Tank pressure

[Cold_Under_Here]

There's a good reason we use the ideal gas law; working in the conversion factors takes too much time.

While we have the neat ideal gas equation, PV=nRT the real gas equation (van der Waals equation) is more complex.

(P(real) + ((n^2)a)/V(real)^2) x (V(real) - nb) = nRT

Where a is a constant for gas reflecting the intermolecular forces and b is constant actual volume of the gas molecules.

Now remember air has two main components, oxygen gas and nitrogen gas, so we would have to do the equation twice for each partial pressure. Yeck.


Let me get some paper...

Not only that, the equation gets icky when you need to calulate an unknown volume when everything else is known! One word:...iteration

love physics :14:

 

[Blackwood]

One word:...iteration

That's what Excel is for

 

[500 PSI]

Excellent Question and answers. I also notice the tank pressure quick drop from 3000 PSI to 2500 PSI in early part of a dive. I do dive in cold water and the drop in temperature from say 80+ to 50 & 60 Degrees Farenheight seems a good answer.

I believe instructors do comment on this physical change in beginners classes. An inexperienced diver would certainly not recognize this.

I also believe that more experienced divers will learn to "sip" air rathert than "gulp" it!!

 

[The Horn]

Given that the volume of the tank is fixed and both nitrogen annd oxygen will experience the same temperature drop:

p1/p2 = t1/t2

so your 204 bar tank (3000 psi) is initially at 80 deg F or 300 Kelvin and you dive to say 50 deg F depth or 283 Kelvin the pressure in the tank will drop down to p2= p1/(t1/t2) =204/(300/283) =192 bar or 2822 PSI

Take into account BC/dry suit inflation, any breathing at the surface for equipment checks or waiting for your buddy and chances are that your first reference to your pressure gauge will show a big drop.

Any one with an AI computer should have a neat graph that shows a big drop as the tank and contents cool and then stabilizes.