What is the density of water vapor at room temperature?

1 Answer
Aug 7, 2016

Answer:

#"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)|)))#