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Pressure under water.

  1. Apr 14, 2009 #1
    Hey, is there an equation to work out what the pressure is at given depths under water?

    Basically, I have a set of results showing the volume of 'some' air increasing as the pressure is changed. I know that pressure x volume is constant. I need the above equation to explain how my results occured. (as well as to see how accurate my experiment was)!

    One thing that I have been told that the pressure increases by 1 atmospheric pressure unit every 10m under water, however I dont know what the magnitude of the atmospheric pressure unit is in terms of pressure (to be used in the pV = nRT equation).

    My guesses from what i've been told is that... Pressure = 1/10 * depth (in meters) * atmospheric pressure unit

    Any help would be greatly appreicated, thanks!
     
    Last edited: Apr 14, 2009
  2. jcsd
  3. Apr 14, 2009 #2

    CompuChip

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    Welcome to PF Kasc.

    For air under water you can to good approximation apply the ideal gas law (PV/T constant, i.e. PV is constant if the temperature doesn't vary appreciably).

    A pressure of 1 atm(osphere) corresponds 101,325 Pa(scal). It is equal to about 1 bar, which is 100,000 Pa (10^5 Pa). It is preferred to give pressure in SI units (e.g. N/m^2 = Pa) although the use of bars is also common. On the surface of the earth at sea level the pressure is usually 1 atm ~ 1 bar. So an atmosphere is the air pressure you normally feel due to the atmosphere of the earth -- hence the name.
     
  4. Apr 14, 2009 #3
    So does that mean my equation works? (Pressure = 1/10 * depth (in meters) * 10^5)
     
  5. Apr 14, 2009 #4
    Pressure p, is defined as a force F exerted on area A, that is p = F / A. The force F is really just the force of gravity of the column of water above you, which can be calculated as mg (where m is the mass of the water and g is Earth's gravitational constant). The mass of the water is give by m = V * ρ, where V is the water's volume and ρ its density. The volume can be expressed as the area (on which the water exerts the Fg) times the height of the water column. Notice that this area is the same as the area A from p = F / A, hence we have:

    [tex]p = \frac{F}{A} = \frac{m \cdot g}{A} = \frac{V \cdot \rho \cdot g}{A} = \frac{h \cdot A \cdot \rho \cdot g}{A} = \rho \cdot g \cdot h[/tex]

    where ρ is the water density, g is the Earth's gravitational constant and h is the height of the water column (e.g. the depth)

    If we take water's density (at 4°C) to be 1,000kg m-3 and the standard atmospheric pressure to be 101,325 Pa, then the depth at which the water pressure is equal to 101,325 Pa is given by

    [tex]h = \frac{101,325 Pa}{\rho \cdot g} = \frac{101,325 N \cdot m^{-2} }{ 1,000 kg \cdot m^{-3} \times 9.8 m \cdot s^{-2} } = 10.34m[/tex]
     
    Last edited: Apr 14, 2009
  6. Apr 14, 2009 #5

    CompuChip

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    Moreover, since the density and gravitational constant are assumed to be a constant, then it is also true that
    [itex]\Delta p = \rho g \Delta h [/itex]
    and by the same reasoning as scibuff's, a difference of about 10 meters in water height corresponds to 1 bar of pressure.
     
  7. Apr 14, 2009 #6
    Ok thanks guys :)

    So as to just check: if you were under 25m of water, you would be experiencing 3.5 bars of pressure?
     
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