Question

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

Short answer

Because wet air has a lower molar mass than dry air, we can argue that wet air has a lower density.

Step-by-step solution

Definition of volume

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

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

"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

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 \( = 18\frac{{{\rm{ gram }}}}{{{\rm{mol}}}}\)

Find which air is denser

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.