C13: Gas Volumes
Higher tier only — using molar gas volume at room temperature and pressure, and the ideal gas equation to calculate volumes, moles and masses of gases.
Molar Gas Volume at RTP
At room temperature and pressure (RTP), defined as 20 °C and 1 atmosphere, one mole of any gas occupies 24 dm³. This is called the molar gas volume.
This means 1 mol of oxygen (O2), 1 mol of carbon dioxide (CO2) and 1 mol of hydrogen (H2) all occupy the same volume at RTP, even though they have different masses.
Volume of gas (dm³) = moles × 24
Moles of gas = volume (dm³) / 24
At RTP, equal volumes of all gases contain the same number of molecules. This follows from Avogadro's law and is why the molar gas volume is the same for all gases.
If you need the volume in cm³, use 24 000 cm³ instead of 24 dm³:
Volume (cm³) = moles × 24 000
Worked Example: Volume from Moles
What volume does 0.5 mol of carbon dioxide occupy at RTP?
Volume = 0.5 × 24 = 12 dm³
Worked Example: Moles from Volume
A gas occupies 6 dm³ at RTP. How many moles of gas are present?
moles = volume / 24 = 6 / 24 = 0.25 mol
Worked Example: Mass from Volume
What mass of oxygen gas (O2) occupies 7.2 dm³ at RTP? (Mr of O2 = 32)
moles = 7.2 / 24 = 0.3 mol
mass = 0.3 × 32 = 9.6 g
Calculating Volumes from Equations
The mole ratio from a balanced equation can be used with molar gas volume to calculate the volume of a gaseous reactant or product. The three-step method is the same as for reacting masses.
Worked Example: Volume of Gas Product
What volume of carbon dioxide is produced when 10 g of calcium carbonate decomposes at RTP?
CaCO3 → CaO + CO2
Mr of CaCO3 = 100
moles of CaCO3 = 10 / 100 = 0.1 mol
Mole ratio CaCO3:CO2 = 1:1, so moles of CO2 = 0.1 mol
Volume of CO2 = 0.1 × 24 = 2.4 dm³
Worked Example: Volume of Gas Reactant
What volume of oxygen is needed for the complete combustion of 4.8 g of magnesium at RTP?
2Mg + O2 → 2MgO
moles of Mg = 4.8 / 24 = 0.2 mol
Mole ratio Mg:O2 = 2:1, so moles of O2 = 0.2 / 2 = 0.1 mol
Volume of O2 = 0.1 × 24 = 2.4 dm³
Only use the molar gas volume of 24 dm³ when the question states "at RTP" or "at room temperature and pressure". At other conditions, use the ideal gas equation.
The Ideal Gas Equation
The ideal gas equation allows you to calculate the volume, pressure or temperature of a gas when conditions are not at RTP. It relates all four gas variables in one equation.
pV = nRT
p = pressure (Pa)
V = volume (m³)
n = number of moles
R = gas constant = 8.314 J/(mol·K)
T = temperature (K)
Units are critical in the ideal gas equation. Pressure must be in pascals (Pa), volume in cubic metres (m³), and temperature in kelvin (K). Getting the units wrong is the most common error.
To convert between units:
- kPa to Pa: multiply by 1000
- dm³ to m³: divide by 1000
- cm³ to m³: divide by 1 000 000
- °C to K: add 273
- atm to Pa: multiply by 101 325
| Quantity | Symbol | Required Unit | Common Conversions |
|---|---|---|---|
| Pressure | p | Pa | 1 kPa = 1000 Pa; 1 atm = 101 325 Pa |
| Volume | V | m³ | 1 dm³ = 0.001 m³; 1 cm³ = 0.000 001 m³ |
| Temperature | T | K | K = °C + 273 |
| Gas constant | R | 8.314 J/(mol·K) | Given in the exam |
Worked Example: Calculating Volume Using pV = nRT
0.2 mol of gas is at 300 K and 100 kPa. Calculate its volume.
