Question

What is the density of water vapour at 25$^{o}C$?

Answer 1

As we know that generally density of an object equals its total mass divided by its total volume and the average density of an object is equal to its total mass divided by its total volume. So here we have to calculate the density of water vapour at 25$^{o}C$.
Formula used:
For ideal gas: PM = DRT

Complete answer:
We can calculate the density of water vapour at 25$^{o}C$ either by considering it as ideal gas or as a real gas.
Let us consider water vapour to be an ideal gas:-
So the ideal gas equation which includes density is: PM = DRT
where,
P = Pressure of the gas in bar
M = molar mass of the gas in g/mol
D = density of the gas
R = gas constant = $0.083145\dfrac{L\cdot bar}{mol\cdot K}$
T= temperature in Kelvin
Let us first rearrange the formula as follows:-
$D=\dfrac{PM}{RT}$
The values of the above terms are as follows:-
P = Pressure of the gas in bar = 1 bar
M = molar mass of the gas in g/mol = 18.015 g/mol
T= temperature in Kelvin = 25$^{o}C$= (273.15 + 25) K = 298.25K
On putting all these values in the given formula we get:-
$\begin{aligned} & \Rightarrow D=\dfrac{1bar\times 18.015g/mol}{0.083145\dfrac{L\cdot bar}{mol\cdot K}\times 298.15K} \\\ & \Rightarrow D=\dfrac{18.015bar\cdot g/mol}{24.789\dfrac{L\cdot bar}{mol}} \\\ \end{aligned}$D = 0.7267 g/L
-Hence the density of water vapour at 25$^{o}C$ is = 0.7267 g/L

Note:
-There was no requirement of calculating the density of water vapour using van der waal equation (considering it as real gas) because the result is approximately almost the same and also it is a lengthy method which should be preferred only when asked.
-Use all the units along the values for errorless calculations and use the gas constant and other terms according to the units.