Molar Gas Volume

RTP is a standard set of laboratory conditions defined as a temperature of $20^\circ\text{C}$ ($293\text{ K}$) and a pressure of $1\text{ atmosphere}$ ($101\text{ kPa}$). The specific volume of $24\text{ dm}^3/\text{mol}$ is known as the constant molar gas volume.

Calculating Gas Volumes

To calculate the volume of a gas, you multiply the number of moles by the molar gas volume. The standard formula is:

$$\text{Volume (dm}^3) = \text{Amount (mol)} \times 24$$

Gas volumes are commonly measured in two different units. A decimetre cubed (dm³) is equivalent to $1\text{ litre}$ or $1000\text{ cm}^3$. A centimetre cubed (cm³) is equivalent to $1\text{ millilitre}$, which is the unit typically read on laboratory equipment like gas syringes.

To convert between these units:

• $\text{cm}^3$ to $\text{dm}^3$: Divide by $1000$
• $\text{dm}^3$ to : Multiply by

If a question specifically asks for the answer in $\text{cm}^3$, you can use the modified formula:

$$\text{Volume (cm}^3) = \text{Amount (mol)} \times 24,000$$

Worked Example: Moles to Volume

Calculate the volume of $0.15\text{ moles}$ of carbon dioxide gas ($CO_2$) at RTP. Give your answer in $\text{cm}^3$.

Step 1: State the formula.

• $\text{Volume (cm}^3) = \text{Amount (mol)} \times 24,000$

Step 2: Substitute the known values into the equation.

• $\text{Volume (cm}^3) = 0.15\text{ mol} \times 24,000\text{ cm}^3/\text{mol}$

Step 3: Calculate the final answer with units.

• $\text{Volume} = \mathbf{3600\text{ cm}^3}$

Multi-Step Calculations: Mass to Volume

Often, you will be given the mass of a substance and asked to find the volume of gas it produces. This requires a two-step method: first converting the mass to moles, and then converting those moles to a volume.

Worked Example: Mass to Volume

Calculate the volume of gas produced at RTP when $1.4\text{ g}$ of nitrogen gas ($N_2$) is collected. ($A_r$: $$)

Step 1: Calculate the relative molecular mass ($M_r$) of the gas.

• $M_r$ of $N_2 = 14 \times 2 = 28$

Step 2: Convert mass to moles using $\text{Moles} = \text{Mass} \div M_r$.

• $\text{Moles} = 1.4\text{ g} \div 28 = 0.05\text{ mol}$

Step 3: Convert moles to volume.

• $\text{Volume (dm}^3) = 0.05\text{ mol} \times 24\text{ dm}^3/\text{mol}$

Step 4: State the final answer.

• $\text{Volume} = \mathbf{1.2\text{ dm}^3}$

Reacting Volumes and Stoichiometry

Stoichiometry refers to the ratio of reactants and products in a balanced chemical equation. If a reaction involves a solid or liquid reacting to form a gas, you must calculate the moles of the known substance first, use the molar ratio to find the moles of the gas, and finally multiply by $24$ to find the volume. Always check the state symbols — the $24\text{ dm}^3$ rule does not apply to solids or liquids; it strictly applies only to substances in the gas $(g)$ state.

For reactions involving only gases at the same temperature and pressure, there is a highly useful direct ratio shortcut based on Avogadro's Law. Because the molar ratios are directly proportional to the volume ratios, the coefficients in the balanced equation can be used as volumes directly, bypassing the need for mole calculations.

Worked Example: Reacting Masses to Gas Volume

What volume of hydrogen gas ($H_2$) is produced at RTP when $1.2\text{ g}$ of magnesium ($Mg$) reacts with excess hydrochloric acid? ($A_r$ of )

Step 1: Calculate the moles of the known solid reactant.

• $\text{Moles of } Mg = 1.2\text{ g} \div 24 = 0.05\text{ mol}$

Step 2: Use the balanced equation to find the molar ratio.

• The ratio of $Mg : H_2$ is $1:1$.
• Therefore, $0.05\text{ mol}$ of $$ produces of .

Step 3: Calculate the volume of the gas.

• $\text{Volume (dm}^3) = 0.05\text{ mol} \times 24\text{ dm}^3/\text{mol}$

Step 4: State the final answer.

• $\text{Volume} = \mathbf{1.2\text{ dm}^3}$