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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}Pa$$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 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
$1 Pa = 1019 \times10^{-5}Pa$


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 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".

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Source Link
Curd
  • 126
  • 5

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
$1 Pa = 1019 \times10^{-5}Pa$


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 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
$1 Pa = 1019 \times10^{-5}Pa$


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 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
$1 Pa = 1019 \times10^{-5}Pa$


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".

added 354 characters in body
Source Link
Curd
  • 126
  • 5

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
$1 Pa = 1019 \times10^{-5}Pa$


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

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

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 $1.000\times 10^n$).

  • 1 at is defined as the pressure of 10m water column
  • The density of water is not by coincidence (but by early definitions of unit of mass kg) almost exactly 1000kg/m³
  • The gravitational acceleration $g$ is not by coincidence (but by early definitions of unit of force N) approximately 10N/kg (more exactly about 9.81N/kg; even more exact value depends on the place on earth (especially latitude)).

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
$1 Pa = 1019 \times10^{-5}Pa$


EDIT:
I'm referring above to the value of 1 at (technical atmosphere).

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
$1 Pa = 1019 \times10^{-5}Pa$


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".

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