The shape of the lesson
By the end of the lesson, learners can
- Pressure, volume and temperature, together: For a fixed amount of ideal gas the three quantities are tied by one relation.
- Two forms, one constant: The equation of state can be written per mole as pV = nRT, with the molar gas constant R, or per molecule as pV = NkT, with the Boltzmann constant k.
- 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.
One picture every idea returns to
An ideal gas obeys pV = nRT = NkT, with T in kelvin. The Boltzmann constant links the two forms: k = R / N₊.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, One equation of state.. A single relation ties pressure, volume and temperature for a fixed amount of ideal gas. 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 min | Teach and check: Pressure, volume and temperature, together | For a fixed amount of ideal gas the three quantities are tied by one relation. Project tab 2 of the lab (Equation of state) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Two forms, one constant | The equation of state can be written per mole as pV = nRT, with the molar gas constant R, or per molecule as pV = NkT, with the Boltzmann constant k. | Frayer Model |
| 30 to 43 min | Check and plenary | The 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 min | Exit and homework | Set the worksheet below (7 questions, 79 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
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 Two forms, one constant, 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.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
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.
The equation of state: topic worksheet
7 questions, 79 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.
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.
(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]
A small bubble of ideal gas rises from the bottom of a lake to the surface, as shown in Fig. 3.1. At the bottom its volume is 1.2 cm³, the pressure is 2.8 × 10⁵ Pa and the temperature is 8 °C. At the surface the pressure is 1.0 × 10⁵ Pa and the temperature is 18 °C.
(a) Write down, in terms of p, V and T, the relationship that holds for a fixed mass of ideal gas between two states. [2]
(b) Calculate the volume of the bubble at the surface. [3]
(c) Explain, in molecular terms, why the pressure of the gas in the bubble is lower at the surface. [3]
(d) Calculate the amount, in mol, of gas in the bubble. [3]
(e) Suggest one reason the real bubble would not behave exactly as predicted. [2]
A fixed mass of gas at a pressure of 1.0 × 10⁵ Pa occupies 250 cm³. It is compressed isothermally, as shown in Fig. 5.1.
(a) State Boyle’s law. [2]
(b) Calculate the pressure when the volume is reduced to 100 cm³. [2]
(c) Explain why the temperature must be constant for Boyle’s law to hold. [1]
(d) Describe the shape of the p-V graph for this change. [2]
(e) Calculate the volume at which the pressure would be 4.0 × 10⁵ Pa, at the same temperature. [2]
(f) State one condition on the gas for Boyle’s law to apply. [1]
A fixed mass of gas is kept at constant volume. At 27 °C its pressure is 1.2 × 10⁵ Pa.
(a) Convert 27 °C to kelvin. [1]
(b) State how the pressure depends on the thermodynamic temperature at constant volume. [2]
(c) Calculate the pressure at 127 °C. [2]
(d) Calculate the temperature, in °C, at which the pressure would be 2.0 × 10⁵ Pa. [3]
(e) Explain why the temperature must be in kelvin in these gas laws. [2]
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]
This question links the equation of state with the kinetic theory. Part (b) connects pV = NkT with pV = ⅓Nm⟨c²⟩.
