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Dalton's Law

Partial Pressures

Throughout our discussions thus far, we have been talking about the effect of pressure on air. Wehave discussed air in balloons, air in a tank, and air in a divers lungs. It is important to point outthat air is a mixture of many different gasses, but mainly nitrogen and oxygen.

The air mixture is approximately 78% nitrogen, and 21% oxygen with the remaining 1% being amix of argon, carbon dioxide, neon, helium and other rare gases. While some recreational divingis done on special mixtures like nitrox, most is done breathing plain air. While the fact thatair is a mixture of gases is important when we deal with the physiology of diving, we will spend afew moments now to understand the physics of gas mixtures.

It was the English scientist John Dalton that studied the properties of gas mixtures as they relateto pressure and developed Dalton's Law. Dalton's Law states: The total pressure of a gasmixture equals the sum of the partial pressures that make up the mixture.

To study this law as itrelates to scuba divers, let's see how this law affects air at different pressures. In order to makeour numbers a little more manageable, we will assume that air is a mixture of just two gases,nitrogen and oxygen. We will also assume that the mixture is comprised of 80% nitrogen and20% oxygen.

If we then look at this mixture as it relates to Dalton's Law, we know that 80% of the pressure ofthe gas is due to the nitrogen in the mixture and 20% of the pressure is due to the oxygen in themixture. We refer to these as partial pressures. This means at the surface, the pressure exertedon us by the nitrogen in the air mixture is 80% of 14.7, or 11.76 pounds per square inch. Thepressure from the oxygen is 2.94 psi. Together, these account for the 14.7 psi of pressure at thesurface.

If we look at pressures at varying depths, we get the following chart:
` Partial Pressures of Compressed Air (assuming air is 80% nitrogen, 20% oxygen) Absolute Oxygen NitrogenDepth Atms Pressure Pressure Pressure 0 1 14.7 2.94 11.76 33 2 29.4 5.88 23.52 66 3 44.1 8.82 35.28 99 4 58.8 11.76 47.04132 5 73.5 14.70 58.80165 6 88.2 17.64 70.56198 7 102.9 20.58 82.32231 8 117.6 23.52 94.08264 9 132.3 26.46 105.84297 10 147.0 29.40 117.60`
We see then that as we increase the pressure on us by descending, we are dealing with increasedpressure of both nitrogen and oxygen in a 80 - 20 ratio.

It is an easy task to determine the partial pressure of any gas at any depth by using the formulaswe have learned thus far. Let's try to determine the partial pressure of oxygen at a depth of 50 feet insea water assuming oxygen is 20% of the gas mixture.

The first thing we must do is determine the ambient pressure for this depth (If unfamiliar with ambient pressure, refer to Intro to Gas Laws. We know that saltwater exerts .445 pounds of pressure per foot, so the water pressure for this depth would be.445 x 50, or 22.25 psi. To this we add the atmospheric pressure of 14.7 for an ambient pressureof 36.95. If we take 20% of this, we have our answer. 36.95 x .20 = 7.39 Thus, the partialpressure of oxygen at a depth of 50 feet in sea water would be 7.39psi.

Determine the partial pressure of nitrogen at a depth of 40 feet in fresh water, while breathing agas mixture that is 79% nitrogen.

Fresh water exerts .432 psi per foot of depth so we multiply .432 by 40 to get 17.28. To this weadd our atmosphere of air pressure, 14.7 to get an ambient pressure of 31.98. If you take 79% ofthis number you get the answer: 31.98 x .79 = 25.2642

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