Water vapour content as air pressure changes.....
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Water vapour content as air pressure changes.....
Had a discussion the other day re how pressure affects the amount of moisture that a certain volume of air can hold. Couldn't come up with a suitable answer. Not much luck with google either, nor pprune search. So here's an example to try and convey my query. Let's just consider "air" ie normal composition of nitrogen, oxygen etc, and water vapour. No dust etc in the samples.
Let's say on Day 1 you were to take a cubic metre of air at Airport A (sample 1), temp and dew point = 15 deg C, altimeter setting 29.50. Then on Day 2 you also take a cubic metre of air from airport A (sample 2), temp and dew point = 15 deg C, but now altimeter setting 30.50. Note; temperature is constant.
Would there be more water vapour in the air in Sample 1 or Sample 2? Or the same?
And if you were to also have a cubic meter where there is no air at all (Sample 3), how much water vapour could it hold?
Explanations, formulas much appreciated. Thanks, Hawk
Let's say on Day 1 you were to take a cubic metre of air at Airport A (sample 1), temp and dew point = 15 deg C, altimeter setting 29.50. Then on Day 2 you also take a cubic metre of air from airport A (sample 2), temp and dew point = 15 deg C, but now altimeter setting 30.50. Note; temperature is constant.
Would there be more water vapour in the air in Sample 1 or Sample 2? Or the same?
And if you were to also have a cubic meter where there is no air at all (Sample 3), how much water vapour could it hold?
Explanations, formulas much appreciated. Thanks, Hawk
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From the dark reaches of past study I think Dalton's Law of Partial Pressures applies here. You will find it (and its application) in any reputable physics textbook.
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Hawk, checkout Relative humidity, dewpoint, frostpoint etc. at Aviation Formulary V1.45
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You will find it (and its application) in any reputable physics textbook.
Dalton's law - Wikipedia, the free encyclopedia
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My recollection from an old air conditioning analysis text is that the saturation humidity (i.e. 100% relative) increases slightly as pressure decreases. It's a pretty small factor, though.
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Barit, can you give an example? I'm not sure what you mean. Do you mean the temperature for 100% relative humidity increases?
From what I gather, it appears that Dalton’s law states that the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture. I can’t really see how this helps me….
But from this link in wikipedia
Relative humidity - Wikipedia, the free encyclopedia
I found the quote below
“If the system at State A is isothermally compressed (compressed with no change in system temperature) then the relative humidity of the system increases because the partial pressure of water in the system increases with increasing system pressure. This is shown in State C.
Therefore a change in relative humidity can be explained by a change in system temperature, a change in the absolute pressure of the system, or change in both of these system properties.”
I ** think ** the first paragraph is the key to my question. At higher pressures (and at the same temperature), one would expect the relative humidity to increase. If the relative humidity increases, but the amount of water stays the same, that would mean the air can hold less water at higher pressures than lower pressures. So…..the answer seems to be that Sample 1, taken on day 1, would have more water vapour than sample 2 on day 2.
So...a cubic metre of air on a low pressure day can hold more water vapour than a cubic metre of air on a higher pressure day.
And...the most water vapour held in a cubic metre would be at a very low air pressure, a near vacuum??
Does this make sense to you gurus? Barit? Brian? Others?
From what I gather, it appears that Dalton’s law states that the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture. I can’t really see how this helps me….
But from this link in wikipedia
Relative humidity - Wikipedia, the free encyclopedia
I found the quote below
“If the system at State A is isothermally compressed (compressed with no change in system temperature) then the relative humidity of the system increases because the partial pressure of water in the system increases with increasing system pressure. This is shown in State C.
Therefore a change in relative humidity can be explained by a change in system temperature, a change in the absolute pressure of the system, or change in both of these system properties.”
I ** think ** the first paragraph is the key to my question. At higher pressures (and at the same temperature), one would expect the relative humidity to increase. If the relative humidity increases, but the amount of water stays the same, that would mean the air can hold less water at higher pressures than lower pressures. So…..the answer seems to be that Sample 1, taken on day 1, would have more water vapour than sample 2 on day 2.
So...a cubic metre of air on a low pressure day can hold more water vapour than a cubic metre of air on a higher pressure day.
And...the most water vapour held in a cubic metre would be at a very low air pressure, a near vacuum??
Does this make sense to you gurus? Barit? Brian? Others?
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here goes, take it steady
Hawk,
Yes it puzzles folks but it is a straightforward density calculation. Density is Pressure divided by the product of a Gas Constant and Absolute Temperature.
Air contains water vapour, a gas and not as many people think a fine mist of water droplets. Air is a mixture of gases and whatever the size of the container each gas fills the whole container uniformly.
