">Moist air is less dense because the molecular weight of water vapor is less than that of other air constituents, such as nitrogen"
I think that justification only works if you assume water vapor is an ideal gas, since we're dealing of temps neat its saturation point it does not behave like an ideal gas afaik.
Do you know the ideal gas law? (Sometimes it is called the perfect gas law.) The ideal gas law states that a given number of gas molecules have a density that only depends only temperature and pressure in a simple manner.
Air behaves very closely like an ideal gas at room temperature and atmospheric pressure. In fact its components also behave like ideal gases (oxygen, nitrogen, water vapor, carbon dioxide, etc...) That means that 1 liter of oxygen has the same number of molecules as 1 liter of nitrogen and 1 liter of water vapor - at the same temperature and pressure. However each water molecule (18 atomic units) is lighter than the oxygen (32 atomic units) and nitrogen molecules (28 atomic units). Thus 1 liter of water vapor weighs less than 1 liter of oxygen or 1 liter of nitrogen.
Thus a given volume of dry air weighs more than the same volume of wet air if they are at the same temperature and pressure.
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I think that justification only works if you assume water vapor is an ideal gas, since we're dealing of temps neat its saturation point it does not behave like an ideal gas afaik. >>
Water vapor being near saturation does nothing to it density. For most practical purposes, atmospheric gases are treated as ideal.
Assuming an ideal gas, dry air is about 61% heavier than an equivalent amount of water vapor.
There are many cases where the ideal gas law fails. Extreme temperatures and extreme pressures are examples.
The ideal gas law also fails when the molecules have attractions toward one another. One oxygen molecule really isn't attracted to another oxygen molecule since they are symmetric and thus have a neutral charge distribution. (Assuming the molecules aren't forced together by the extremes listed above and also ignoring the small Van der Walls forces). Similarly a neutral oxygen molecule isn't attracted to neutral nitrogen or carbon dioxide molecules.
Water is different, it has a dipole (one end is slighly negatively charged and one end is slightly postively charged). Thus if you have lots of water molecules, they will attract one another - leading to a higher density. So when you add more and more water vapor to the air it will become more dense. The key is will it ever reach the 60% more weight needed to surpass the weight of the dry air? Under normal temp and pressure ranges, this 60% increase in density doesn't occur.
Capn is ABSOLUTELY correct. water vapor does not behave like an ideal gas anywhere in the vicinity of the liquid-vapor dome (applicable to most practical and engineering applications). This becomes extremely important when analyzing state changes.
Just for chits and giggles:
the specific volume of pure saturated water vapor @ 1bar 100 degrees C is (from steam tables):
v=1.696 m^3/kg
the specific volume of air @ 1bar 100 degrees C is (from Pv=RuT/molecular weight):
v=1.0704 m^3/kg
Therfore, at this state (and it actually works out at other states as well) that the density of water is less than that of air, so water is "lighter"
Just look at how a barometer works. The barometer level drops when it rains or it's about to rain (less atmospheric pressure), and it rises when the weather is good (increasing atmospheric pressure). That should answer your question.
From my environmental engineering book: "Ironically, although we refer to hot, humid air as 'heavy,' it is in fact less dense than cool, dry air." They go on to prove this through mole fractions, but I don't feel like summarizing it.
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