(a) Write the expression for the mean translational kinetic energy of a molecule in terms of temperature. [1]
(b) Show that this follows from pV = NkT and pV = ⅓Nm⟨c²⟩. [3]
(c) Calculate the mean translational kinetic energy of a molecule at 300 K. [2]
(d) An oxygen molecule has mass 5.3 × 10⁻²⁶ kg. Calculate its r.m.s. speed at 300 K. [3]
(e) State how the mean translational kinetic energy at a given temperature depends on the type of gas. [2]
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
- (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 ✓
- (a)For a fixed mass of ideal gas, pV / T is constant ✓, so p₁V₁ / T₁ = p₂V₂ / T₂ (T in kelvin) ✓
- (b)T₁ = 281 K, T₂ = 291 K ✓; V₂ = p₁V₁T₂ / (T₁p₂) = (2.8 × 10⁵ × 1.2 × 291) / (281 × 1.0 × 10⁵) ✓ = 3.5 cm³ ✓
- (c)As the bubble rises the surrounding water pressure falls, so the bubble expands ✓; with a larger volume the molecules are more spread out and strike unit area of the surface less often ✓, so the pressure they exert is lower ✓
- (d)Using surface values, n = pV / RT = (1.0 × 10⁵ × 3.5 × 10⁻⁶) / (8.31 × 291) ✓ = 0.35 / 2418 ✓ = 1.4 × 10⁻⁴ mol ✓
- (e)Any one: water vapour evaporates into the bubble, adding gas ✓; the temperature may not be uniform, or the gas is not perfectly ideal ✓
- (a)For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume (pV = constant) ✓✓
- (b)p₂ = p₁V₁/V₂ = 1.0 × 10⁵ × 250/100 ✓ = 2.5 × 10⁵ Pa ✓
- (c)From pV = nRT, pV is constant only if T (and n) are constant ✓
- (d)A smooth curve (a rectangular hyperbola): p falls as V rises ✓✓
- (e)V = p₁V₁/p = 1.0 × 10⁵ × 250 / 4.0 × 10⁵ ✓ = 62.5 cm³ ✓
- (f)A fixed mass of ideal gas at constant temperature ✓
- (a)300 K ✓
- (b)At constant volume the pressure is proportional to the thermodynamic temperature: p/T = constant ✓✓
- (c)p₂ = p₁ × T₂/T₁ = 1.2 × 10⁵ × 400/300 ✓ = 1.6 × 10⁵ Pa ✓
- (d)T = T₁ × p₂/p₁ = 300 × (2.0/1.2) = 500 K ✓✓ = 227 °C ✓
- (e)The proportionality is measured from absolute zero; the Celsius zero is arbitrary, so p ∝ T fails in °C ✓✓
- (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 ✓✓
- (a)½m⟨c²⟩ = ₃⁄₂ kT ✓
- (b)Equating: ⅓Nm⟨c²⟩ = NkT ✓; so m⟨c²⟩ = 3kT ✓; hence ½m⟨c²⟩ = ₃⁄₂ kT ✓
- (c)₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 300 ✓ = 6.2 × 10⁻²¹ J ✓
- (d)½m c² = 6.21 × 10⁻²¹, so c² = 2 × 6.21 × 10⁻²¹ / 5.3 × 10⁻²⁶ ✓✓ = 2.34 × 10⁵; c_rms = 4.8 × 10² m s⁻¹ ✓
- (e)It does not depend on the type of gas; it depends only on the temperature ✓✓
- (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 ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
For a fixed mass of ideal gas at constant volume, the pressure is proportional to:
- Athe thermodynamic temperature in kelvin
- Bthe Celsius temperature
- Cthe square of the temperature
- Dthe volume
The equation pV = NkT is written in terms of:
- Athe number of molecules N and the Boltzmann constant k
- Bthe number of moles n and R
- Cthe mass and the molar mass
- Dthe r.m.s. speed
The Boltzmann constant k is related to the molar gas constant R and the Avogadro constant N₊ by:
- Ak = R / N₊
- Bk = R N₊
- Ck = N₊ / R
- Dk = R + N₊
0.50 mol of an ideal gas at 300 K occupies 0.020 m³. Its pressure pV = nRT is about:
- A6.2 × 10⁴ Pa
- B1.2 × 10⁵ Pa
- C3.1 × 10⁴ Pa
- D2.5 × 10³ Pa
Quiz key
From pV = nRT at fixed V and n, p ∝ T, with T in kelvin.
pV = NkT uses the molecule count N and k; the per-mole form pV = nRT uses n and R.
k = R / N₊ = 8.31 / 6.02 × 10²³ = 1.38 × 10⁻²³ J K⁻¹.
p = nRT / V = 0.50 × 8.31 × 300 / 0.020 = 6.2 × 10⁴ Pa.
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 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.