# Density of dry vs. humid air

by bobbykilkenny
Tags: density, humid
 P: 4 How do I explain to my Engineer friends that a ball will actually travel further in humid conditions b/c humid air has less density than drier air? Please help us come to a conclusion. Thanks
 Sci Advisor HW Helper PF Gold P: 12,016 Why do you believe this? Because a ball in less dense air experience less air resistance? Note that a ball in denser air will experience a stronger buouyancy force, and hence "float" for a longer time. Anyways, these concerns are really about measuring unmeasurable negligibilities..
 P: 4 My assumption is an object will travel further with less resistance. I'm not so sure about your bouyancy theory though. Everyone knows baseballs fly further in the less dense air of high elevation ball parks. I am more concerned with the density (and related resistance) of dry vs. humid air
 Sci Advisor HW Helper PF Gold P: 12,016 Density of dry vs. humid air You're probably right; but the important issue is whether the density difference is large enough to matter. I don't think so, but then I don't know the usual, relevant density values. possibly someone else knows, or have a reference table nearby
 P: 4,777 Your logic seems to make sense to me. Humid air weighs less than dry air becuase water vapor is less dense. The thing that will cause your ball to slow down the most soonest will be air resistance. Which is primarily a function of density. $$f ~= Dv^2$$ where f is the fluid resistance v is the object velocity and D is a proportionality constant that depends on the shape and size of the body and the properties of the fluid. The shape and size of the ball never change, so the only variable there would be the properties of the fluid. If you decrease the density, then I would likely think that D should increase in value. Because the fluid resistance should decrease with decreasing density. But for your situation you are going so slow, and the effective change in density is so small, that you would not detect it. You would need a major change in density to have noticable effects on the hight the ball obtains. The bouyancy part of what he says would probably mean that on the way down, the ball would take longer for more dense air, than it would for less dense air, since the air cushions the ball as it falls down. Again, this would be a very small undetectable difference though.
HW Helper
P: 2,280
USA Today explains this very well:
http://www.usatoday.com/weather/wdensity.htm

 Quote by USA Today To see why humid air is less dense than dry air, we need to turn to one of the laws of nature the Italian physicist Amadeo Avogadro discovered in the early 1800s. In simple terms, he found that a fixed volume of gas, say one cubic meter, at the same temperature and pressure, would always have the same number of molecules no matter what gas is in the container. Most beginning chemistry books explain how this works. Imagine a cubic foot of perfectly dry air. It contains about 78% nitrogen molecules, which each have a molecular weight of 28 (2 atoms with atomic weight 14) . Another 21% of the air is oxygen, with each molecule having a molecular weight of 32 (2 atoms with atomic weight 16). The final one percent is a mixture of other gases, which we won't worry about. Molecules are free to move in and out of our cubic foot of air. What Avogadro discovered leads us to conclude that if we added water vapor molecules to our cubic foot of air, some of the nitrogen and oxygen molecules would leave — remember, the total number of molecules in our cubic foot of air stays the same. The water molecules, which replace nitrogen or oxygen, have a molecular weight of 18. (One oxygen atom with atomic weight of 16, and two hudrogen atoms each with atomic weight of 1). This is lighter than both nitrogen and oxygen. In other words, replacing nitrogen and oxygen with water vapor decreases the weight of the air in the cubic foot; that is, it's density decreases.
This is going to make a very small difference, though. Pressure and temperature will have a much greater impact.

Ideal would be hot, humid day, with a storm front moving in.
 P: 4 How would you respond to someone who says: That all of the computations that shows humid air is less than dry air rely on the ideal gas law, which fails to account for the polar nature of water molecules (and therefore the REAL nature of the gas). If an alternative equation that accounts for the non-ideal (REAL) effect of water vapor were applied (such as the Van der Waal) equation, it would become evident that, in fact, there was a higher molecular density associated with air having high humidity. It is not disputed that the mass/volume nature of dry air is greater than humid air – it is. The point Camp B is trying to make is that there is actually a higher moles/volume number associated with humid air compared to dry air. Therefore, because the ball has to flow through more molecules (even if they are lighter molecules) the frictional resistance is higher with wet air. RESULT: the ball will fly further when levels of humidity are low. Again, all of the analysis we have been able to find apply the ideal gas law (which is incorrect). Additionally, all practical inspections into the frictional resistances rely on simple Newtonian Mechanics associated with dry air having a greater mass/volume density than wet air. Because one side believes the effect to be on a quantum level (i.e. frictional resistances with individual atoms because wet air has a greater moles/volume number), Newtonian Mechanics arguments are invalid.
 Sci Advisor PF Gold P: 9,435 The density of a mixture of dry air molecules and water vapor molecules is: $$D =\frac{P_d}{R_dT} + \frac{P_v}{R_vT}$$ D = density, kg/m3 Pd = pressure of dry air, Pascals Pv = pressure of water vapor, Pascals Rd = gas constant for dry air, J/(kg*degK) = 287.05 Rv = gas constant for water vapor, J/(kg*degK) = 461.495 T = temperature, degK = deg C + 273.15 As you can see, the largest value for D will result when Pv is zero.