Riddle: Two containers of air have the same Pressure and Temperature. Container 1 has dry air with 0% water vapor and 100% air while container 2 has 2% water vapor and 98% air. Which container is less dense and how much less dense is it?
Answer to Riddle: The average atomic weight of air is 28.97 atomic units. This comes from the air being 78% Nitrogen (N2 has 28.02 atomic units), 21% Oxygen (O2 has 32.00 atomic units) and 1% Argon (Ar has 39.94 atomic units). Using a weighted average the average atomic weight of air is (0.78*28.02 + 0.21*32 + 0.01*39.94 = 28.97 atomic units.
If the air is 2% water vapor then the air is 98% air with an atomic weight of 28.97 and 2% water vapor in which the water vapor has 18 atomic units. Water vapor is H20. Each Hydrogen is 1 unit and a solo Oxygen is 16 units for a total of 18 units. Using a weighted average the average atomic weight of the air with 2% water vapor is (0.98*28.97 + 0.02*18 = 28.75 atomic units.
Since the air with the moisture has a lower average atomic weight then Container 2 is less dense than Container 1. The amount of change is equal to the difference divided by the original value. The change is equal to 28.75 - 28.97 = -0.22. The percent change is (-0.22 / 28.97) * 100% = -0.76%. The change is less than 1 percent but the change is very important when you consider how these density differences influence the weather and bouyancy in the atmosphere.