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
$P\times V=\left({\displaystyle \frac{m}{MM}}\right)\times R\times T$
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 ($$0.0002{\text{m}}^3/{\text{m}}^3$$degrees C) and multiply it by the temperature difference, which is $$10$$ degrees C in this example. Work out$$0.0002x10=0.002.$$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.