1. What percent of the atmosphere is water vapour? 2. The lapse rate and latent heat of evaporation

There are many sources that say it’s on average 2 percent.

I had long assumed this was correct, it turns it out that it’s complete rubbish, and it’s actually about one-tenth of that.

Here is a good overview:

all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

This number seems to be widely accepted and seems very plausible, although some say as much as 1.5 inches.

I like to reduce things down to an average m2. The total mass of air pressing down is about ten tonnes, a 2.5cm deep x 1 m2 puddle would be about 25 litres = 25 kg. So that means about 0.25 per cent, or 2,500 ppm, about six times as much as there is CO2.
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The lapse rate and latent heat of evaporation.

We know that the dry lapse rate should be about 10 degrees/km, that’s easy, it’s acceleration due to gravity ÷ specific heat capacity = 10 degrees/km. Energy is conserved, it just changes from one form to another; thermal energy is converted to potential energy as air rises and vice versa. But the observed lapse rate is only 6.5 degrees/km. So there are 3.5 degrees/km ‘missing’.

That means if 1 m3 of air at sea level contains X grams of water vapour, then 1 m3 at 1 km altitude must have X – 1.5 grams*.

I spent about two hours scribbling calculations to reconcile all this before I noticed that the original estimate is out by a factor of ten. The answer is about 9 grams of water per m3 at sea level, falling by 1.5 grams per m3 per kilometre altitude up to six km altitude. The air is pretty dry above that (all water vapour has condensed out as clouds). Add them all up, and you get back to the total mass of 2.5 kg in the lowest six km (average = 4.5g x 6,000m = 2.5 kg).

* Entropy says the total amount of energy will spread out evenly. Using rounded figures…
1 m3 at sea level has:
- potential energy = 0 J
- 288K x 1,000 J/K/kg thermal energy = 288,000 J
- 9g water vapour x 2,257 J/g latent heat = 20,000 J
Total 308,000 J.
1 m3 at 1 km altitude has:
- 1 kg x 9.81 N/s2 x 1,000m = 10,000 J potential energy
– 281.5K x 1,000 J/K/kg thermal energy = 282,000 J
- 7.5g water vapour x 2,257 J/g latent heat = 17,000 J
Total = 308,000 = the same.
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Then there’s the thorny issue of splitting that 2.5kg into gaseous water vapour and tiny droplets in clouds (which have lost their latent heat). It turns out that the amount actually in clouds is only a very small part of the 2.5 kg total, so can be ignored for these purposes.
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Which is all rather surprising really; of all the water on Earth, only about 1/10 of 1% of 1% is in the atmosphere, and barely any of that is in clouds. Yet they have a dramatic effect on the weather, and if you overlook them, your in/out radiation calculations will be wildly wrong. For some reason, AGW Theories rely on the wildly wrong calculations, but hey.