p = 100 000 Pa, n = 0.2, R = 8.314, T = 300 K
V = nRT / p = (0.2 × 8.314 × 300) / 100 000
V = 498.84 / 100 000 = 0.004 988 m³
V = 0.004 988 × 1000 = 4.99 dm³
Worked Example: Calculating Moles Using pV = nRT
A 500 cm³ container holds a gas at 200 kPa and 25 °C. How many moles of gas are present?
p = 200 000 Pa, V = 500 / 1 000 000 = 0.0005 m³, T = 25 + 273 = 298 K
n = pV / RT = (200 000 × 0.0005) / (8.314 × 298)
n = 100 / 2477.6 = 0.0404 mol
Worked Example: Calculating Mass from pV = nRT
A gas cylinder contains oxygen at 150 kPa and 20 °C. The volume is 2 dm³. Calculate the mass of oxygen. (Mr of O2 = 32)
p = 150 000 Pa, V = 0.002 m³, T = 293 K
n = pV / RT = (150 000 × 0.002) / (8.314 × 293)
n = 300 / 2436 = 0.1232 mol
mass = 0.1232 × 32 = 3.94 g
Write down the values of p, V, n, R and T with their units BEFORE substituting into pV = nRT. This helps you spot unit conversion errors and shows the examiner your working clearly.
Comparing RTP Method and Ideal Gas Equation
| Feature | Molar Gas Volume (RTP) | Ideal Gas Equation (pV = nRT) |
|---|---|---|
| When to use | At RTP only (20 °C, 1 atm) | At any temperature and pressure |
| Key value | 24 dm³/mol | R = 8.314 J/(mol·K) |
| Units needed | Volume in dm³ | Volume in m³, pressure in Pa, temperature in K |
| Accuracy | Approximate (assumes RTP) | More accurate for non-standard conditions |
| Ease of use | Simpler — direct multiplication or division | More complex — requires unit conversions |
Practice Questions
What volume does 3 mol of nitrogen gas occupy at RTP?
Volume = 3 × 24 = 72 dm³
8.1 g of hydrogen gas (H2) is collected at RTP. What volume does it occupy? (H = 1)
Mr of H2 = 2
moles = 8.1 / 2 = 4.05 mol
Volume = 4.05 × 24 = 97.2 dm³
A 10 dm³ sample of gas at RTP contains 0.4 mol. Calculate the Mr of the gas if its mass is 11.2 g.
Mr = mass / moles = 11.2 / 0.4 = 28
(This could be N2 or CO.)
Using pV = nRT, calculate the volume of 0.5 mol of gas at 250 kPa and 50 °C.
p = 250 000 Pa, T = 50 + 273 = 323 K, n = 0.5, R = 8.314
V = nRT / p = (0.5 × 8.314 × 323) / 250 000
V = 1342.7 / 250 000 = 0.005 371 m³
V = 5.37 dm³
0.15 mol of chlorine gas is in a container at 100 kPa and a temperature of 27 °C. Use the ideal gas equation to find the volume of the container in dm³.
p = 100 000 Pa, T = 27 + 273 = 300 K, n = 0.15
V = nRT / p = (0.15 × 8.314 × 300) / 100 000
V = 374.13 / 100 000 = 0.003 741 m³
V = 3.74 dm³
Maths Skills
Molar Gas Volume at RTP Higher
At RTP (20 °C, 1 atm), one mole of any gas occupies 24 dm³. Use Volume = moles × 24 to find the volume, or moles = Volume / 24 to find the number of moles. If the volume is given in cm³, use 24 000 instead: Volume (cm³) = moles × 24 000.
When using the three-step method with gas volumes, Steps 1 and 2 are the same as for reacting masses. In Step 3, instead of converting moles to mass, convert moles of gas to volume using Volume = moles × 24. For reactions where all reactants and products are gases, the mole ratio directly gives the volume ratio (Gay-Lussac's law of combining volumes).
