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Ok, you asked for this !!!
I have made a few assumptions, simplifications and rounding in the following - all calculations are at standard Temperature (273K), density of water is taken as 1kg/l (sea water is denser), and it is assumed that you know what a mole is (though that is not essential).
1 mole of air at STP has a volume of 22.4 litres, (imagine 22 cartons of orange juice) and a mass of 0.029 kg.
In order to increase the density of air, we must compress it into a smaller volume (ie apply a pressure). For it to sink, it must have a density greater than water.
To have the same density as water (for ease of calculation), we must compress our 0.029 kg of air into 0.029 litres (that's about 2 tablespoons)
So that; density = mass / volume = 0.029 / 0.029 = 1kg / litre
The pressure, P, required to compress our mole of air into this small volume is given by the Ideal Gas Law,
PV = nRT
where P = Pressure, V = Volume, n = no. of moles T = Temperature (in Kelvin) and R is the gas constant = 8.3145 (for pressures in kPa)
So;
PV = nRT => P = nRT/V
=> P = {1 x 8.3145 x 273} / 0.029
=> P = 78271 kPa (kiloPascals, a unit of pressure)
[Cont�.]
I have made a few assumptions, simplifications and rounding in the following - all calculations are at standard Temperature (273K), density of water is taken as 1kg/l (sea water is denser), and it is assumed that you know what a mole is (though that is not essential).
1 mole of air at STP has a volume of 22.4 litres, (imagine 22 cartons of orange juice) and a mass of 0.029 kg.
In order to increase the density of air, we must compress it into a smaller volume (ie apply a pressure). For it to sink, it must have a density greater than water.
To have the same density as water (for ease of calculation), we must compress our 0.029 kg of air into 0.029 litres (that's about 2 tablespoons)
So that; density = mass / volume = 0.029 / 0.029 = 1kg / litre
The pressure, P, required to compress our mole of air into this small volume is given by the Ideal Gas Law,
PV = nRT
where P = Pressure, V = Volume, n = no. of moles T = Temperature (in Kelvin) and R is the gas constant = 8.3145 (for pressures in kPa)
So;
PV = nRT => P = nRT/V
=> P = {1 x 8.3145 x 273} / 0.029
=> P = 78271 kPa (kiloPascals, a unit of pressure)
[Cont�.]
[Cont�.]
We must now work out the depth of water required to impart this pressure.
A metre depth of water imparts a pressure of 9.8 kPa, so we must divide our Pressure required by 9.8, having first subtracted the pressure imparted by the atmosphere;
Pressure required, P = 78271 kPa - 101 kPa (atmospheric pressure)
= 78170 kPa
This equates to a depth of water of 78170 / 9.8 = 7976 m (= 7.976 km or about 5 miles)
The average depth of the Atlantic Ocean is about 3300 m, so you would have to find the deepest part (8605m) to have sufficient pressure to squeeze your air into the required density.
[Cont�.]
We must now work out the depth of water required to impart this pressure.
A metre depth of water imparts a pressure of 9.8 kPa, so we must divide our Pressure required by 9.8, having first subtracted the pressure imparted by the atmosphere;
Pressure required, P = 78271 kPa - 101 kPa (atmospheric pressure)
= 78170 kPa
This equates to a depth of water of 78170 / 9.8 = 7976 m (= 7.976 km or about 5 miles)
The average depth of the Atlantic Ocean is about 3300 m, so you would have to find the deepest part (8605m) to have sufficient pressure to squeeze your air into the required density.
[Cont�.]
[Cont�.]
There is, however, one final and quite important point.
Solubility of gases increases with pressure - (think of your bottle of fizzy drink, when you release the pressure by opening the cap, the carbon dioxide gas becomes less soluble in the water and so comes out of solution, forming the bubbles that we see as 'fizziness') - so, at that depth, it is most likely that your bubble would simply dissolve into the water as soon as you release it.
You could try Henry's Law (for gas solubility) to work out what would happen, but finding out the gas solubility constant and partial pressures for an air mixture, and dissolved air concentrations for Atlantic seawater at 8000m would involve more than a couple of minutes of calculation / googling !!
Still, it's the first time I've used the Ideal gas Law in about twenty-ah-hem years - and it's certainly passed a quiet morning at work!
There is, however, one final and quite important point.
Solubility of gases increases with pressure - (think of your bottle of fizzy drink, when you release the pressure by opening the cap, the carbon dioxide gas becomes less soluble in the water and so comes out of solution, forming the bubbles that we see as 'fizziness') - so, at that depth, it is most likely that your bubble would simply dissolve into the water as soon as you release it.
You could try Henry's Law (for gas solubility) to work out what would happen, but finding out the gas solubility constant and partial pressures for an air mixture, and dissolved air concentrations for Atlantic seawater at 8000m would involve more than a couple of minutes of calculation / googling !!
Still, it's the first time I've used the Ideal gas Law in about twenty-ah-hem years - and it's certainly passed a quiet morning at work!
brachiopod, I would appreciate your evaluation of what would have been my answer to this question:
Liquids tend to be highly resistant to further compression as the molecules are already in contact with each other. Air compressed to a liquid state has a lower density (approximately 870 kg/m3) than water (greater than 1000 kg/m3) and so would remain buoyant until dissolved.
Also, I have been unable to find a table or graph with a pressure/temperature curve for the transition of air between a liquid and a gas. Do you have any idea at what pressure this transition would take place, say, at zero or four degrees celcius?
And thanks for the great information you already provided!
I did find this if anyone is interested:
Bubbles sink in a pint of Guinness
Liquids tend to be highly resistant to further compression as the molecules are already in contact with each other. Air compressed to a liquid state has a lower density (approximately 870 kg/m3) than water (greater than 1000 kg/m3) and so would remain buoyant until dissolved.
Also, I have been unable to find a table or graph with a pressure/temperature curve for the transition of air between a liquid and a gas. Do you have any idea at what pressure this transition would take place, say, at zero or four degrees celcius?
And thanks for the great information you already provided!
I did find this if anyone is interested:
Bubbles sink in a pint of Guinness
Thanks Brachiopod. Thats exactly the kind of answer I was looking for, though I know it is only theoretical and I agree that the bubble would probably dissolve first. What if the air was inside a very large balloon, as MIBN2CWEUS
( whatever that means ) says, liquid air is still not as dense as water / sea water, but has anyone ever tried this for real over one of the sea trenches etc??
( whatever that means ) says, liquid air is still not as dense as water / sea water, but has anyone ever tried this for real over one of the sea trenches etc??
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