Molar Volume Calculations
A focused revision guide to molar volume calculations for Edexcel A Level Chemistry. This page covers Avogadro’s law, molar gas volume at room temperature and pressure, reacting gas volume calculations and the ideal gas equation pV = nRT with full worked examples for Topic 5.
Avogadro’s Law
Avogadro’s law states that equal volumes of any gases, measured under the same conditions of temperature and pressure, contain equal numbers of molecules (or atoms, if the gas is monatomic).
This means that for gases, the volume ratio in a reaction is the same as the mole ratio. If 1 mol of gas A occupies a certain volume under set conditions, then 2 mol of any other gas will occupy double that volume under the same conditions.
Avogadro’s law: equal volumes of gases at the same temperature and pressure contain equal numbers of particles.
Why it matters: Avogadro’s law lets you treat gas volumes in a balanced equation as if they were moles, provided all the gases are measured under the same conditions.
Different gases at the same temperature and pressure pack the same number of particles into the same volume, regardless of the size of the molecules.
The Molar Volume of a Gas
The molar volume is the volume occupied by one mole of a gas at a specified temperature and pressure. Because Avogadro’s law applies, every gas has the same molar volume under the same conditions.
For Edexcel A Level Chemistry, you should know the two standard values:
| Conditions | Symbol | Molar volume | Equivalent in cm3 |
|---|---|---|---|
| Room temperature (25 °C) and pressure (1 atm) | r.t.p. | 24 dm3 mol−1 | 24 000 cm3 mol−1 |
| Standard temperature (0 °C) and pressure (1 atm) | s.t.p. | 22.4 dm3 mol−1 | 22 400 cm3 mol−1 |
The two key relationships you will use repeatedly are:
Unit check: if the volume is in cm3, divide by 24 000 instead of 24. Always check whether the question gives dm3 or cm3 before substituting.
One mole of any gas occupies the same volume under fixed conditions, which is what makes converting between moles and gas volumes so straightforward.
Worked Example: 1 Mole of Oxygen
Calculate the volume occupied by 1 mole of oxygen gas at r.t.p.
Setting up the calculation
At r.t.p., the molar volume of any gas is 24 dm3 mol−1. This value applies to every gas, including oxygen, because of Avogadro’s law.
Answer: 1 mol of O2 occupies 24 dm3 at r.t.p., which equals 24 000 cm3. The identity of the gas does not matter; nitrogen, hydrogen and carbon dioxide would all give the same volume under the same conditions.
Because the molar volume value already accounts for the conditions, the calculation reduces to a single multiplication by 24.
Simple Calculations Using the Molar Volume
The molar volume can be used in two directions: to find the volume from a known mass, or to find the mass from a known volume. The strategy is always the same; convert to moles first.
Example A : Volume from mass
Calculate the volume of 0.01 g of hydrogen at r.t.p. (H = 1, so 1 mol H2 = 2 g).
Example B : Mass from volume
Calculate the mass of 100 cm3 of CO2 at r.t.p. (C = 12, O = 16, so 1 mol CO2 = 44 g).
Exam tip: convert cm3 values into dm3 by dividing by 1000, or use 24 000 cm3 mol−1 directly. Mixing units is the most common source of error in molar volume questions.
Check Your Understanding: Molar Volume Basics
Use this activity to practise converting between moles, mass and gas volumes using the molar volume at r.t.p.
Calculations From Equations Involving Gases
When a balanced equation involves gases, you can use the molar volume together with the stoichiometric ratio to convert between gas volumes and masses, or between volumes of different gases.
The general strategy has three steps:
Step 1 : Convert the given quantity into moles
Use mass → moles (divide by Mr) or volume → moles (divide by 24 dm3 mol−1 at r.t.p.).
Step 2 : Apply the mole ratio
Use the balancing numbers from the balanced equation to convert the moles of the known substance into moles of the substance you need.
Step 3 : Convert moles into the required answer
Use moles → mass (multiply by Mr) or moles → volume (multiply by 24 dm3 mol−1).
