# Is the fact that 100 kPa equals about 1 atmosphere accidental?

Typical atmosphere near sea level, in ambient conditions is around 100,000 pascals.

But the pascal, as the unit, is not defined through Earth atmospheric pressure. It's defined as one newton per square meter. The newton is $\rm{kg \: m}\over s^2$. So, $\rm[Pa] = [ {kg \over {m \: s^2}} ]$.

Nowadays, definitions of units are often fixed to various natural phenomena, but it wasn't quite so when they were being created.

The Second is an ancient unit, derived from a fraction of day, 1/86400 of synodic day on Earth. The meter is derived from circumference of Earth, $10^{-7}$ the distance from north pole to equator. The kilogram came to be as mass of a cubic decimeter of water.

100,000 pascals, or 1 bar, though, is about the average atmospheric pressure at sea level. That's an awfully round number - while Earth atmosphere pressure doesn't seem to have anything in common with the rest of the "sources" of the other units.

Is this "round" value accidental, or am I missing some hidden relation?

• It is not a completely round number, though: $$1\;\mathrm{atm}=101325\;\mathrm{Pa}\approx 101\;\mathrm {kPa}$$ Aug 30, 2017 at 17:12
• Maybe I'm missing something, but does "round" (in the sense of exactly divisible by some power of 10) actually have any meaning in the context of "explaining" the natural world? I can certainly see that such a concept might be relevant if we were talking about some power of 2, but apart from it's (pre-)historical significance as "most common number of digits on two (front) legs/arms in vertebrates", why should 10 be any more important than, say, 9 or 11? Aug 31, 2017 at 13:20
• @FumbleFingers: Units are usually chosen such that they are powers of ten of "parent" units, or otherwise "round numbers" - like 10,000,000 meters being the distance from pole to equator; 100 Celsius degrees being the difference between freezing and boiling. It doesn't explain the natural world, it's just convenient to use. So most of "round" values of derived units you encounter in results of calculations are actually artifacts of these choices. But that's not always the case, plus when it is the case, the path the artifact arrived there is often non-obvious.
– SF.
Aug 31, 2017 at 13:41
• @FumbleFingers: "Most long-established "everyday" units are chosen on the basis of ratios / bases other than 10" - Maybe in your USA. Most of the world is metric, and uses metric system for "everyday" units.
– SF.
Aug 31, 2017 at 22:59
• @Lucian: One nanocentury is $\pi$ seconds, to within 0.5%.
– SF.
Aug 31, 2017 at 23:01

This is a coincidence. There's nothing about the atmosphere that would make it have a nice relationship with the Earth's rotation or diameter, or the fact that water is plentiful on the surface.

On the other hand, it's important to note that the coincidence isn't quite as remarkable as you note, because of a version of Benford's law. Given absolutely zero prior knowledge about how much air there is in the atmosphere, our guess about the value of the atmospheric pressure would have to be evenly distributed over many orders of magnitude. This is akin to throwing a dart at a piece of log-scale graph paper: Note that the squares in which the coordinates start with $1.\:{{.}{.}{.}}$ are bigger than the others, so they're rather more likely to catch the dart. A similar (weaker) effect makes the probability of the second digit being 0 be 12% instead of the naive 10%.

• Comments are not for extended discussion; this conversation has been moved to chat.
– rob
Aug 31, 2017 at 12:58
• I'd upvote, but I don't want to break the 42 votes :)p May 13, 2019 at 19:25

To expand on @EmilioPisanty's answer. The original definition of the meter was:

one ten-millionth of the distance between the North Pole and the Equator[.]

The word gramme was adopted by the French National Convention in its 1795 decree revising the metric system as replacing the gravet introduced in 1793. Its definition remained that of the weight (poids) of a cubic centimetre of water.

And the original definition of a second can be traced back to the sexagesimal counting system of the Babylonians and the length of a day. Thus the original definition of a second is $\frac{1\operatorname{day}}{24 \times 60 \times 60}$, on average. Roughly speaking, then, we can trace the value of $1$ Pascal back to the radius of the Earth, the rotational period of the Earth, and the density of water. There is no reason to expect the pressure of the atmosphere at sea level to be particularly closely related to any of those, since that will depend on the composition of the Earth (the mass dictated by the Earth's material composition [balance of silicates vs iron/nickel]), its radius, and the temperature at its surface (which depends on the sun, planet's albedo, etc.), the strength of the Earth's magnetic field, and how energetic the solar winds stripping the atmosphere away are.

At any rate, in terms of the original definitions, the Pascal is given by: $$1\operatorname{Pa} \equiv \left(\frac{729 \pi^2}{39,062,500,000}\right)\frac{\rho_{\mathrm{water}} R_{\mathrm{Earth}}^2}{T_{\mathrm{day}}^2},$$ and we have no reason to expect that the atmospheric pressure at sea level to be particularly closely related to any of those variables, especially given the interference of other factors in fixing the air pressure at sea level.