Each gas has its own Gas Constant which for dry air is 287.053 joules/kg-deg K and for water vapour, a gas remember, it is 461.495 joules/kg-deg K.
So let’s begin with a cubic metre of dry air at 15 deg C which is 288.15 Kelvin and ISA conditions so the pressure is 101325 Pascal.
Density is therefore 101325/(287.053 x 288.15) = 1.225 kg/cu m.
Now imagine a planet whose atmosphere is solely water vapour. Each molecule of vapour has two hydrogen atoms and one of oxygen, much lighter than a molecule of either nitrogen or oxygen, the main constituents of air. Density = 101325/(461.95 x 288.15) = 0.7612 kg/cu m, so water vapour is lighter than air and 761.2 grams of vaporised water are in that cubic metre.
Ordinary air is a mixture of gases including water vapour which is, we can never tire of repeating is a gas, not a mist of water droplets. Water vapour is not the same as precipitation!
In that cubic metre of yours there is therefore a fraction of dry air co-existing with a fraction of water vapour and being gases each fills the cubic metre perfectly. If we wish to know how much is air and how much is vapour we must discover the partial pressure that each fraction contributes to the total pressure. It is a mistake to imagine that each fraction in a mixture of gases must be at the same pressure – some molecules are more massive than others.
Thus the total air density is the sum of two calculations of P/RT for dry air and a different P/RT for water vapour.
Assuming sea-level for elevation, 29.5” Hg and 15*C for both ambient temperature and dew point my calculator gives an air density of 1.1999 kg/cu m. Increasing the pressure to 30.5” Hg gives an air density of 1.2409 kg/cu m. How?
At 29.50” Hg the vapour pressure from tables is 1704 Pascal which incidentally is also the saturated vapour pressure because dew point and ambient temperature are the same. So the density of the water vapour is 1704/(461.95 x 288.15) = 0.0128 kg/cu m. To put it another way your cubic metre has 12.8 grams of vaporised water.
At 30.50” Hg the vapour pressure is still also 1704 Pascal because dew point and ambient temperature are unchanged. So the density of the water vapour is still the same and the cubic metre still has 12.8 grams of vaporised water in it.
29.50” Hg total pressure is 99898 Pascal so subtracting the 1704 Pascal due to the partial pressure of water vapour gives the partial pressure due to the dry air fraction as 98194 Pascal. Therefore the density of the dry air fraction is 98194/(287.053 x 288.15) = 1.1871 kg/cu m. Now add the two densities together because each gas completely fills the cubic metre and you get 1.1871 + 0.0128 = 1.1999 kg/cu m as the calculator said.
Repeating this at 30.50” Hg total pressure is 103285 Pascal; subtract 1704 Pascal due to saturated vapour pressure and the partial pressure of the dry air fraction is now 101581 Pascal giving a dry air density of 101581/(287.053 x 288.15) = 1.2281 kg/cu m. Add the two densities as before and we have 1.2281 + 0.0128 = 1.2409 kg/cu m, again agreeing with the calculator.
Notice also that at 30.50” Hg if we could dry the air completely the air density would be 103285/(287.053 x 288.15) = 1.2487 kg/cu m.
Google Richard Shelquist + Wahiduddin + Calculator too.
Yes it puzzles folks but it is a straightforward density calculation. Density is Pressure divided by the product of a Gas Constant and Absolute Temperature.
Air contains water vapour, a gas and not as many people think a fine mist of water droplets. Air is a mixture of gases and whatever the size of the container each gas fills the whole container uniformly.
Each gas has its own Gas Constant which for dry air is 287.053 joules/kg-deg K and for water vapour, a gas remember, it is 461.495 joules/kg-deg K.
So let’s begin with a cubic metre of dry air at 15 deg C which is 288.15 Kelvin and ISA conditions so the pressure is 101325 Pascal.
Density is therefore 101325/(287.053 x 288.15) = 1.225 kg/cu m.
Now imagine a planet whose atmosphere is solely water vapour. Each molecule of vapour has two hydrogen atoms and one of oxygen, much lighter than a molecule of either nitrogen or oxygen, the main constituents of air. Density = 101325/(461.95 x 288.15) = 0.7612 kg/cu m, so water vapour is lighter than air and 761.2 grams of vaporised water are in that cubic metre.
Ordinary air is a mixture of gases including water vapour which is, we can never tire of repeating is a gas, not a mist of water droplets. Water vapour is not the same as precipitation!
In that cubic metre of yours there is therefore a fraction of dry air co-existing with a fraction of water vapour and being gases each fills the cubic metre perfectly. If we wish to know how much is air and how much is vapour we must discover the partial pressure that each fraction contributes to the total pressure. It is a mistake to imagine that each fraction in a mixture of gases must be at the same pressure – some molecules are more massive than others.
