## 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.

#### Tip:

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: \[\boxed{V_{g} = \text{22.4}n_{g}}\]

#### Tip:

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

### Question

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*}