Note that the air pressure at sea level can also change. Consider Venus, for example. It's a smaller planet, but the combination of being closer to the sun and a strong greenhouse effect means its atmosphere is incredibly heavy. The pressure at the surface of Venus is $9.3\operatorname{MPa}$, or about $92$ times greater than atmospheric pressure here on Earth.

Looking at how the temperature of Earth, the composition of its atmosphere, and the solar output have all changed over time, I would be surprised to learn that the sea level pressure of the atmosphere has been stable to better than a factor of $2$ over its whole history.

Edit: I found a reference to one reconstruction of ancient atmospheric pressure.

Here, we calculate absolute Archaean barometric pressure using the size distribution of gas bubbles in basaltic lava flows that solidified at sea level $\sim 2.7 \operatorname{Gyr}$ in the Pilbara Craton, Australia. Our data indicate a surprisingly low surface atmospheric pressure of $P_{\mathrm{atm}} = 0.23 \pm 0.23\ (2\sigma) \operatorname{bar}$, and combined with previous studies suggests $\sim 0.5 \operatorname{bar}$ as an upper limit to late Archaean $P_{\mathrm{atm}}$

• I have asked a question over at the Earth Science StackExchange for a paleoclimate reconstruction of the Earth's atmospheric pressure at sea level. earthscience.stackexchange.com/questions/12186/… Aug 30, 2017 at 22:08
• Originally, they wanted to define the meter from the length of the seconds pendulum. So that would mean $1\operatorname{m} \equiv g \left(\frac{1\operatorname{s}}{\pi}\right) ^2$ where $g$ is the standard gravity. You can write your answer again using that definition of the meter. The fact that the old definition with the seconds pendulum gives $0.994$ times the other definition with the North-Pole-to-Equator distance, must be another coincidence. I think they changed to the latter definition because $0.994$ is so close to $1$. Aug 30, 2017 at 22:35
• @JeppeStigNielsen Interesting. That would remove one of the coincidences, and pressure is more tightly tied to $g$ than any of the other quantities mentioned. I think that the temperature coincidence can't be avoided, though it can be argued that temperature and pressure near water's triple point is essential for life, but that leaves a wide range of pressures, as this graph shows. Aug 30, 2017 at 22:41
• Actually, that definition of the meter just piles on more coincidences, including two more factors of $R_{\mathrm{Earth}}$, two factors of $M_{\mathrm{Earth}}$, and two factors of $G$. Aug 30, 2017 at 23:26
• Hm, you should express the Venus athmosphere pressure in terms of Venus-Pascals, i.e., density of the most prevalent liquid, radius of the planet, length of a venusian day Aug 31, 2017 at 21:09

No it is not coincidence:

It is because units for mass, force and pressure have been choosen such that common ratios of those units (density of water $\rho_{Water}$, gravitational acceleration on earth $g$; pressure unit $1 at$) have values that are powers of 10. (Further improvements in measurement changed those values a little bit later on; especially $g$; that's why the values are not exactly $1.000\times 10^n$).

Putting that together you get:

roughly:
$1 at \approx 10 \times 1000kg/m³ \times 10N/kg = 100000 N/m² = 100000Pa$

more exact:
$1 at = 10 \times 1000kg/m³ \times 9.81N/kg = 98100 N/m² = 98100Pa$
or
$1 Pa = 1019 \times10^{-5}at$

EDIT:
I'm referring above to the value of 1 at (called "technical atmosphere") not the standard atmosphere.

Note that the title of the original question is ambiguous as it just mentions "1 atmosphere".

• No, the atmosphere was defined as the average air pressure at sea level at the latitude of Paris. It happens to be about 10 meters of water.
– user137289
Aug 30, 2017 at 20:45
• @Curd Why'd they choose $10\operatorname{m}$ and not 11, 9, or 1? Aug 30, 2017 at 20:56
• I'm interested in knowing the references which support your not by coincidence statements above. Aug 30, 2017 at 21:42
• I've dealt with pressure quite a bit in school. I've never heard "atmosphere" as a unit given as anything besides a standard atmosphere. TBH this is the first I've ever heard of a "technical atmosphere". It's not nearly as common as a standard atmosphere. If hey meant technical atmosphere, the question should ask about it. (Wikipedia for example has the page Atmosphere(unit) which is about atm, not at.
– JMac
Aug 31, 2017 at 10:20
• @Curd I've seen things referred to as "2 atm" (usually called "two atmospheres"), 10 atm, etc, in a lot of engineering texts. I've not once seen a "technical atmosphere". Maybe it's a localization thing; but it's never used in the North American or International versions of the textbooks I had.
– JMac
Aug 31, 2017 at 11:02