Volume & Density at surface ?

[namerg]

I suck in math and I am new on this...

Question: if you have 1 litre of air at 30m/99'. What would the volume and density be at the surface ?

So, why the pressure at the surface is 1/4 the pressure at 30mts/99' ?

And why will expand to 4 liters at the surface ?

And the air density would be 1/4 the density at 30mts/99' ?

I have the depth, pressure, air volume and air density but I can not figure it out....math

Thanks for your help,

 

[victorzamora]

So, Boyle's Law is what's governing the behavior you're talking about. (Remember it like this: Breathe Or Your Lungs Explode). Basically, it just states that as the pressure goes up, the volume goes down. Specifically, Pressure is inversely proportional to Volume. The equation to become familiar with (if you want to look at the math) is that P1*V1=P2*V2....or the pressure times volume at one condition is equal to the pressure times volume at all other conditions for a closed system.

If you have a 4L balloon at the surface (1 atmosphere absolute), you have 1ata*4L. At 30m (30/10+1=4ata) you have 1L of volume. 1ata*4L=4ata*1L. The opposite is true for your balloon.

What this basically means is that as you take your balloon deeper, it gets smaller because the pressure goes up. The pressure inside the ballon is (theoreticall) the same as the pressure OUTSIDE of the balloon. Pressure is what you're REALLY changing when you change depth. So as the pressure goes up, the volume goes down (inversely proportional)....and as the pressure goes down, the volume goes up.

Now let's talk about density. Density is just mass/volume. If you have the same mass in the balloon (which is true for a closed system), then the only thing that changes is volume. When your volume goes up 4x, you have the same mass divided by 4 times the volume....or 1/4 of the density at 30m (99' or 4ata).

 

[chrisch]

Surface pressure is 1ATM. Every 10m plus is another ATM so 30m is 4 ATM (3+1). 1 is one quarter of 4. 'One quarter' is '1/4' (one divided by four) to get the inverse reverse the figures '4/1' four divided by one is four.

Density increases with depth. Same amount of air molecules but less space for them. Halve the space, double the density, quarter the space four times the density etc.

Everyone tends to forget surface pressure if you are on those lovely DM exams

 

[namerg]

Ohh okay. I will read through this slowly, a pen and a paper. Thank you very much. If I have questions I will post them here.

 

[RJP]

I suck in math and I am new on this...

Question: if you have 1 litre of air at 30m/99'. What would the volume and density be at the surface ?

It also depends on the container the air is in. If it's in a rigid container (like a tank) the density and volume are the same at surface and depth. If it's in a balloon, the pressure/density changes as depth changes.

If you're taking a PADI course there MAY be a question on one of your knowledge reviews or OW exam that requires you to read the question carefully and understand this distinction. (Hint, hint, hint.)

 

[nolatom]

I really like "Breathe Or Your Lungs Explode", for "BOYLE".

I (an aging English major) teach sailing as a sideline, and to my engineer-type students I will mention Bernoulli's Law (faster fluid over a surface equals lower pressure) as what allows sailboats to go "against" the wind, at an angle between sails and wind which you have to be careful about so the flow stays laminar (= horizontal yarn telltales on both sides of the sail).

Do we have any acronyms for "Bernoulli"? I have "But every racer needs usable upwind lift, like 'INTREPID'", which shows only that I suck at acronyms, and am old enough to remember the America's Cup defenders from the '60s ;-)

 

[Keith.M]

I think on these types of questions, it is easier to bring it back to the surface and then go down.

Surface 1 ATM 1 ltr

33' 2 ATM 1/2 ltr
66' 3 ATM 1/3 ltr
99' 4 ATM 1/4 ltr

So to go back to the surface

99' = 4X
66' = 3X
33' = 2X

So on your question 1 ltr at 99' 4 ATM will be 4 ltr at the surface 1 ATM

 

[namerg]

If you have a 4L balloon at the surface (1 atmosphere absolute), you have 1ata*4L. At 30m (30/10+1=4ata) you have 1L of volume. 1ata*4L=4ata*1L. The opposite is true for your balloon..

How do you come up with the 1L of volume ?
And,

When your volume goes up 4x, you have the same mass divided by 4 times the volume....or 1/4 of the density at 30m (99' or 4ata).

On this statement and according to my example what is the mass so i can divided by 4 ? and the volumes is ? so mass/4 * volume ? Sorry, sorry i got lost here...

---------- Post added November 14th, 2014 at 10:30 AM ----------

So to go back to the surface

99' = 4X
66' = 3X
33' = 2X

So on your question 1 ltr at 99' 4 ATM will be 4 ltr at the surface 1 ATM

I like your explanation, Keith.M. But, on my way back, the Xs are ?

 

[Keith.M]

I like your explanation, Keith.M. But, on my way back, the Xs are ?

The multiplier of whatever volume you have.

So at 99' if you have 1 ltr of volume to get to the surface, you use 4X. 4 X 1ltr = 4 ltr

 

[namerg]

Ohh ok...is the known volume stated at the beginning....got it got it....thanks,