Imagine you have one mole of golf balls and one mole of beach balls. They both contain the same number of objects — but they take up very different amounts of space. Now imagine the opposite: one mole of hydrogen gas and one mole of oxygen gas. At the same temperature and pressure, they take up exactly the same volume. This is one of the most surprising facts in chemistry — and it's the foundation of all gas calculations.
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One mole of helium gas (mass ≈ 4 g) and one mole of carbon dioxide gas (mass ≈ 44 g) are both placed in separate containers at room temperature and pressure. Would you expect them to take up the same volume of space or different volumes? Justify your answer before you begin this lesson.
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📚 Core Content
Wrong: The mole is a measure of mass.
Right: The mole is a measure of amount of substance; one mole contains Avogadro's number of particles.
At the same temperature and pressure, one mole of any ideal gas occupies the same volume. This is Avogadro's law. It seems strange at first — surely a mole of large CO₂ molecules takes more space than a mole of tiny He atoms?
The key insight is that in a gas, the molecules are so far apart that the actual size of the molecule barely matters. The volume of a gas is almost entirely empty space between particles. What determines the volume is the number of particles (which determines how hard they collectively push on the container walls) and the temperature and pressure. Since one mole always means the same number of particles (NA), one mole of any gas at the same conditions occupies the same volume.
Molar volume only has a fixed value at a defined temperature and pressure. Two standard conditions are used in HSC Chemistry:
This formula works exactly like n = m ÷ MM from Lesson 2, but for gases. Instead of converting between mass and moles using molar mass, you convert between volume and moles using molar volume.
🧮 Worked Examples
🧪 Activities
1 Calculate the volume occupied by 2.50 mol of oxygen gas (O₂) at SATP.
2 A 11.2 L sample of hydrogen gas (H₂) is collected at STP. How many moles does it contain?
3 A balloon contains 8.40 g of methane gas (CH₄) at SATP. Calculate the volume of the balloon. (C = 12.011, H = 1.008)
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| Gas | Formula | Molar Mass (g mol⁻¹) | Amount (mol) | Measured Volume (L) |
|---|---|---|---|---|
| Helium | He | 4.003 | 1.00 | 24.8 |
| Nitrogen | N₂ | 28.014 | 1.00 | 24.8 |
| Carbon dioxide | CO₂ | 44.009 | 1.00 | 24.8 |
| Sulfur dioxide | SO₂ | 64.058 | 1.00 | 24.8 |
| Xenon | Xe | 131.29 | 1.00 | 24.8 |
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At the start of this lesson, you predicted whether one mole of helium and one mole of carbon dioxide would take up the same volume or different volumes.
The answer: they occupy exactly the same volume — 24.8 L at SATP. This is Avogadro's law in action. Gas volume depends on the number of molecules (moles), not their mass. At the same temperature and pressure, all ideal gases have the same molar volume because gas is mostly empty space — the tiny size difference between a helium atom and a CO₂ molecule is negligible compared to the distances between them.
Reflect: how did your prediction compare to what you now know?
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5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. State Avogadro's law and use it to explain why one mole of helium gas (M = 4.003 g mol⁻¹) and one mole of sulfur hexafluoride gas (SF₆, M = 146.06 g mol⁻¹) occupy the same volume at the same temperature and pressure. 3 MARKS
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7. A laboratory produces 4.40 g of carbon dioxide gas (CO₂) during a reaction at SATP. Calculate: (a) the number of moles of CO₂ produced, and (b) the volume this gas occupies at SATP. (C = 12.011, O = 15.999) 4 MARKS
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8. A student collects 3.72 L of an unknown gas at STP and determines it has a mass of 7.44 g. Calculate the molar mass of the gas and suggest what the gas might be. Show all working. 4 MARKS
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9. A chemist needs to prepare exactly 0.500 mol of nitrogen gas (N₂) in the lab at SATP. Two students propose different methods: Student A says "Measure out 14.014 g of N₂ and collect it." Student B says "Collect 12.4 L of N₂ gas at SATP." Evaluate both proposals — identify any errors and determine which, if any, is correct. (N = 14.007 g mol⁻¹) 5 MARKS
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10. A research team needs to identify an unknown gas that is produced during a chemical reaction. They have access to: a gas syringe (to collect a known volume), a balance (to measure mass), and a periodic table. Design a step-by-step procedure to identify the gas using molar volume concepts, and explain how you would use the data collected to calculate the molar mass and identify the gas. 6 MARKS
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A. Avogadro's law states that equal amounts (moles) of any ideal gas at the same temperature and pressure occupy the same volume. All five samples contain exactly NA particles, so they exert the same pressure on their container at the same temperature, and thus occupy the same volume. The mass (and thus the molar mass) of each molecule is irrelevant because gas volume is dominated by empty space between particles, not the size of the molecules themselves.
