Molar Volumes of Gases

Gases and solutions: Molar volumes of gases

It is possible to calculate the volume of one mole of gas at standard temperature and pressure (STP) using what we now know about gases.


STP is a temperature of \(\text{273}\) \(\text{K}\) and a pressure of \(\text{101.3}\) \(\text{kPa}\). The amount of gas is usually \(\text{1}\) \(\text{mol}\).

We write down all the values that we know about one mole of gas at STP:

\begin{align*} p & = \text{101.3}\text{ kPa} = \text{101 300}\text{ Pa} \\ n & = \text{1}\text{ mol} \\ R & = \text{8.31}\text{ J·K$^{-1}$·mol$^{-1}$} \\ T & = \text{273}\text{ K} \end{align*}

Now we can substitute these values into the ideal gas equation:

\begin{align*} pV & = nRT \\ (\text{101 300})V & = (\text{1})(\text{8.31})(\text{273}) \\ (\text{101 300})V & = \text{2 265.9} \\ V & = \text{0.0224}\text{ m$^{3}$} \\ V & = \text{22.4}\text{ dm$^{3}$} \end{align*}

The volume of 1 mole of gas at STP is \(\text{22.4}\) \(\text{dm$^{3}$}\).

And if we had any number of moles of gas, not just one mole then we would get: \[\boxed{V_{g} = \text{22.4}n_{g}}\]


The standard units used for this equation are p in \(\text{Pa}\), V in \(\text{m$^{3}$}\) and T in \(\text{K}\). Remember also that \(\text{1 000}\) \(\text{cm$^{3}$}\) \(=\) \(\text{1}\) \(\text{dm$^{3}$}\) and \(\text{1 000}\) \(\text{dm$^{3}$}\) \(=\) \(\text{1}\) \(\text{m$^{3}$}\).

Example: Molar Gas Volume


What is the volume of \(\text{2.3}\) \(\text{mol}\) of hydrogen gas at STP?

Step 1: Find the volume

\begin{align*} V_{g} & = (\text{22.4})n_{g} \\ & = (\text{22.4})(\text{2.3})\\ & = \text{51.52}\text{ dm$^{3}$} \end{align*}

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