plus 10.

What an altimeter setting is doing, and what QNH means

An altimeter is not a height-measuring instrument. It is a barometer — an instrument that measures air pressure, of the kind described in the matter and fluids module — whose dial happens to be marked in feet instead of hectopascals, with the markings placed according to the standard atmosphere's relationship between pressure and height. It therefore reads a pressure and reports the altitude at which that pressure would be found in the ISA. It has no way of knowing anything else about the air around it.

That creates an obvious difficulty. The conversion from pressure to height needs a starting point — a pressure to call zero feet — and today's sea-level pressure is almost never 1013 hectopascals. If the instrument assumes 1013 while the real figure is 1003, every reading is out by ten hectopascals' worth of altitude, which near the ground is about ten times 27 feet, or 270 feet. So an altimeter carries a small adjustable subscale on which a reference pressure is set, and setting it shifts the whole scale to match the datum you have chosen.

QNH is the name of one of those settings, and specifically it is the pressure value which, when set on the subscale, makes the altimeter read altitude above mean sea level — so that an aircraft standing on the ground at an airfield reads that airfield's elevation. It is obtained by measuring the actual pressure at the station and correcting it to what the pressure would be at mean sea level, using the standard atmosphere to do the correction. That is why QNH is a number in hectopascals rather than a height, and why it changes through the day as the real pressure changes. Which setting is appropriate at which stage of a flight, the other settings that exist and the rules governing them are operational procedure, and they belong to a flying course rather than to this one.

Density altitude

Density altitude is the last idea in the lesson and it is nothing more than the density relationship above given a convenient name. Since the density of air depends on both its pressure and its temperature, two places at the same pressure can have quite different densities if one of them is hotter. Density altitude is defined as the altitude in the standard atmosphere at which the air density equals the density you have actually got. It is not where you are; it is where the standard atmosphere would have to put you to be this thin.

Both of its ingredients push the same way on a hot, high day. Being high means a lower pressure, which lowers the density directly. Being hot means a higher absolute temperature, which lowers the density again. Combine them and the air can easily be as thin as the standard atmosphere's air several thousand feet above the ground you are standing on — which is exactly what the name is saying. The physics consequence is simply that every cubic metre of air contains fewer particles, so anything that has to work on a mass of air per second, from a propeller to an engine's air intake, has less mass available to work on.

Example 1 — the halving rule, both ways

Standard sea-level pressure is 1013 hectopascals, and pressure roughly halves every 18,000 feet. Estimate the pressure at 18,000 feet and at 36,000 feet. Then estimate the altitude at which the pressure is about 125 hectopascals.

At 18,000 feet the pressure has halved once: $\frac{1013}{2}$ is about 506, so roughly 500 hectopascals.
At 36,000 feet it has halved twice, because 36,000 is two lots of 18,000: $\frac{500}{2} = 250\ \text{hPa}$
For the reverse, count the halvings needed to get from about 1000 down to 125. 1000 to 500 is one, 500 to 250 is two, 250 to 125 is three.
Three halvings at 18,000 feet each is 3 × 18,000 = 54,000 feet.
Notice how much altitude the later halvings buy you. The first 18,000 feet cost about 500 hectopascals and the third only about 125 — the same fraction each time, but a much smaller amount, which is the non-linearity in one sentence.

Example 2 — the standard lapse rate, both ways

Using the standard atmosphere, what is the temperature at 20,000 feet? And at what altitude does the standard temperature reach −15 °C?

Use standard temperature = 15 − 2 × (altitude in thousands of feet). At 20,000 feet the altitude figure is 20.
2 × 20 = 40, and 15 − 40 = −25 °C.
For the reverse, set the temperature to −15 and solve for the altitude: $-15 = 15 - 2h$ Subtract 15 from both sides to get −30 = −2h, then divide both sides by −2, which gives h = 15.
So the standard temperature is −15 °C at 15,000 feet.
Both answers are below the tropopause at 36,090 feet, so the lapse rate applies. Above that height the standard temperature stops falling and this arithmetic no longer works — at 45,000 feet it is still −56.5 °C, not the −75 °C the formula would claim.

Example 3 — heat the air and the density falls

Air at a fixed pressure is at the standard sea-level temperature of 15 °C, with a density of 1.225 kilograms per cubic metre. The temperature rises to 45 °C with no change in pressure. What happens to the density?

Convert both temperatures to kelvin, because the relationship is a proportionality: $15 + 273 = 288\ \text{K} \quad\text{and}\quad 45 + 273 = 318\ \text{K}$
Density is proportional to pressure divided by absolute temperature. The pressure is unchanged, so the density is simply inversely proportional to the temperature: the new density is the old one multiplied by $\frac{288}{318}$ .
$\frac{288}{318}$ is about 0.906, so the new density is 1.225 × 0.906, which is about 1.11 kilograms per cubic metre.
The air is therefore about 9 per cent less dense than the standard value, at exactly the same pressure — nothing but the temperature changed.
That single line is the whole of density altitude. Hold the pressure and raise the temperature, and the density must fall, so the air behaves like the air of a higher altitude.

Example 4 — putting both effects together

An airfield sits high enough that the pressure there is 810 hectopascals, about 80 per cent of the sea-level value. The temperature is 27 °C. Standard sea-level air is 1013 hectopascals, 288 K and 1.225 kilograms per cubic metre. Find the actual air density, and say roughly what standard altitude has that density, given that the standard atmosphere has about 0.94 kilograms per cubic metre at 8,500 feet.

Convert the temperature: $T = 27 + 273 = 300\ \text{K}$
Density is proportional to pressure divided by absolute temperature, so scale the sea-level density by the pressure ratio and by the inverse temperature ratio: $\rho = 1.225 \times \frac{810}{1013} \times \frac{288}{300}$
Take the two ratios one at a time. $\frac{810}{1013}$ is about 0.80, and $\frac{288}{300}$ is 0.96.
1.225 × 0.80 = 0.98, and 0.98 × 0.96 is about 0.94 kilograms per cubic metre.
The standard atmosphere reaches that density at about 8,500 feet, whereas 810 hectopascals corresponds to a pressure altitude of only about 6,200 feet. So the density altitude is somewhere near 8,500 feet — well over two thousand feet above where the aircraft actually is.
Both factors contributed, and it is worth separating them: the low pressure alone accounted for a 20 per cent reduction in density, and the temperature being 12 kelvin above the sea-level standard figure of 288 K added a further 4 per cent.

What the test asks

Two misconceptions carry this topic. The first is linearity: an item gives the pressure at sea level and at some altitude and invites you to scale straight upwards, so that if the pressure fell by about 500 hectopascals in the first 18,000 feet it must reach zero by 36,000. It does not, because air is compressible and each slice of altitude removes the same fraction rather than the same amount — the fall in a liquid is linear precisely because a liquid is not compressible. The second is treating the standard atmosphere as a description of the real one, so that an option quotes 15 °C as the temperature at sea level today or −56.5 °C as the temperature at the tropopause this afternoon. ISA is an agreed reference and the real atmosphere departs from it constantly. Expect also a density item where only one of the two ingredients is varied and the other is quietly assumed: density needs both the pressure and the temperature, so an option claiming that two places at the same pressure must have the same air density has forgotten half of the relationship. And watch for the sign of the temperature effect, since hotter air is thinner, not thicker.