Help with density

[richarddean]

Can someone show me how you would work this out or at least explain how to please

if I take a gas volume of 2 litres in a flexible container from the surface to a depth of 17m in seawater, What would the density be compared to the surface.


I know 17m of seawater is 17/10 =1.7ata + 1ata = 2.7ata but not sure where to go from there. I thought it would be 2.7ata/2litres=1.35 density
is that right or wrong

thanks

 

[pcornell]

Almost had it

2 litres / 2.7 ata = .74 litres

 

[ClayJar]

Density is mass per volume. The mass of the gas in the container will be the same, of course. You should be able to find out how the volume changes from the surface pressure to the pressure at that depth. Just take the inverse of that ratio (one divided by whatever it is), and that is how the density changes.

6 volumes of gas in a flexible container at 1 ata will be 3 volumes of gas when you bring that container with you to a pressure of 2 ata. The ratio of the final volume to the initial volume was 2/6, i.e. 1/3. The ratio of the final density to the initial density is 1/(2/6) == 1/(1/3) == 6/2 == 3 (however you want to write it).

As you can see, you were incorrect. If you don't follow well enough to find the answer, I can explain somewhat more thoroughly.

(pcornell: Your math is impeccable, but it does not answer the question.)

 

[texdiveguy]

ClayJar.... in looking at your SB bio.... I like your thinking here, "Dive History: Someday I might dive clear, warm water, but why spend the cash when there's so much cold, dark, murky stuff around."

 

[pcornell]

(pcornell: Your math is impeccable, but it does not answer the question.)

Thanks, I'll take any part of impeccable I can get. I just left out the inverse proportion relationship between pressure and volume........

 

[rakkis]

It's really more of a conceptual question. Not very much math involved. To answer it, you need to know that pressure and density are directly related.

Double one..... you double the other.
Reduce one to 10% its original value.... the other does the same.

So let's say we have 5 liters in a flexible container at the surface and we bring the container to 38 meters, all you have to do is know how much (absolute) pressure you are under at your depth.

Pay closer attention to what is asked: "What would the density be compared to the surface?" So.. they are asking for a ratio... not a density value. Is it the same density? It is double? Is it 1/5th?


So:
38 meters: 3.8 ata (from water) + 1 ata (from atmosphere) = 4.8 ata

Since pressure and density are directly related, if you x4.8 pressure, then you x4.8 density.

Answer: The density of the air of the container is 4.8 times the density at the surface.

 

[pcornell]

my bad

 

[rakkis]

... take the inverse of that ratio (one divided by whatever it is), and that is how the density changes.

Not quite, that's how volume changes. Pressure and density have a direct relationship. Pressure and volume are inverse.


For your example of 6 volumes of gas at the surface, as you descend to 10 meters (2 ata absolute), volume halves, density doubles.

 

[ClayJar]

.... take the inverse of that ratio (one divided by whatever it is), and that is how the density changes.

Not quite, that's how volume changes. Pressure and density have a direct relationship. Pressure and volume are inverse.

You misrepresent what I said (perhaps due to not following what I said). Pressure and density are directly proportional. Pressure and volume are inversely proportional. Therefore, volume and density are inversely proportional, which is what I said:

You should be able to find out how the volume changes[...] Just take the inverse of that [volume] ratio (one divided by whatever it is), and that is how the density changes.

For your example of 6 volumes of gas at the surface, as you descend to 10 meters (2 ata absolute), volume halves, density doubles.

Volume halves. Density 1/halves, i.e. doubles, which, again, is precisely what I said. (The day I get ideal gas laws wrong is likely to be the day *after* the day I start drinking.)

 

[Steve50]

6 volumes of gas in a flexible container at 1 ata will be 3 volumes of gas when you bring that container with you to a pressure of 2 ata. The ratio of the final volume to the initial volume was 2/6, i.e. 1/3. The ratio of the final density to the initial density is 1/(2/6) == 1/(1/3) == 6/2 == 3 (however you want to write it).

.)

Wouldn't that be 3/6 ie 1/2 and 6/3 = 2 ?