Molar Gas Volume

One mole of gas at room temperature and pressure

A gas consists of particles separated by large gaps. Its volume therefore depends strongly on temperature and pressure: heating a gas tends to make it expand, while increasing the pressure tends to compress it. A gas volume is meaningful only when its conditions are known.

For these OCR calculations, at room temperature and pressure (RTP), one mole of any gas is assumed to occupy:

$24\,\text{dm}^3$

The molar gas volume is therefore $24\,\text{dm}^3\,\text{mol}^{-1}$. The gas may be hydrogen, oxygen, carbon dioxide or another gas: at the same temperature and pressure, equal numbers of gas particles occupy equal volumes. Their masses can differ even though their molar volumes are the same. For example, one mole of hydrogen has a much smaller mass than one mole of carbon dioxide, but each occupies at RTP.

This relationship gives:

$V=n\times24$

where $V$ is the gas volume in $\text{dm}^3$ and $n$ is the amount of gas in moles. Rearranging gives $n=V/24$.

Gas volume calculations from chemical reactions

A balanced chemical equation connects substances through mole ratios. To find a gas volume from an amount of reactant:

1. Write or use the balanced equation.
2. Calculate the amount of the given substance in moles if necessary.
3. Use the equation coefficients to find the moles of the required gas.
4. Multiply the gas amount by $24\,\text{dm}^3\,\text{mol}^{-1}$.

For example, calcium carbonate decomposes as:

$\mathrm{CaCO_3(s)\rightarrow CaO(s)+CO_2(g)}$

The equation shows a $1:1$ ratio between calcium carbonate and carbon dioxide. The relative formula mass of $\mathrm{CaCO_3}$ is 100, so $50\,\text{g}$ is $50\mathrm{/}100=$. It therefore produces of carbon dioxide. At RTP, its volume is:

$0.50\times24=12\,\text{dm}^3$

The result means that complete decomposition of the stated amount produces $12\,\text{dm}^3$ of carbon dioxide at RTP.

If the starting information is a mass, the whole route can be represented as:

$\text{mass}\rightarrow\text{moles of given substance}\rightarrow\text{moles of gas}\rightarrow\text{gas volume}$

Use $n=m/M_r$ for the first step and $V=24n$ for the last. The balanced equation supplies the middle mole ratio.

Using gas volumes directly

At the same temperature and pressure, the volumes of reacting gases follow the same ratio as their coefficients in the balanced equation. Consider:

$\mathrm{N_2(g)+3H_2(g)\rightarrow2NH_3(g)}$

The gaseous ratio is $1:3:2$. Consequently, $500\,\text{cm}^3$ of nitrogen can react with $1500\,\text{cm}^3$ of hydrogen to form of ammonia, provided the gases are compared under the same conditions and react completely.

There is no need to convert through moles in this case because every gas volume would be divided and multiplied by the same molar gas volume. Keep all the volumes in the same unit and apply the coefficient ratio directly.

This shortcut works only for gases. A coefficient belonging to a solid, liquid or aqueous substance cannot be treated as a gas-volume ratio.

Converting the volume units

The RTP value is expressed in $\text{dm}^3$, so volumes must use compatible units:

$1\,\text{dm}^3=1000\,\text{cm}^3$

Therefore:

• convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000;
• convert $\text{dm}^3$ to $$ by multiplying by 1000.

For example, $600\,\text{cm}^3=0.600\,\text{dm}^3$. This represents $0.600/24=0.025\,\text{mol}$ of gas at RTP. Equivalently, one occupies at RTP.

Deciding which amount controls the answer

If the question supplies more than one reactant, the amount of gas produced is controlled by the reactant that runs out first. A reactant described as being in excess is not the one that limits the product, so calculate from the other reactant. Where amounts of both reactants are given, compare how many reaction-ratio units each can provide before calculating the gas volume.

Also inspect the state symbols. Count a product in the gas volume only when it is gaseous under the stated conditions. A substance shown as $(l)$ does not contribute to the calculated gas volume, and a vapour produced while hot may condense when cooled to RTP.

For OCR Chemistry A, calculations using the $24\,\text{dm}^3$ molar gas volume are Higher Tier content.

RTP relationship

At room temperature and pressure:

$1\text{ mol of any gas}=24\,\text{dm}^3$

$V=24nn=\frac{V}{}$