# What is the density of wet air with 75% relative humidity at 1atm and 300K? Given : vapour pressure of H_2O is 30 torr and average molar mass of air is 29g/mol?

## a) $\text{1.174 g/L}$ b) $\text{1.156 g/L}$ c) $\text{1.178 g/L}$ d) $\text{1.143 g/L}$

Sep 1, 2017

It should be $\text{1.165 g/L}$. There is a typo in the given answers...

The relative humidity $\phi$ of an air-water mixture is given by:

$\phi = {P}_{{H}_{2} O} / \left({P}_{{H}_{2} O}^{\text{*}}\right) = 0.75$

where ${P}_{{H}_{2} O}$ is the partial vapor pressure of water in the air and $\text{*}$ indicates the substance in isolation.

In this case we have

${P}_{{H}_{2} O}^{\text{*" = "30 torr}}$
${P}_{{H}_{2} O} = 0.75 \times \text{30 torr" = "22.5 torr}$

at P_("wet air") = "1 atm" = "760 torr" and $T = \text{300 K}$.

Since we know the pressure of water vapor in the air is $\text{22.5 torr}$, the vapor pressure of the dry air is found by subtraction:

${P}_{\text{dry air" = "760 torr" - "22.5 torr" = "737.5 torr}}$

or $\text{0.970 atm}$. And thus, the vapor pressure of water in the air mixture is $\text{0.030 atm}$.

This variant on the ideal gas law, assuming air is an ideal gas, can be used to find its density:

$P M = D R T$

where:

• $P$ is the pressure in $\text{atm}$ of the ideal gas. Here we treat wet air as an ideal gas.
• $M$ is the molar mass of a given component in the sample in $\text{g/mol}$,
• $D$ is the density of the ideal gas in $\text{g/L}$.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant.
• $T$ is the temperature in $\text{K}$.

In a given volume of air at a given temperature and total pressure, the density for ideal gases is additive, i.e.

$D = {D}_{1} + {D}_{2}$

Thus, the density of the wet air is given by:

color(blue)(D_"wet air") = D_"dry air" + D_(H_2O)

$= \frac{{P}_{\text{dry air"M_"air}}}{R T} + \frac{{P}_{{H}_{2} O} {M}_{{H}_{2} O}}{R T}$

$= \frac{{P}_{\text{dry air"M_"air}} + {P}_{{H}_{2} O} {M}_{{H}_{2} O}}{R T}$

$= \left(0.970 \cancel{\text{atm" cdot "29 g/"cancel"mol" + 0.030 cancel"atm" cdot "18.015 g/"cancel"mol")/("0.082057 L"cdotcancel"atm""/"cancel"mol"cdotcancel"K" cdot 300 cancel"K}}\right)$

$=$ $\textcolor{b l u e}{\text{1.165 g/L}}$