Question

How do you determine the density of water vapour at different temperatures?

Answer 1

Hint: you can consider water vapor as an ideal gas , so you can write the ideal gases law by the statement that product of pressure and volume is equal to the product of temperature, gas constant and number of moles.
\[P\times V=n\times R\times T\]
where \[P=\text{ pressure }V=\text{volum }n=\text{number}\]
of mol ($n=$ mass in g/ MM)
$R=$ universal constant of ideal gases
$T=$ temperature in K

Complete step by step answer:
So from above formula you can write the ideal gases law in this way
$P \times V = \left( {\dfrac{m}{{MM}}} \right) \times R \times T$
density is mass divided volume\[d=mV\]
and rearrange the equation you have
\[d=mV=P\times MMR\times T\] where you can see that increasing temperature, d decreases.
Moreover as the MM of water \[\left( 18\text{g/mol} \right)\] is less than MM of air (about\[29\text{ g}/\text{mol}\]), vapor has a less density than air
For water, use its volumetric temperature expansion coefficient (\[0.0002\text{ }{{\text{m}}^{3}}/{{\text{m}}^{3}}\]degrees C) and multiply it by the temperature difference, which is \[10\] degrees C in this example. Work out\[0.0002\text{ }x\text{ }10=0.002.\]Add one to this number to get: \[~1+0.002=1.002.\] It can be used to calculate exact quantity of water vapor in the air from a relative humidity (\[\text{RH}=%\] local air humidity measured / local total air humidity possible ) Given an RH percentage, the density of water in the air is given by \[\text{RH}\times \text{SVD}=\text{Actual}\] Vapor Density.

Note: Heating a substance causes molecules to speed up and spread slightly further apart, occupying a larger volume that results in a decrease in density. ... Hot water is less dense and will float on room-temperature water. Cold water is more dense and will sink in room temperature water.
This question might look tricky but helps to build up concepts , so try and understand it carefully.