Question

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

Answer 1

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.