rakuenso Posted October 18, 2005 Posted October 18, 2005 My physics text asks "At what depth below the surface of a lake will you experience a pressure twice the pressure you experience on the surface?" I'm guess if at sea level, the amount of pressure in atm exerted on me is exactly 1 atm. So if its at 2 atm, there should be 2(760) torrs as well? So 1520 torrs = 1560mm Hg? Where do I go from here, I guess I can't use pV=nRT because it only applies to ideal GASES
swansont Posted October 18, 2005 Posted October 18, 2005 [math]\rho gh[/math] Remember that the atmospheric pressure is still present.
ecoli Posted October 18, 2005 Posted October 18, 2005 take a look at this site. http://www.engineersedge.com/fluid_flow/relationship_depth_pressure.htm roughly, for every 33 ft of water you decend, there is an additional 1 atm of pressure (plus 1 for the atmosphere above the water)
BhavinB Posted October 23, 2005 Posted October 23, 2005 [math]\rho gh[/math] Remember that the atmospheric pressure is still present. The rule of thumb used in fluid mechanics is that the pressure experienced underwater at some depth h, is equal to the weight of the water directly above you. So this formula is correct.
insane_alien Posted October 23, 2005 Posted October 23, 2005 yes but the fluid is already under pressure from the atmosphere unless you are talking about gauge pressure in which the atmosphere is defined as 0
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