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Lesson pack · A-Level Physics 9702 · 15.1 · A Level (A2)

The mole

Counting by the mole. A mole is simply a count: a fixed, huge number of particles.

Student lesson Lab, full screen
At a glance

The shape of the lesson

Topic
Ideal gases
Syllabus reference
Cambridge International AS and A Level Physics 9702, 15.1
Level
A Level (A2)
Duration
45 minutes, lab-led; stretches to a 60 minute block (see the contingency notes)
Position in topic
Lesson 1 of 3 in Ideal gases; leads into The equation of state
Central model
Fig. 1: One mole is the amount containing the Avogadro number of particles; N = nNₐ
Cooperative structure
Assessment
The quick checks from the student lesson, plus the worksheet as the exit task
Learning objectives

By the end of the lesson, learners can

Section by section
  • Why chemists count in moles: Atoms and molecules come in numbers far too large to count one by one, so we group them: one mole is a fixed, enormous number of particles, the Avogadro number.
  • The Avogadro constant links the two: Amount of substance is an SI base quantity, with the mole as its unit.
Key points to leave on the board
  • Counting particles: one mole holds N₊ particles, N = nN₊, and k = R / N₊.
  • Equation of state: pV = nRT = NkT, with T in kelvin.
  • Kinetic theory: pV = ⅓Nm⟨c²⟩, and the mean translational KE per molecule is ½m⟨c²⟩ = ₃⁄₂ kT.
The core visual model

One picture every idea returns to

Amount of substance, in moles, is an SI base quantity. One mole contains the Avogadro number of particles, N₊ = 6.02 × 10²³, so the number of particles is N = nN₊.

1 mol contains Nₐ = 6.02 × 10²³ particles number of particles N = n Nₐ
Fig. 1: One mole is the amount containing the Avogadro number of particles; N = nNₐ
Lesson sequence

Forty-five minutes, phase by phase

TimePhaseWhat happens in the roomGrouping
0 to 5 minHook: predictOpen with the question in the title, Counting by the mole.. A mole is simply a count: a fixed, huge number of particles. Put the figure on the board: it is the one picture every idea in the lesson returns to.Whole class, then pairs
5 to 17 minTeach and check: Why chemists count in molesAtoms and molecules come in numbers far too large to count one by one, so we group them: one mole is a fixed, enormous number of particles, the Avogadro number. Project tab 1 of the lab (The mole) and let a student drive it while the class predicts.Whole class, then pairs
17 to 30 minTeach and check: The Avogadro constant links the twoAmount of substance is an SI base quantity, with the mole as its unit.Corners
30 to 43 minCheck and plenaryThe quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last.Individual, then whole class
43 to 45 minExit and homeworkSet the worksheet below (4 questions, 45 marks) from the topic practice set. Learners self-assess against the objectives before they leave.Individual
Timing and contingency

Protect the parts that carry the learning

The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.

Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.

Cut first if the build overruns: the last check in The Avogadro constant links the two, never the check or the plenary.

In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.

Running the cooperative task

Corners

Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.

The full structure, with its variations and the other twelve, is on the cooperative-learning page.

Examiner traps to pre-empt

What to head off, and how

No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.

Worksheet

The mole: topic worksheet

4 questions, 45 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.

NameClassDate
Q1 · analysis[12 marks]

A cylinder contains 0.40 mol of an ideal gas, as shown in Fig. 1.1, at a temperature of 27 °C and a pressure of 1.5 × 10⁵ Pa.

ideal gas, 0.40 mol piston
Fig. 1.1

(a) State what is meant by one mole of a substance. [1]

(b) Calculate the number of molecules in the cylinder. [1]

(c) Calculate the volume occupied by the gas. [3]

(d) The gas is heated at constant volume until its temperature is 177 °C. Calculate the new pressure. [3]

(e) Show that the Boltzmann constant k = R / N₊ is about 1.38 × 10⁻²³ J K⁻¹, and state the equation of state written in terms of N and k. [2]

(f) State two assumptions about the molecules of an ideal gas. [2]

Q2 · analysis[10 marks]

A container holds 64 g of oxygen gas, which has a molar mass of 32 g mol⁻¹.

(a) Define the mole. [2]

(b) State the relation between the number of molecules N, the amount of substance n and the Avogadro constant. [1]

(c) Calculate the amount of substance, in moles. [2]

(d) Calculate the number of oxygen molecules. [2]

(e) Calculate the number of oxygen atoms. [1]

(f) State the value and unit of the Avogadro constant. [2]

Q3 · analysis[11 marks]

A cylinder contains 0.16 kg of nitrogen gas (molar mass 28 g mol⁻¹) at 300 K and a pressure of 2.0 × 10⁵ Pa. Parts (a) and (b) use the mole; parts (c) and (d) use the equation of state.

(a) Calculate the amount of substance, in moles. [2]

(b) Calculate the number of nitrogen molecules. [2]

(c) Using pV = nRT, calculate the volume occupied by the gas. [3]

(d) The gas is heated at constant pressure to 450 K. Calculate the new volume. [2]

(e) State one assumption of the ideal-gas model used here. [2]

Q4 · analysis[12 marks]

A sealed flask of volume 2.0 × 10⁻³ m³ contains helium gas (molar mass 4.0 g mol⁻¹) at 1.5 × 10⁵ Pa and 300 K. This question uses all three lessons of the topic in turn.

