If the atmosphere is assumed to be composed of a layer of air of uniform density, 1.23 kg m-3, calculate its height if it produces a pressure of 1.01*10^5 Pa at the Earth’s surface.

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We know that we can express the pressure by the following equation.

`P = hrhog` where;

P = pressure

h = height

`rho` = density of air

g = gravitational acceleration

For our question;

`P = 1.01xx10^5pa`

`h = ?`

`rho = 1.23(kg)/(m^3)`

`g = 9.81m/(s^2)`

`h = P/(rhog)`

`h = (1.01xx10^5)/(1.23xx9.81)`

`h = 8.37xx10^3`

*So the height of the atmosphere that produces `1.01xx10^5pa` pressure at earth surface is 8370m.*

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