Example C : Mass of aluminium for a fixed gas volume
A student produces hydrogen by reacting aluminium with excess dilute hydrochloric acid. The hydrogen is collected in a 100 cm3 gas syringe at r.t.p. What is the maximum mass of aluminium that can be used without exceeding the syringe capacity? (Al = 27, molar volume = 24 000 cm3 at r.t.p.)
Balanced equation: 2Al + 6HCl → 2AlCl3 + 3H2
Example D : Reacting volumes of gases
500 cm3 of methane is burned at 1 atm and 300 K. Calculate the volume of oxygen needed and the volume of CO2 produced under the same conditions.
Balanced equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Because Avogadro’s law applies, the volume ratio matches the mole ratio for gases at the same conditions. CH4 : O2 : CO2 is 1 : 2 : 1.
Water is a liquid under these conditions and so does not contribute to the gas volume.
Reacting volumes shortcut: for gas-only reactions, you can skip the conversion to moles entirely and just apply the volume ratio directly, as long as all gases are at the same temperature and pressure.
The stoichiometric numbers in a balanced equation map directly onto gas volumes when every reacting species is in the gas phase.
Check Your Understanding: Reacting Gas Volumes
Practise using balanced equations with the molar volume to find masses and volumes from gas reaction data.
The Ideal Gas Equation
The molar volume value of 24 dm3 mol−1 only works at room temperature and pressure. For any other temperature or pressure, use the ideal gas equation:
Where the quantities and their SI units are:
| Symbol | Quantity | SI unit | Conversion notes |
|---|---|---|---|
| p | Pressure | Pa (pascal) | 1 kPa = 1000 Pa; 1 atm = 101 325 Pa. |
| V | Volume | m3 | 1 dm3 = 1 × 10−3 m3; 1 cm3 = 1 × 10−6 m3. |
| n | Amount of gas | mol | Use mol directly. |
| R | Gas constant | J K−1 mol−1 | R = 8.31 J K−1 mol−1. |
| T | Temperature | K (kelvin) | Add 273 to convert °C to K. |
Key idea: the ideal gas equation applies to all gases and to mixtures of gases. For a mixture, n is the total moles of all gases present.
Unit warning: almost every mark lost in ideal gas questions comes from unit conversion errors. Convert pressure to Pa, volume to m3 and temperature to K before substituting any numbers.
Memorising the equation is the easy part; converting every quantity into SI units before substituting is what separates a correct answer from a wrong one.
Ideal Gas Equation Worked Examples
Example E : Finding mass of a gas
Calculate the mass of Cl2 gas at a pressure of 100 kPa, temperature 20 °C and volume 500 cm3. (Cl = 35.5; R = 8.31)
n = pV / RT = (100 000 × 5 × 10−4) / (8.31 × 293) = 0.0205 mol
Mass = n × Mr = 0.0205 × 71 = 1.46 g
Example F : Finding the relative formula mass of a volatile liquid
0.150 g of a volatile liquid was injected into a sealed gas syringe placed in an oven at 70 °C. The pressure was 100 kPa and the measured gas volume was 80 cm3. Calculate the Mr of the liquid. (R = 8.31)
n = pV / RT = (100 000 × 8 × 10−5) / (8.31 × 343) = 0.00281 mol
Mr = mass / amount = 0.150 / 0.00281 = 53.4 g mol−1
Example G : Changing the conditions of a gas
40 cm3 of oxygen and 60 cm3 of carbon dioxide, each at 298 K and 100 kPa, were placed into an evacuated flask of volume 0.50 dm3. Calculate the pressure of the gas mixture in the flask at 298 K.
Because temperature is constant, the combined gas law simplifies to p1V1 = p2V2.
Total starting volume of gas at 100 kPa = 40 + 60 = 100 cm3; new volume = 0.50 dm3 = 500 cm3.
p2 = (p1V1) / V2 = (100 000 × 100) / 500 = 20 000 Pa
Approach for changing-conditions questions: either work out the amount in moles using pV = nRT and then put it back into the equation with the new conditions, or use the combined form p1V1/T1 = p2V2/T2. The combined form is faster when the amount of gas is unchanged.
