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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?

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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: $$29 \mathrm{g/mol} \times0.98+18\mathrm{g/mol}\times0.02=28.78\mathrm{g/mol}.$$

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  • $\begingroup$ pilots know this too; on a humid day, the plane's wings produce less lift and the takeoff run for a given plane must be extended accordingly. $\endgroup$
    – niels nielsen
    Commented Apr 24, 2018 at 21:16
  • $\begingroup$ How does one use the Clausius-Clapeyron relation? Whereever I look it requires a lot more than just the relative humidity and temperature $\endgroup$
    – andrew
    Commented Apr 27, 2018 at 19:19
  • $\begingroup$ @andrew To do it properly you need to numerically integrate the equation. For rough estimates, there is a closed form solution if you approximate the latent heat of vaporization as constant with temperature. Alternatively, you can get approximate values off a plot, like this $\endgroup$
    – Ben51
    Commented Apr 30, 2018 at 16:23
  • $\begingroup$ Can you show how to get that 2% figure? $\endgroup$
    – fab
    Commented May 23, 2023 at 2:36

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