Thus the total air density is the sum of two calculations of P/RT for dry air and a different P/RT for water vapour.
Assuming sea-level for elevation, 29.5” Hg and 15*C for both ambient temperature and dew point my calculator gives an air density of 1.1999 kg/cu m. Increasing the pressure to 30.5” Hg gives an air density of 1.2409 kg/cu m. How?
At 29.50” Hg the vapour pressure from tables is 1704 Pascal which incidentally is also the saturated vapour pressure because dew point and ambient temperature are the same. So the density of the water vapour is 1704/(461.95 x 288.15) = 0.0128 kg/cu m. To put it another way your cubic metre has 12.8 grams of vaporised water.
At 30.50” Hg the vapour pressure is still also 1704 Pascal because dew point and ambient temperature are unchanged. So the density of the water vapour is still the same and the cubic metre still has 12.8 grams of vaporised water in it.
29.50” Hg total pressure is 99898 Pascal so subtracting the 1704 Pascal due to the partial pressure of water vapour gives the partial pressure due to the dry air fraction as 98194 Pascal. Therefore the density of the dry air fraction is 98194/(287.053 x 288.15) = 1.1871 kg/cu m. Now add the two densities together because each gas completely fills the cubic metre and you get 1.1871 + 0.0128 = 1.1999 kg/cu m as the calculator said.
Repeating this at 30.50” Hg total pressure is 103285 Pascal; subtract 1704 Pascal due to saturated vapour pressure and the partial pressure of the dry air fraction is now 101581 Pascal giving a dry air density of 101581/(287.053 x 288.15) = 1.2281 kg/cu m. Add the two densities as before and we have 1.2281 + 0.0128 = 1.2409 kg/cu m, again agreeing with the calculator.
Notice also that at 30.50” Hg if we could dry the air completely the air density would be 103285/(287.053 x 288.15) = 1.2487 kg/cu m.
Google Richard Shelquist + Wahiduddin + Calculator too.
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The relatitonship I have used is found in Chart A of "Air Conditioning Analysis", a book by W. Goodman. I do not have the book, but took notes when I was writing a humidity calculation program 25 years ago.
The gist of it: a 1% decrease in atmospheric pressure yields a 4.03% increase in saturation humidity (as a mass ratio, water vapor to dry air) or 100% relative humidity.
Since the book was written for terrestrial application, I wouldn't expect you could extend this relationship into the stratosphere.
The gist of it: a 1% decrease in atmospheric pressure yields a 4.03% increase in saturation humidity (as a mass ratio, water vapor to dry air) or 100% relative humidity.
Since the book was written for terrestrial application, I wouldn't expect you could extend this relationship into the stratosphere.
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“If the system at State A is isothermally compressed [...] then the relative humidity of the system increases because the partial pressure of water in the system increases [...]”
I ** think ** the first paragraph is the key to my question. At higher pressures (and at the same temperature), one would expect the relative humidity to increase. If the relative humidity increases, but the amount of water stays the same, that would mean the air can hold less water at higher pressures than lower pressures [...]
So...a cubic metre of air on a low pressure day can hold more water vapour than a cubic metre of air on a higher pressure day.
I ** think ** the first paragraph is the key to my question. At higher pressures (and at the same temperature), one would expect the relative humidity to increase. If the relative humidity increases, but the amount of water stays the same, that would mean the air can hold less water at higher pressures than lower pressures [...]
So...a cubic metre of air on a low pressure day can hold more water vapour than a cubic metre of air on a higher pressure day.
If the total pressure is higher in sample 2, it is because it contains more air, as you cannot stuff any more water vapour in it or it will condense into liquid.
In other words, once you reach saturation, you have the water vapour at its maximum partial pressure possible for that temperature and since the temperature is the same in all three cases, so is the partial pressure of the saturated water vapour, its density and ultimately the total amount of water. In short, cubic meters 1, 2 and 3 contain all three the same amount of water.
In the Wikipedia example, on the other hand, it is understood that the vapour in the system is not saturated yet (i.e. partial pressure < saturated vapour pressure) so if you compress the system keeping the temperature constant, the partial pressure of water vapour can and will increase, giving a higher relative humidity. If you keep compressing, at some point the vapour's partial pressure will equal the maximum value possible at that temperature (= saturated vapour pressure) and won't increase any more (you just got to 100% relative humidity). Any further compression will not increase the water content of a gas particle, it will only cause more vapour to condense.
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Thanks Mathy, it took me some time but I was able to follow your explanation and re create your numbers on my own. You didn’t specifically say so, but since I believe I understand what you have said I surmise that the amount of vapour a certain volume can hold is independent of the pressure. Ergo, as DeeDay said in his post, the answer is that all 3 samples hold the same amount of water vapour.