Gas syringes are a common apparatus for collecting and measuring gas volumes at known temperature and pressure, which links practical work directly to molar volume and ideal gas calculations.
Check Your Understanding: The Ideal Gas Equation
Apply pV = nRT to find moles, masses, volumes and relative formula masses of gases under any set of conditions.
Common Exam Points
- Avogadro’s law lets you treat gas volume ratios as mole ratios, provided every gas is measured at the same temperature and pressure.
- The molar volume at r.t.p. is 24 dm3 mol−1, which equals 24 000 cm3 mol−1.
- The molar volume at s.t.p. is 22.4 dm3 mol−1. Use whichever value the question specifies.
- Always convert temperature to kelvin by adding 273 before using pV = nRT.
- Convert pressure to pascals: 1 kPa = 1000 Pa.
- Convert volume to m3: divide cm3 by 1 × 106, or dm3 by 1000.
- R = 8.31 J K−1 mol−1 in SI units. This value only works with Pa, m3 and K.
- For mixtures, n in pV = nRT is the total moles of every gas in the mixture.
- When conditions change but the moles of gas are constant, use p1V1/T1 = p2V2/T2 directly.
- Show every unit conversion clearly on your script; markers award method marks even if the final number is wrong.
Extension Questions
These extension questions go beyond the standard examples and stretch your application of molar volume and the ideal gas equation in more demanding contexts.
QuickSnap
This text summary condenses the page into the essential exam ideas.
- Avogadro’s law: equal volumes of gases at the same temperature and pressure contain equal numbers of particles.
- Molar volume at r.t.p.: 24 dm3 mol−1 = 24 000 cm3 mol−1.
- Moles of gas at r.t.p.: volume in dm3 ÷ 24, or volume in cm3 ÷ 24 000.
- Reacting volumes: for gas-only reactions, volume ratio = mole ratio from the balanced equation.
- Ideal gas equation: pV = nRT, using Pa, m3, mol, J K−1 mol−1 and K.
- R = 8.31 J K−1 mol−1.
- Temperature: always convert °C to K by adding 273.
- Changing conditions: p1V1/T1 = p2V2/T2 when the amount of gas is unchanged.
- Finding Mr of a gas: use pV = nRT to find moles, then Mr = mass / moles.
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A compound contains carbon, hydrogen and oxygen only. Its percentage composition by mass is shown.
| Element | Percentage by mass |
|---|---|
| C | 40.0% |
| H | 6.7% |
| O | 53.3% |
What is the empirical formula of the compound?
Explain how the volume of hydrogen can be used to calculate the amount of magnesium that reacted.
FAQs
These questions address the most common points of confusion students encounter when working with molar volume and the ideal gas equation.
What is the molar volume of a gas at r.t.p.?
At room temperature and pressure (25 °C and 1 atm), one mole of any gas occupies 24 dm3, which is equivalent to 24 000 cm3. This value applies to every gas because of Avogadro’s law.
Why do all gases have the same molar volume?
Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. The size of the particles makes almost no difference compared with the space between them, so all gases occupy the same volume per mole under the same conditions.
When do I use pV = nRT instead of the molar volume value?
Use the molar volume value (24 dm3 mol−1) only at r.t.p. For any other temperature or pressure, use the ideal gas equation pV = nRT with consistent SI units of Pa, m3, mol and K.
What units must I use in the ideal gas equation?
Pressure in pascals (Pa), volume in cubic metres (m3), amount in moles (mol), R in J K−1 mol−1 (8.31), and temperature in kelvin (K). Convert °C to K by adding 273, kPa to Pa by multiplying by 1000, and cm3 to m3 by multiplying by 1 × 10−6.
How do I handle a mixture of gases in pV = nRT?
In a mixture, n is the total number of moles of all gases combined. The equation applies to the mixture as a whole, provided the gases do not react with each other.
How do I find the relative formula mass of a volatile liquid?
Vaporise a known mass of the liquid into a sealed gas syringe at a known temperature and pressure. Use pV = nRT to find the moles of vapour produced, then calculate Mr using Mr = mass / moles.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.