Why does moist air have lesser density than dry air? When the amount of vapor inside increases, the mass increases and hence won't the density as well increase?
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2 Answers
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The composition of dry air is about $78\%$ $\ce{N2}$, $21\%$ $\ce{O2}$ and $1\%$ $\ce{Ar}$. The molecular weights of these compounds are:
$\ce{N2} = \pu{28 g/mol}$
$\ce{O2} = \pu{32 g/mol}$
$\ce{Ar} = \pu{40 g/mol}$
So, the average molecular weight of dry air is given as:
$\pu{(0.78 * 28 g/mol) + ( 0.21 * 32 g/mol) + ( 0.01 * 40 g/mol) g/mol = 29 g/mol}$
The molecular weight of water is only $\pu{18 g/mol}$.
So, a molecule of air being replaced by a molecule of water results in an air molecule with an average molecular weight of $\pu{29 g/mol}$ being replaced by a water molecule having a molecular weight of only $\pu{18 g/mol}$.
This scenario assumes a constant temperature and pressure when humidifying the air.
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$\begingroup$ Important starting principle is that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or atoms in the case of monatomic gases)... $\endgroup$– DJohnMCommented Feb 28, 2017 at 2:22
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$\begingroup$ @DJohnM , of course very true. Worth pointing out, thanks. $\endgroup$– airhuffCommented Feb 28, 2017 at 2:31
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If you considered the air within a given volume in two rooms, one with moist air and one with dry air, moist air will have some water vapor in place of other molecules that make up air (e.g. oxygen and nitrogen). A Water molecule has a lower mass than either oxygen or nitrogen, thus the volume of moist air will have a lower mass than the volume of dry air and therefore a lower density.