Dee Day too, thanks for your explanation, you and Mathy seem to be on the same page for sure.
I still have trouble visualizing that two different columns of gas, each exerting the same pressure, will have different densities. I will just have to accept that.
In my previous post I said:
Mathy and Dee Day, in my original question I think I could have used the term “isothermally compressed” (and not have changed the question) to get from sample 1 to sample 2. Wikipedia seems to state that this would require an increase in relative humidity. Since sample one was already at 100% relative humidity, this would mean, as DeeDay says, condensation and thus less vapour in the sample. Yet your explanations seem to say that there is the still the same amount of vapour in all 3 samples. Does this make sense?
Thanks, Hawk
Dee Day too, thanks for your explanation, you and Mathy seem to be on the same page for sure.
I still have trouble visualizing that two different columns of gas, each exerting the same pressure, will have different densities. I will just have to accept that.
In my previous post I said:
Relative humidity - Wikipedia, the free encyclopedia
I found the quote below
“If the system at State A is isothermally compressed (compressed with no change in system temperature) then the relative humidity of the system increases because the partial pressure of water in the system increases with increasing system pressure. This is shown in State C.
Therefore a change in relative humidity can be explained by a change in system temperature, a change in the absolute pressure of the system, or change in both of these system properties.”
I ** think ** the first paragraph is the key to my question. At higher pressures (and at the same temperature), one would expect the relative humidity to increase. If the relative humidity increases, but the amount of water stays the same, that would mean the air can hold less water at higher pressures than lower pressures. So…..the answer seems to be that Sample 1, taken on day 1, would have more water vapour than sample 2 on day 2.
I found the quote below
“If the system at State A is isothermally compressed (compressed with no change in system temperature) then the relative humidity of the system increases because the partial pressure of water in the system increases with increasing system pressure. This is shown in State C.
Therefore a change in relative humidity can be explained by a change in system temperature, a change in the absolute pressure of the system, or change in both of these system properties.”
I ** think ** the first paragraph is the key to my question. At higher pressures (and at the same temperature), one would expect the relative humidity to increase. If the relative humidity increases, but the amount of water stays the same, that would mean the air can hold less water at higher pressures than lower pressures. So…..the answer seems to be that Sample 1, taken on day 1, would have more water vapour than sample 2 on day 2.
Thanks, Hawk
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For simplicity, take a cylinder-piston system containing water vapour only.
If you are at saturation and start compressing isothermally, liquid water will condense at the bottom of the cylinder, while the vapour above it will stay exactly in the same state (pressure, temperature and density).
So the cylinder will contain less water vapour in absolute terms, but the amount of vapour of a sample of the gaseous phase (say 1 cm^3) will be exactly the same (= the vapour density is the same).
Repeat the experiment with a mixture of air and vapour and you'll get the same result for the vapour part, but since this time you've got a mass of air as well, which does not condense and is forced into a smaller volume, its density - and therefore the density of the mixture air+vapour - will have to increase, but the density of the water vapour itself will not.
If that can help, the two columns (assuming same volume and temperature as well) will have the same number of gas molecules, but since the two gases are chemically different, the weight of an individual molecule will be different for the two gases, and so will the total weight - and density - of the column.
If you are at saturation and start compressing isothermally, liquid water will condense at the bottom of the cylinder, while the vapour above it will stay exactly in the same state (pressure, temperature and density).
So the cylinder will contain less water vapour in absolute terms, but the amount of vapour of a sample of the gaseous phase (say 1 cm^3) will be exactly the same (= the vapour density is the same).
Repeat the experiment with a mixture of air and vapour and you'll get the same result for the vapour part, but since this time you've got a mass of air as well, which does not condense and is forced into a smaller volume, its density - and therefore the density of the mixture air+vapour - will have to increase, but the density of the water vapour itself will not.
I still have trouble visualizing that two different columns of gas, each exerting the same pressure, will have different densities.
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Just to clarify, in my previous post, when talking about different gases, I was referring to the ideal gas law, which applies to 'small' systems, in which the density is the same throughout the gas.
If you take a 'column' of air so tall that the density inside it varies with height, then things get more complicated. I'd have to think about it.
If you take a 'column' of air so tall that the density inside it varies with height, then things get more complicated. I'd have to think about it.
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Ok thanks DeeDay, I'm happy with my understanding now.
I said in my last post
But this is not correct, as now the compressed sample 1 would not be the same 1 cu metre volume as sample 2. So the wikipedia quote makes sense to me.
Thanks to Mathy too.
I said in my last post
I could have used the term “isothermally compressed” (and not have changed the question) to get from sample 1 to sample 2.
Thanks to Mathy too.
