Question

What is the density of water vapor at room temperature?

Answer 1

`"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^@"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^@"C"`. You'll find it listed as

`P = "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.

`color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "`

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` `"g"` of water vapor. Use the molar mass of water, `M_("M H"_2"O")`, to express the number of moles in terms of this mass `m`

`n = m/M_("M H"_2"O")`

Plug this into the ideal gas law equation to get

`PV = m/M_("M H"_2"O") * RT`

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

`P = m/M_("M H"_2"O") * 1/V * RT`

`P * M_("M H"_2"O") = m/V * RT`

`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 `"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"))))`

`color(green)(|bar(ul(color(white)(a/a)color(black)(rho = "0.0172 g L"^(-1))color(white)(a/a)|)))`