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p = ρ k B T m (atmosphere),

where k B is Boltzmann’s constant, which has a value of 1.38 × 10 −23 J/K .

You may have encountered the ideal gas law in the form p V = n R T , where n is the number of moles and R is the gas constant. Here, the same law has been written in a different form, using the density ρ instead of volume V . Therefore, if pressure p changes with height, so does the density ρ . Using density from the ideal gas law, the rate of variation of pressure with height is given as

d p d y = p ( m g k B T ) ,

where constant quantities have been collected inside the parentheses. Replacing these constants with a single symbol α , the equation looks much simpler:

d p d y = α p d p p = α d y p 0 p ( y ) d p p = 0 y α d y [ ln ( p ) ] p 0 p ( y ) = [ α y ] 0 y ln ( p ) ln ( p 0 ) = α y ln ( p p 0 ) = α y

This gives the solution

p ( y ) = p 0 exp ( α y ) .

Thus, atmospheric pressure drops exponentially with height, since the y -axis is pointed up from the ground and y has positive values in the atmosphere above sea level. The pressure drops by a factor of 1 e when the height is 1 α , which gives us a physical interpretation for α : The constant 1 α is a length scale that characterizes how pressure varies with height and is often referred to as the pressure scale height.

We can obtain an approximate value of α by using the mass of a nitrogen molecule as a proxy for an air molecule. At temperature 27 °C, or 300 K, we find

α = m g k B T = 4.8 × 10 −26 kg × 9.81 m/s 2 1.38 × 10 −23 J/K × 300 K = 1 8800 m .

Therefore, for every 8800 meters, the air pressure drops by a factor 1/ e , or approximately one-third of its value. This gives us only a rough estimate of the actual situation, since we have assumed both a constant temperature and a constant g over such great distances from Earth, neither of which is correct in reality.

Direction of pressure in a fluid

Fluid pressure has no direction, being a scalar quantity, whereas the forces due to pressure have well-defined directions: They are always exerted perpendicular to any surface. The reason is that fluids cannot withstand or exert shearing forces. Thus, in a static fluid enclosed in a tank, the force exerted on the walls of the tank is exerted perpendicular to the inside surface. Likewise, pressure is exerted perpendicular to the surfaces of any object within the fluid. [link] illustrates the pressure exerted by air on the walls of a tire and by water on the body of a swimmer.

Figure A is a schematic drawing of a tire filled with air. Pressure inside this tire exerts forces perpendicular to all surfaces it contacts, including the valve. Figure B is a schematic drawing of a swimmer under the water. Pressure is exerted perpendicular to all sides of this swimmer with a larger net buoyant force underneath the swimmer.
(a) Pressure inside this tire exerts forces perpendicular to all surfaces it contacts. The arrows represent directions and magnitudes of the forces exerted at various points. (b) Pressure is exerted perpendicular to all sides of this swimmer, since the water would flow into the space he occupies if he were not there. The arrows represent the directions and magnitudes of the forces exerted at various points on the swimmer. Note that the forces are larger underneath, due to greater depth, giving a net upward or buoyant force. The net vertical force on the swimmer is equal to the sum of the buoyant force and the weight of the swimmer.

Summary

  • A fluid is a state of matter that yields to sideways or shearing forces. Liquids and gases are both fluids. Fluid statics is the physics of stationary fluids.
  • Density is the mass per unit volume of a substance or object, defined as ρ = m / V . The SI unit of density is kg/m 3 .
  • Pressure is the force per unit perpendicular area over which the force is applied, p = F / A . The SI unit of pressure is the pascal: 1 Pa = 1 N/m 2 .
  • Pressure due to the weight of a liquid of constant density is given by p = ρ g h , where p is the pressure, h is the depth of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity.
Practice Key Terms 4

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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