Calculation of VOCs concentration (ppm) in the headspace of a bottle What is the right way to calculate the concentration of volatile compounds in the air in a closed container?

Suppose we have a $\pu{50 mL}$ bottle with $\pu{25 mL}$ of $100\%$ ethanol $(\ce{EtOH})$ in it.

In this case, will it be okay to use $PV = nRT?$

Doing so, it will be:

$$
\begin{align}
p(\ce{EtOH}) &amp;= \pu{0.05872 atm}\\
V &amp;= \pu{0.025 L}\\
R &amp;= \pu{0.08206 L atm mol-1 K-1}\\
T &amp;= \pu{298 K}\\
\end{align}
$$

giving us $n = \pu{6.00314E-5 mol}.$

The ppm: $\pu{1 ppm} = \pu{1 mg l-1}$

$$M(\ce{EtOH}) = \pu{46.07 g mol-1}$$

$$\pu{6.00314E-5 mol} × \pu{46.07 g mol-1} × 1000 =  \pu{2.766 mg}$$

This lead to $$[\pu{ppm}] = \frac{\pu{2.766 mg}}{\pu{0.025 l}} = \pu{110.6 ppm}$$
Usage of ppm/ppb etc units is discouraged, as they are ambiguous without the explicit context. As generally, 1 ppm (w/w) &lt;> 1 ppm (V/V) &lt;> 1 ppm (n/n) &lt;> 1 ppm (w/V).

It is recommended to use explicit units, like e.g mg/L.

Expecting ideal gas behaviour $pV = nRT$ is not reasonable.   Ideal gases are not in equilibrium with their liquid phase.

There should be rather used a real gas equation.The easiest to use is the van Der Waals equation, but even that has significant error near condensation state. 

Additionally, the sample(!) temperature  has to be strictly controlled, otherwise $n$ would be variable.

Why not rather to use small volume of ethanol to be known to fully evaporate in given volume ? Or perhaps other solvent, unless ethanol is waterless.
Note that ethanol has significant deviation from the Raoult's law.



I prefer $R = \pu{8.31446 J K-1 mol-1}$

Note that by May 2019 redefinition, 
 It's exact value is $\pu{8.31446261815324 J K-1 mol-1}$