The shape of the lesson
By the end of the lesson, learners can
- The ideal gas model: Change the temperature, volume and amount of gas in the simulator piston and watch the pressure respond.
- Using pV = nRT: Almost every error in this section is a unit error.
- Finding M r: Inject a known mass of a volatile liquid into a gas syringe held above its boiling point, and measure the volume of vapour.
- p in Pa, V in m3, T in K.
- R = 8.31 J K−1 mol−1.
- M = mRT ÷ pV with m in g gives g mol−1.
- Ideal: low pressure, high temperature, weak forces.
One picture every idea returns to
An ideal gas obeys pV = nRT exactly. The kinetic model assumes the molecules are in rapid random motion, have negligible volume compared with the container, have no intermolecular forces, and collide elastically. Real gases behave most ideally at low pressure and high temperature, where the molecules are far apart and move fast. In pV = nRT use SI units: p in Pa, V in m3, T in K and R = 8.31 J K−1 mol−1. Rearranged with n = m ÷ M, the equation gives M = mRT ÷ pV, a way to find Mr of a gas or volatile liquid.
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 for every gas.. Hydrogen, oxygen and carbon dioxide are very different molecules, yet at ordinary pressures they all obey one equation. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The ideal gas model | Change the temperature, volume and amount of gas in the simulator piston and watch the pressure respond. Project tab 2 of the lab (States of matter lab, ideal gas) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Using pV = nRT | Almost every error in this section is a unit error. | Pairs |
| 21 to 30 min | Teach and check: Finding M r | Inject a known mass of a volatile liquid into a gas syringe held above its boiling point, and measure the volume of vapour. Project tab 3 of the lab (States of matter lab, pressure and real gases) and let a student drive it while the class predicts. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 13 marks). 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 Finding M r, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Keep M in g mol −1 by leaving the mass in grams. | And check your answer is sensible: an M r of 0.058 or 58 000 for a simple liquid means a volume or pressure was not converted. A temperature in °C instead of K is the other classic slip. |
Ideal gases and pV = nRT: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
State two assumptions of the kinetic theory that apply to an ideal gas.
Calculate the pressure, in kPa, of 0.0200 mol of gas in a 500 cm3 container at 350 K. (R = 8.31 J K−1 mol−1)
1.00 g of a gaseous element occupies 0.600 dm3 at 20 °C and 100 kPa. Calculate its Mr and suggest its identity.
A gas has a density of 1.25 g dm−3 at 273 K and 101 kPa. Calculate its Mr.
0.400 mol of gas occupies 5.00 dm3 at 200 kPa. Calculate the temperature in K.
Explain why real gases deviate from ideal behaviour at high pressure and at low temperature.
Answer key and marking notes
Any two of: the molecules have negligible volume; there are no intermolecular forces; collisions are perfectly elastic; the molecules move randomly and rapidly.
- V = 500 × 10−6 = 5.00 × 10−4 m3.
- p = nRT ÷ V = 0.0200 × 8.31 × 350 ÷ 5.00 × 10−4 = 1.16 × 105 Pa = 116 kPa.
- M = mRT ÷ pV = 1.00 × 8.31 × 293 ÷ (1.00 × 105 × 6.00 × 10−4) = 40.6.
- An element with Mr about 40 that is a gas: argon.
- Density = 1250 g m−3; M = ρRT ÷ p (from M = mRT ÷ pV with m ÷ V = ρ).
- M = 1250 × 8.31 × 273 ÷ 101 000 = 28.1 (for example N2 or CO).
- T = pV ÷ nR = 2.00 × 105 × 5.00 × 10−3 ÷ (0.400 × 8.31).
- T = 1000 ÷ 3.324 = 301 K.
- At high pressure the molecules are close together, so their own volume is not negligible compared with the container volume.
- At low temperature the molecules move slowly, so intermolecular attractions become significant.
- Both contradict assumptions of the ideal gas model, so pV = nRT no longer holds exactly.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which pair of statements are both assumptions of the kinetic theory for an ideal gas?