The Ideal Gas Equation: pV = nRT Higher
When conditions are not at RTP, use pV = nRT. The key challenge is unit conversions: pressure must be in Pa (multiply kPa by 1000, multiply atm by 101 325), volume must be in m³ (divide dm³ by 1000, divide cm³ by 1 000 000), and temperature must be in K (add 273 to °C).
To rearrange pV = nRT: to find V, use V = nRT/p. To find n, use n = pV/RT. To find p, use p = nRT/V. To find T, use T = pV/nR. Always write down the values with units before substituting, as this helps catch conversion errors.
Unit Conversions Higher
Common conversion errors: forgetting to convert kPa to Pa (multiply by 1000), forgetting to convert dm³ to m³ (divide by 1000), and forgetting to convert °C to K (add 273). These are the most frequent sources of error in ideal gas calculations. One useful check: at RTP, 1 mol of gas should occupy roughly 0.024 m³ — use this to verify your answer is in the right order of magnitude.
Common Misconceptions
Gas Volume and Gas Type Higher
Wrong: Gas volume depends on the type of gas at RTP Correct: At RTP, one mole of any gas occupies the same volume (24 dm³), regardless of the gas type. This is because gas volume depends on the spacing between molecules, not the size of the molecules themselves, and at the same temperature and pressure the spacing is the same for all gases (Avogadro's law)
Pressure and Volume Higher
Wrong: Increasing pressure always reduces gas volume linearly Correct: Boyle's law states that pressure × volume = constant, but only at a constant temperature and for ideal gases. The relationship is inversely proportional, not linear: doubling the pressure halves the volume. Real gases deviate from ideal behaviour at high pressures and low temperatures
Wrong: The molar gas volume of 24 dm³ applies at all temperatures and pressures Correct: 24 dm³ only applies at RTP (20 °C, 1 atm). At different conditions, you must use the ideal gas equation pV = nRT. The molar gas volume changes with temperature and pressure
6-Mark Extended Question
Calculating Gas Volume from a Reaction Higher
Explain how to calculate the volume of a gas produced in a reaction using the molar gas volume. Use an example in your answer.
First, write the balanced equation for the reaction and identify the mole ratio. For example, when calcium carbonate decomposes: CaCO3 → CaO + CO2. The mole ratio of CaCO3:CO2 is 1:1. [2 marks]
Next, calculate the moles of the reactant using moles = mass / Mr. For 10 g of CaCO3 (Mr = 100): moles = 10 / 100 = 0.1 mol. From the 1:1 mole ratio, moles of CO2 = 0.1 mol. [2 marks]
Finally, convert moles of gas to volume using the molar gas volume at RTP: Volume = moles × 24 = 0.1 × 24 = 2.4 dm³. This method only works at RTP. At other conditions, use pV = nRT instead, making sure to convert all units correctly (pressure to Pa, volume to m³, temperature to K). [2 marks]
AO3: Analyse and Evaluate
Using the Ideal Gas Equation with Experimental Data Higher
A student collects a gas produced in a reaction. The gas occupies 300 cm³ at a pressure of 120 kPa and a temperature of 50 °C. (a) Calculate the number of moles of gas collected. (b) The gas is carbon dioxide from the reaction CaCO3 → CaO + CO2. Calculate the mass of calcium carbonate that must have reacted. (Mr: CaCO3 = 100, R = 8.314 J/(mol·K))
(a) p = 120 000 Pa, V = 300 / 1 000 000 = 0.0003 m³, T = 50 + 273 = 323 K. n = pV/RT = (120 000 × 0.0003) / (8.314 × 323) = 36 / 2685.4 = 0.0134 mol.
(b) From the 1:1 mole ratio, moles of CaCO3 = 0.0134 mol. Mass = 0.0134 × 100 = 1.34 g. Note that if the student had used the RTP method instead (assuming 24 dm³/mol), they would have got 0.3 / 24 = 0.0125 mol, which is a different answer because the conditions are not at RTP. This shows the importance of using the ideal gas equation when conditions differ from RTP.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Chemistry exams?
Get the best revision books and guides to boost your grades.