airatmospheric sciencehumidity
The higher the relative humidity percentage, the higher the molar mass of air, right? So then how do I add the higher amount of water in the air to the molar mass of air?
For example, let's take a cubic meter of air, with the molar mass of air being 28,96 g/mol, with a relative humidity of 85 %. Surely the humidity has added to the molar mass of the air within that cubic meter? How do I calculate this?
You are right. Field manual is wrong. Water has a lower molecular mass, which reduces air density.
But is the difference significant enough to notice?
Terminal velocity is the speed at which a falling body reaches a stable velocity when gravity and air resistance meet in a stable equilibrium. Let's assume the bullet travels close to terminal velocity so we can use its model.
Air density is part of the terminal velocity equation. Air density is inverse square-root proportional to terminal velocity. This means terminal velocity goes down, but only as a function of the square root of air density.
Air density is charted above (credit to jaffer@MIT). You can see that in a hot climate of 30 deg C, air density has about a 3 percent from 0-100% humidity. But remember we have to take the square root of that difference - about 1.5%. So the most humidity can affect the speed of the bullet is 1.5%.
I don't know the math for how that affects a ballistic trajectory. Sorry.
It could be insignificant, which is why it they got it wrong. It's probably based on what people think intuitively, and backed up by confirmation bias. If it had a significant impact, you think they would know this by know. This scientific knowledge is not exactly new.
The field manual is wrong.
The evaporation rate can be calculated but with a quite complex diffusion differential equation.
@Samuel is correct that you cannot assume a uniform distribution of humidity. There is water density variation in space. The gradient is the driving force to move water vapor away from the water surface and to maintain the evaporation process.
The water density is higher close to the water surface. On the surface, we may assume it to be at vapor pressure in most case. If it is less than the vapor pressure, the evaporation is aggressive and if it is more than the vapor pressure, we will see condensation.
When it is at water vapor pressure, the water vapor is then driven out into space under diffusion. If the space humidity is high, the driven force is small that the less water vapor moves into space and vice versa.
There is also heat transfer associated to this evaporation process making it more complicated.
Best Answer
Water weighs about 18 g/mol, which is less than 29, so adding water vapor to air actually decreases the density. The amount that the density (or molar mass, if you like) decreases depends on the change in the absolute humidity--you can't calculate based on relative humidity alone without knowing the temperature. If the temperature and relative humidity are known, you can determine the absolute humidity from the Clausius-Clapeyron relation. Let's say your 85%-relative-humidity air is at about 20 degrees Celsius, and you find, from Clausius-Clapeyron, that this equates to an absolute humidity of 2% (this is roughly right--for the illustration purposes, I'm not bothering with precision). Then your sample consists of 98% dry air molecules (average molar mass 29) and 2% water molecules (molar mass 18). The molar mass of the combined sample is: $$29 \mathrm{g/mol} \times0.98+18\mathrm{g/mol}\times0.02=28.78\mathrm{g/mol}.$$