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I am studying meteorology ( a required course ) and am having difficulty understanding the affect of pressure in the atmosphere on molecules. Any information or websites that help explain this would be much appreciated.

2006-10-03 14:09:37 · 2 answers · asked by Anonymous in Science & Mathematics Weather

2 answers

First of all, very rarely will a meteorologist care about individual molecules in the atmosphere. In general, they will refer to "parcels of air" or "particles" or any myriad of other things that all mean the same thing.

Second, most basically the relation between pressure, temperature, and density can be described by the ideal gas law: p=rho*R*T where p is pressure (in pascals I think,) rho (a Greek letter I can't type here) is density (in kg per meter cubed,) R is the ideal gas constant (287 joules per kilogram per kelvin,) and T is temperature (in kelvin.) Looking at this equation, it is easy to see that if pressure increases, so will the temperature and density of the particles.

Finally, you will learn that practically everything that happens in the atmosphere is thrown into equations involving pressure and temperature, so if you are looking for an answer to your question with regard to a specific case, you need to be more specific in asking.

2006-10-03 17:55:08 · answer #1 · answered by wdmc 4 · 0 0

in case you benefit a million environment of tension for each 34 feet then you extremely may be under 4 atmospheres of tension at 102 feet. (3 * 34) for the water and a million for the air. The regulator which you breathe from on scuba keeps your lungs on the comparable tension as your intensity so the strain interior you is the comparable through fact the exterior. that's the comparable for once you're on dry land. The tension interior you is the comparable as exterior. the genuine question is why can unfastened divers bypass so a ways without being overwhelmed. i think of the checklist for "no limits" unfastened dive is 687 feet!. that's over 20 atmospheres of tension.

2016-12-08 07:57:15 · answer #2 · answered by Anonymous · 0 0

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