B. All gases would occupy 2 × 24.8 = 49.6 L. For example, for N₂: V = 2.00 × 24.8 = 49.6 L.
C. The student is incorrect. The molar volume of 24.8 L mol⁻¹ applies only to gases. Liquid water at 25°C has a density of 1.00 g mL⁻¹, so 1.00 mol (18.015 g) of water occupies only 18.015 mL = 0.018 L — nearly 1400 times less than 24.8 L. In liquids, molecules are densely packed with no significant empty space between them.
1. B — 24.8 L mol⁻¹ at SATP (25°C, 100 kPa). 22.71 L mol⁻¹ is for NESA STP (0°C, 100 kPa).
2. C — n = 49.6 ÷ 24.8 = 2.00 mol.
3. A — Avogadro's law: equal moles of ideal gases → equal volumes at same T and P.
4. D — MM(Cl₂) = 2 × 35.45 = 70.90 g mol⁻¹. n = 35.45 ÷ 70.90 = 0.500 mol. V = 0.500 × 24.8 = 12.4 L.
5. C — Convert: 620 mL = 0.620 L. n = 0.620 ÷ 24.8 = 0.025 mol.
6. B — The student is incorrect. Equal volumes apply to equal moles, not equal masses. MM(O₂) = 32.00 g mol⁻¹, so 32 g = 1.00 mol → 24.8 L. MM(SO₂) = 64.06 g mol⁻¹, so 32 g = 0.500 mol → 12.4 L.
7. C — n(Fe) = 2.793 ÷ 55.85 = 0.0500 mol. Mole ratio Fe:H₂ = 1:1, so n(H₂) = 0.0500 mol. V = 0.0500 × 24.8 = 1.24 L = 1240 mL.
Q6 (3 marks): Avogadro's law states that equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules [1]. Both 1 mol He and 1 mol SF₆ contain exactly 6.022 × 10²³ molecules [1]. Gas volume is determined by the number of particle–wall collisions per second (pressure) and temperature, not by the mass of individual molecules — so despite SF₆ being ~36× heavier, both samples occupy the same volume at the same T and P [1].
Q7 (4 marks):
(a) MM(CO₂) = 12.011 + 2(15.999) = 44.009 g mol⁻¹ n = m ÷ MM = 4.40 ÷ 44.009 = 0.100 mol (b) V = n × Vm = 0.100 × 24.8 = 2.48 LQ8 (4 marks):
STP → Vm = 22.71 L mol⁻¹ n = V ÷ Vm = 3.72 ÷ 22.71 = 0.164 mol MM = m ÷ n = 7.44 ÷ 0.1661 = 44.8 g mol⁻¹ ≈ 44 g mol⁻¹A molar mass of ~44 g mol⁻¹ is consistent with carbon dioxide (CO₂, MM = 44.009 g mol⁻¹) or propane (C₃H₈, MM = 44.097 g mol⁻¹). At STP, CO₂ is the more likely candidate in a typical lab context.
Q9 (5 marks):
Evaluating Student A: MM(N₂) = 2 × 14.007 = 28.014 g mol⁻¹ [1]. Mass for 0.500 mol = 0.500 × 28.014 = 14.007 g, not 14.014 g [1]. Student A's mass is incorrect — they appear to have used only one nitrogen atom's mass instead of the diatomic molecule [1].
Evaluating Student B: V = n × Vm = 0.500 × 24.8 = 12.4 L [1]. Student B is correct [1].
Q10 (6 marks):
Step 1: Weigh the gas syringe empty using the balance and record its mass (m₁) [1].
Step 2: Collect a known volume of the gas at SATP (read temperature as 25°C and pressure as 100 kPa from lab conditions). Record the volume V in litres [1].
Step 3: Seal the syringe and weigh it again with the gas inside. Record mass m₂ [1].
Step 4: Calculate the mass of gas: m = m₂ − m₁ [1].
Calculation: Calculate moles: n = V ÷ 24.8. Calculate molar mass: MM = m ÷ n [1].
Identification: Compare the calculated molar mass to known molecular masses on the periodic table. For example, if MM ≈ 32 g mol⁻¹ → likely O₂ or S (but S is solid); if MM ≈ 44 g mol⁻¹ → likely CO₂ [1].
Gases & Molar Volume
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