What is the density of water vapor at room temperature?

Aug 7, 2016

${\text{0.0172 g L}}^{- 1}$

Explanation:

The first thing to do here is to pick a temperature that can be considered room temperature. I'll go with ${20}^{\circ} \text{C}$, which is equivalent to

T["K"] = 20^@"C" + 273.15 = "293.15 K"

Next, look up the partial pressure of water vapor at ${20}^{\circ} \text{C}$. You'll find it listed as

$P = \text{0.0230 atm}$

http://www.endmemo.com/chem/vaporpressurewater.php

Now, that you have a pressure and a temperature to work with, use the ideal gas law equation to find the density of water vapor.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Here you have

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant
$T$ - the absolute temperature of the gas

Let's assume that your room contains a mass $m$ $\text{g}$ of water vapor. Use the molar mass of water, ${M}_{\text{M H"_2"O}}$, to express the number of moles in terms of this mass $m$

$n = \frac{m}{M} _ \left(\text{M H"_2"O}\right)$

Plug this into the ideal gas law equation to get

$P V = \frac{m}{M} _ \left(\text{M H"_2"O}\right) \cdot R T$

As you know, density, $\rho$, is defined as mass per unit of volume. Rearrange the above equation to get $\frac{m}{v} = \rho$ isolated on one side

$P = \frac{m}{M} _ \left(\text{M H"_2"O}\right) \cdot \frac{1}{V} \cdot R T$

$P \cdot {M}_{\text{M H"_2"O}} = \frac{m}{V} \cdot R T$

m/V = (P * M_ ("M H"_ 2"O"))/(RT) implies color(purple)(|bar(ul(color(white)(a/a)color(black)(rho = M_ ("M H"_2"O") * P/(RT))color(white)(a/a)|)))

Water has a molar mass of ${\text{18.015 g mol}}^{- 1}$. Plug in your values to find

rho = "18.015 g"color(red)(cancel(color(black)("mol"^(-1)))) * (0.0230 color(red)(cancel(color(black)("atm"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 293.15 color(red)(cancel(color(black)("K"))))

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\rho = {\text{0.0172 g L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$