Which is denser at the same temperature and pressure, dry air or air saturated with water vapour? Explain.

A substance's volume is the amount of space it takes up.

"Dry" air contains no water vapor, and is mostly a mixture of molecular nitrogen \(\left( {{{\rm{N}}_2}} \right)\) and molecular oxygen \(\left( {{{\rm{O}}_2}} \right)\).

Molar mass \((\mu )\) of dry air:

Atomic weight of Carbon \(({\rm{N}}) = 14\) grams.

Atomic weight of Chlorine \(({\rm{O}}) = 16{\rm{gram}}\).

So, molecular weight of dry air \( = (14) + (16)\) grams.

\( = 30\)grams.

Molar mass of dry air \( = 30\frac{{{\rm{ gram }}}}{{{\rm{ mol }}}}\).

Hence, as "Wet" air ( or ) air saturated with water vapour mostly contains water \(\left( {{{\rm{H}}_2}{\rm{O}}} \right)\).

Molar mass \((\mu )\)of wet air:

Atomic weight of Carbon (H)\( = 1\)grams.

Atomic weight of Chlorine\(({\rm{O}}) = 16\)gram.

So, molecular weight of wet air\( = (2) \cdot (1) + (16)\)grams.

\( = 2 + 16\)

\( = 18\)grams.

Therefore, molar mass of wet air \( = 18\frac{{{\rm{ gram }}}}{{{\rm{mol}}}}\)

We have

Density=\((\rho ) = \frac{{P \cdot (\mu )}}{{RT}}\)

Here, molar mass\((\mu )\)is directly proportional to density\((\rho )\).

Therefore, molar mass of wet air is less than dry air, we can say that density of wet air is less than dry air.