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, which says:
... if 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 of surface or 1 m3 of air. The total mass of air pressing down on 1 m2 is about ten tonnes, a 2.5cm deep x 1 m2 puddle would 25 litres = 25 kg. So that means about 0.25 per cent by mass, or nearly 5,000 ppm by number of molecules, about ten times as much as there is CO2.
------------------------------------------
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 when air rises and vice versa. But the observed lapse rate is only 6.5 degrees/km. So there are 3.5 degrees/km 'missing'. How does that tie in with the amount of water vapour, which holds 2,257 J/g of latent heat?.
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; after that it took five minutes. The answer is about 9 grams of water per m3 at sea level, falling by 1.5 grams per m3 per kilometer altitude up to 6 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 about 2.5 kg in the lowest 6 km (average = 4.5g x 6,000m = 2.7 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 kJ
- 9g water vapour x 2,257 J/g latent heat = 20 kJ
Total 308 kJ.
1 m3 at 1 km altitude has:
- 1 kg x 9.81 N/s2 x 1,000m = 10 kJ potential energy
- 281.5K x 1,000 J/K/kg thermal energy = 281 kJ
- 7.5g water vapour x 2,257 J/g latent heat = 17 kJ
Total = 308 kJ = the same as at sea level.
Reality check: 9 grams/m3 = 7g/kg of air = 65% relative humidity at 288K. That seems a bit high, but that is probably largely due to my crass rounding, approximations and the fact that sea level temps are between 'below freezing' and 'very warm' and RH is not linear with temperature. Above the Poles at minus 20C and 50% RH, there's only 0.6g/kg; over the tropical oceans at 30C and 75% RH, there's 20 g/kg. Or something like that.
------------------------------------------
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 few grams out of the 2.5 kg total, so can be ignored for these purposes.
------------------------------------------
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 only a small fraction of that is in clouds. Yet clouds 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.