Chemistry • Year 11 • Module 2 • Lesson 4
Gases & Molar Volume
Lock in the core vocabulary, Avogadro’s law, the two standard conditions, and the molar-volume formula before tackling harder problems.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: molar volume, STP, SATP, Avogadro’s law, ideal gas, V = n × Vm, n = V ÷ Vm, mole, standard conditions, empty space. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The volume occupied by one mole of any gas at a specified temperature and pressure; units L mol−1. | |
| 1.2 | The principle that equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules. | |
| 1.3 | Standard conditions of 0 °C and 100 kPa at which Vm = 22.71 L mol−1 (NESA standard). | |
| 1.4 | Standard conditions of 25 °C and 100 kPa at which Vm = 24.8 L mol−1. | |
| 1.5 | A theoretical gas with no particle volume and no intermolecular forces; real gases approximate this at low pressure and high temperature. | |
| 1.6 | The formula used to calculate the volume of a gas when the amount in moles is known. | |
| 1.7 | The rearrangement of the molar-volume formula used to find the amount of gas from its volume. | |
| 1.8 | 6.022 × 1023 particles of a substance; the SI unit of amount of substance. | |
| 1.9 | The reason all ideal gases have the same molar volume: a gas is almost entirely this between its particles. | |
| 1.10 | A defined temperature and pressure used to allow fair comparison of gas volumes between experiments. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 At the same temperature and pressure, one mole of CO2 gas occupies a larger volume than one mole of He gas because CO2 molecules are much larger. T / F
2.2 At STP (NESA standard), the molar volume of an ideal gas is 22.71 L mol−1. T / F
2.3 The formula V = n × Vm can be used to find the volume of one mole of liquid water at SATP. T / F
2.4 SATP is defined as 25 °C and 100 kPa, giving a molar volume of 24.8 L mol−1. T / F
2.5 The old value of 22.4 L mol−1 applies at 0 °C and 100 kPa, and is the value recommended by NESA for NSW HSC Chemistry. T / F
2.6 If a question states “standard laboratory conditions” without specifying temperature, the default molar volume to use in NSW HSC Chemistry is 24.8 L mol−1. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word or phrase is used once. 8 marks (1 per blank)
Word bank:
24.8 · 22.71 · Avogadro’s law · collisions · empty space · litres · molar volume · temperature
The ___________ of a gas is the volume occupied by one mole of that gas at a specified temperature and pressure. ___________ states that equal volumes of all ideal gases at the same temperature and pressure contain equal numbers of molecules. This is possible because gas volume is almost entirely ___________ between particles; what matters is the number of particle–wall ___________ per second, not the size of individual molecules. At SATP (25 °C, 100 kPa), the molar volume is ___________ L mol−1. At STP (0 °C, 100 kPa), the NESA standard molar volume is ___________ L mol−1. At lower ___________, gas particles move more slowly and press together more closely, so the same number of moles occupies fewer ___________.
4. Formula recall and unit checks
Answer each question in 1–2 sentences or a calculation. 8 marks (2 each)
4.1 Write the three forms of the molar-volume formula (find V; find n; find Vm) and state the units of each quantity.
4.2 Show the unit cancellation that confirms n = V ÷ Vm gives the correct unit for amount of substance.
4.3 A volume of 500 mL of gas is recorded in an experiment. What value (in litres) should be substituted into V = n × Vm? Explain why the conversion is necessary.
4.4 State the condition under which you would use 22.71 L mol−1 rather than 24.8 L mol−1 in a NSW HSC Chemistry calculation.
5. Build a concept map
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “is calculated using”, “applies to”, “assumes”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: molar volume · Avogadro’s law · ideal gas · STP · SATP · V = n × Vm.
6. Complete the conditions comparison table
Fill in every empty cell in the table. 6 marks (1 per cell)
| Feature | STP | SATP |
|---|---|---|
| Temperature | 25 °C (298.15 K) | |
| Pressure | 100 kPa | |
| Molar volume (L mol−1) | ||
| NESA default for NSW HSC? | No | |
| Which is colder? | — | |
| Key phrase in question | “0 °C, 100 kPa” or “STP” |
Q1 — Term–definition match
1.1 molar volume • 1.2 Avogadro’s law • 1.3 STP • 1.4 SATP • 1.5 ideal gas • 1.6 V = n × Vm • 1.7 n = V ÷ Vm • 1.8 mole • 1.9 empty space • 1.10 standard conditions.
Q2 — True / false with correction
2.1 False. At the same temperature and pressure, one mole of any ideal gas occupies the same volume (Avogadro’s law). Gas volume is dominated by empty space; molecular size is negligible.
2.2 True.
2.3 False. V = n × Vm applies only to gases. One mole of liquid water occupies approximately 18 mL (0.018 L), not 24.8 L.
2.4 True.
2.5 False. 22.4 L mol−1 applies at 0 °C and 1 atm (101.325 kPa), not 100 kPa. The NESA standard for NSW HSC is 22.71 L mol−1 at 0 °C and 100 kPa (STP).
2.6 True.
Q3 — Cloze paragraph
In order: molar volume / Avogadro’s law / empty space / collisions / 24.8 / 22.71 / temperature / litres.
Q4.1 — Three formula forms
Find V: V = n × Vm (V in L, n in mol, Vm in L mol−1). Find n: n = V ÷ Vm. Find Vm: Vm = V ÷ n.
Q4.2 — Unit cancellation
n = V ÷ Vm ⇒ L ÷ (L mol−1) = L × mol L−1 = mol ✓
Q4.3 — Unit conversion
500 mL ÷ 1000 = 0.500 L. The conversion is necessary because Vm is in L mol−1; substituting mL would make the unit L cancel incorrectly and give an answer 1000× too large.
Q4.4 — When to use 22.71 L mol−1
Use 22.71 L mol−1 when the question states the conditions are STP, i.e. 0 °C (273.15 K) and 100 kPa. Use 24.8 L mol−1 for SATP (25 °C, 100 kPa) or when no conditions are specified in a NSW HSC question.
Q5 — Sample concept map
Valid arrows include: Avogadro’s law → explains → molar volume; ideal gas → is assumed by → Avogadro’s law; STP → gives Vm = 22.71 L mol−1 for → molar volume; SATP → gives Vm = 24.8 L mol−1 for → molar volume; molar volume → is the key value in → V = n × Vm; V = n × Vm → only applies to → ideal gas. Award 1 mark per valid labelled arrow.
Q6 — Conditions comparison table
Temperature: STP = 0 °C (273.15 K); SATP = 25 °C (given). Pressure: both 100 kPa (STP given; SATP = 100 kPa). Molar volume: STP = 22.71 L mol−1; SATP = 24.8 L mol−1. NESA default: STP = No; SATP = Yes. Which is colder: STP (0 °C). Key phrase: SATP = “25 °C, 100 kPa”, “standard laboratory conditions”, or “SATP”.