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:

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

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 $$[\pu{ppm}] = \frac{\pu{2.766 mg}}{\pu{0.025 l}} = \pu{110.6 ppm}$$

  • $\begingroup$ You have calculated the number of moles of ethanol. Do you need something else to find the concentration? $\endgroup$ Commented Sep 23, 2019 at 16:30
  • $\begingroup$ I just want to verify that it is the right way to calculate concentration of volatile liquids in the air? Thank you! $\endgroup$
    – VGranin
    Commented Sep 23, 2019 at 16:32
  • $\begingroup$ Just a couple more steps and you're done. $\endgroup$ Commented Sep 23, 2019 at 16:43
  • $\begingroup$ Thank you, made them, is it right? As this calculator giving me completely different answer madur.com/index.php?page=/partial_pressure $\endgroup$
    – VGranin
    Commented Sep 23, 2019 at 16:54
  • $\begingroup$ Don't you think approx.0.06 atm should lead to approx.60000 ppm v/v, even without any calculation ? $\endgroup$
    – Poutnik
    Commented Sep 24, 2019 at 7:33

1 Answer 1


Usage of ppm/ppb etc units is discouraged, as they are ambiguous without the explicit context. As generally, 1 ppm (w/w) <> 1 ppm (V/V) <> 1 ppm (n/n) <> 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 $\pu{8.31446261815324 J K-1 mol-1}$

  • $\begingroup$ Thank you for your answer! I will follow your advises to try to calculate a more reasonable number. Could you please also advise on the ppm calculation in that case? Thank you! $\endgroup$
    – VGranin
    Commented Sep 24, 2019 at 7:15
  • $\begingroup$ @VGranin ppm is ambiguous: w/w, V/V, n/n ? $\endgroup$
    – Poutnik
    Commented Sep 24, 2019 at 14:40
  • $\begingroup$ Pure liquid water would be 1,000,000 ppm w/w, V/V, n/n, and m/V. Pure steam at normal pressure and a temperature a bit above the normal boiling point on the other hand would be 1,000,000 ppm w/w, V/V, n/n, but ~800 w/V because the density of steam is much lower than that of liquid water. $\endgroup$
    – Karsten
    Commented Sep 24, 2019 at 15:23
  • 1
    $\begingroup$ Using ppm is a recipe for disaster... $\endgroup$
    – Karsten
    Commented Sep 24, 2019 at 15:24
  • $\begingroup$ Engineer here. Using the vapour pressure and the ideal gas law (IGL) will usually give a sufficiently accurate solution for gases for which $\dfrac{R\cdot T}{P} > 20 \frac{L}{g\ mol}$. That said, if you want to correlate vapour-liquid equilibrium, using the IGL is not a good idea. $\endgroup$ Commented Sep 26, 2019 at 9:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.