(a) Using pV = nRT, calculate the amount of substance. [2]

(b) Calculate the number of helium atoms. [2]

(c) Calculate the mass of helium in the flask. [2]

(d) Calculate the mean translational kinetic energy of a helium atom. [2]

(e) The mass of a helium atom is the molar mass divided by N₊. Calculate the r.m.s. speed of the atoms. [3]

(f) State whether argon atoms (heavier) at the same temperature would have a greater or smaller r.m.s. speed. [1]

Answer key and marking notes
Q1 · 12 marks
  • (a)One mole is the amount of substance that contains N₊ (6.02 × 10²³) particles ✓
  • (b)N = nN₊ = 0.40 × 6.02 × 10²³ = 2.4 × 10²³ molecules ✓
  • (c)pV = nRT, T = 300 K ✓; V = nRT / p = (0.40 × 8.31 × 300) / (1.5 × 10⁵) ✓ = 6.6 × 10⁻³ m³ ✓
  • (d)At constant volume p / T is constant ✓; T₂ = 450 K, so p₂ = p₁ × T₂ / T₁ = 1.5 × 10⁵ × 450 / 300 ✓ = 2.25 × 10⁵ Pa ✓
  • (e)k = R / N₊ = 8.31 / (6.02 × 10²³) = 1.38 × 10⁻²³ J K⁻¹ ✓; the equation of state is pV = NkT ✓
  • (f)Any two: the molecules have negligible volume compared with the container ✓; there are no intermolecular forces except during collisions; collisions are perfectly elastic; the motion is random ✓
Q2 · 10 marks
  • (a)The amount of substance containing as many particles as there are atoms in 12 g of carbon-12, that is N₊ particles ✓✓
  • (b)N = nN₊ ✓
  • (c)n = 64/32 = 2.0 mol ✓✓
  • (d)N = nN₊ = 2.0 × 6.02 × 10²³ ✓ = 1.2 × 10²⁴ ✓
  • (e)Each O₂ molecule has 2 atoms, so 2.4 × 10²⁴ atoms ✓
  • (f)6.02 × 10²³ mol⁻¹ ✓✓
Q3 · 11 marks
  • (a)n = 160/28 = 5.7 mol ✓✓
  • (b)N = nN₊ = 5.71 × 6.02 × 10²³ ✓ = 3.4 × 10²⁴ ✓
  • (c)V = nRT/p = 5.71 × 8.31 × 300 / (2.0 × 10⁵) ✓✓ = 7.1 × 10⁻² m³ ✓
  • (d)At constant p, V ∝ T: V₂ = 0.0712 × 450/300 ✓ = 1.1 × 10⁻¹ m³ ✓
  • (e)Any one: molecules have negligible volume; no intermolecular forces; collisions are elastic; motion is random ✓✓
Q4 · 12 marks
  • (a)n = pV/RT = 1.5 × 10⁵ × 2.0 × 10⁻³ / (8.31 × 300) ✓ = 0.12 mol ✓
  • (b)N = nN₊ = 0.120 × 6.02 × 10²³ ✓ = 7.2 × 10²² ✓
  • (c)mass = nM = 0.120 × 4.0 = 0.48 g = 4.8 × 10⁻⁴ kg ✓✓
  • (d)₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 300 ✓ = 6.2 × 10⁻²¹ J ✓
  • (e)m = 4.0 × 10⁻³ / 6.02 × 10²³ = 6.6 × 10⁻²⁷ kg ✓; c² = 2 × 6.21 × 10⁻²¹ / 6.64 × 10⁻²⁷ = 1.87 × 10⁶ ✓; c_rms = 1.4 × 10³ m s⁻¹ ✓
  • (f)Smaller: the mean kinetic energy is the same, so a larger mass gives a smaller speed ✓
Quiz

The lesson's own checks as a quiz

The quick-check and exam-style questions from the student lesson, printable with the key.

Q11 mark

Amount of substance is an SI base quantity whose unit is the:

  • Amole
  • Bkilogram
  • Cmolecule
  • Dgram
Q21 mark

The Avogadro constant N₊ is approximately:

  • A6.02 × 10²³ mol⁻¹
  • B6.02 × 10²³ kg
  • C1.38 × 10⁻²³ mol⁻¹
  • D8.31 mol⁻¹
Q31 mark

The number of molecules in 3.0 mol of a gas is:

  • A1.8 × 10²⁴
  • B6.0 × 10²³
  • C2.0 × 10²³
  • D3.0 × 10²³
Q41 mark

A 1.0 mol sample of helium and a 1.0 mol sample of oxygen have:

  • Athe same number of molecules but different masses
  • Bthe same mass
  • Cthe same number of molecules and the same mass
  • Ddifferent numbers of molecules
Quiz key
Q1: A

Amount of substance is a base quantity measured in moles; mass (kg) is a separate base quantity.

Q2: A

N₊ = 6.02 × 10²³ per mole is the number of particles in one mole of any substance.

Q3: A

N = nN₊ = 3.0 × 6.02 × 10²³ = 1.8 × 10²⁴ molecules.

Q4: A

Equal amounts contain equal numbers of molecules (N = nN₊), but the masses differ because the molar masses differ.

Lab and demo

What to project

The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.

Slides

Slides for this lesson

A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.

Also for this topic

Everything else you can use

Original work by the TheLucidSTEM team. The worksheet items are written in the style of the papers; no past paper question is reproduced.