# What is the density of water vapor at room temperature?

##### 1 Answer

#### Answer:

#### Explanation:

The first thing to do here is to pick a temperature that can be considered *room temperature*. I'll go with

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

Next, look up the **partial pressure** of water vapor at

#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# - theuniversal gas constant

#T# - theabsolute temperatureof the gas

Let's assume that your room contains a mass **molar mass** of water, *number of moles* in terms of this mass

#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**,

#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

#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)|)))#