- Athe molecules have negligible volume, and there are no intermolecular forces between them
- Bthe molecules have negligible mass, and they all move at the same speed
- Cthe molecules attract each other weakly, and collisions between them are elastic
- Dthe molecules move in straight lines in one direction, and they never collide with each other
Under which conditions does a real gas behave most like an ideal gas?
- Ahigh temperature and low pressure
- Blow temperature and high pressure
- Clow temperature and low pressure
- Dhigh temperature and high pressure
Calculate the amount of gas, in mol, in a 250 cm3 sample at 100 kPa and 25 °C. (R = 8.31 J K−1 mol−1)
- A0.0101 mol
- B1.01 × 104 mol
- C0.120 mol
- D1.01 × 10−5 mol
What volume, in dm3, is occupied by 0.500 mol of an ideal gas at 101 kPa and 20 °C? (R = 8.31 J K−1 mol−1)
- A12.1 dm3
- B0.823 dm3
- C1.21 × 104 dm3
- D0.0121 dm3
At the same temperature and pressure, which gas shows the greatest deviation from ideal behaviour?
- Aammonia, NH3
- Bhelium, He
- Chydrogen, H2
- Dneon, Ne
0.200 g of a volatile liquid is injected into a gas syringe at 100 °C and 101 kPa. It vaporises completely and the vapour occupies 80.0 cm3. What is the relative molecular mass, Mr, of the liquid? (R = 8.31 J K−1 mol−1)
- A76.7
- B7.67 × 10−5
- C20.6
- D7.67 × 104
The density of a gas is 1.25 g dm−3 at 0 °C and 101 kPa. What is its Mr? (R = 8.31 J K−1 mol−1)
- A28.1
- B0.0281
- C2.81 × 104
- D0.0356
The graph shows pV/nRT against pressure for an ideal gas (dashed) and a real gas X at constant temperature. What is the main reason why pV/nRT becomes greater than 1 for gas X at very high pressure (region 2)?
- Athe molecules are so close together that their own volume is a significant part of the gas volume, so the gas is harder to compress than an ideal gas
- Bthe intermolecular attractions become stronger at high pressure, which increases the volume of the gas
- Cthe collisions between molecules become inelastic at high pressure, so the pressure increases
- Dthe molecules move faster at high pressure, so the temperature of the gas rises
Quiz key
The ideal gas model assumes the molecules themselves take up a negligible volume compared with the container, that there are no intermolecular forces, that collisions are perfectly elastic, and that the molecules move randomly. Molecules have mass (otherwise they would exert no pressure) and move with a range of speeds, not one speed. Weak attractions are exactly what the model leaves out; real gases have them. Motion is random in all directions, and molecules do collide with each other and with the walls; the collisions are simply elastic.
At low pressure the molecules are far apart, so their own volume is a tiny fraction of the container volume and they are rarely close enough to attract. At high temperature they have a large kinetic energy, so any attractions have little effect. Low temperature and high pressure is the worst case: the molecules are crowded together and slow, so both their volume and their attractions matter, and the gas may liquefy. Low temperature on its own lets attractions become significant. High pressure on its own makes the molecular volume significant.
p = 100 kPa = 1.00 × 105 Pa; V = 250 cm3 = 250 × 10−6 m3 = 2.50 × 10−4 m3; T = 25 + 273 = 298 K. n = pV/RT = (1.00 × 105 × 2.50 × 10−4) ÷ (8.31 × 298) = 25.0 ÷ 2476 = 0.0101 mol. 1.01 × 104 mol comes from leaving the volume in cm3 (a factor of 106 too big). 0.120 mol uses T = 25 instead of 298 K. 1.01 × 10−5 mol leaves the pressure in kPa (a factor of 1000 too small).
T = 20 + 273 = 293 K and p = 1.01 × 105 Pa. V = nRT/p = (0.500 × 8.31 × 293) ÷ 1.01 × 105 = 0.0121 m3 = 12.1 dm3. 0.823 dm3 uses T = 20 instead of 293 K. 1.21 × 104 dm3 uses p = 101 (kPa) instead of 1.01 × 105 Pa. 0.0121 is the correct volume in m3 but has not been converted to dm3.
Ammonia molecules are polar and form hydrogen bonds with each other, so the intermolecular attractions are much stronger than in the other gases, and the gas deviates most from the ideal assumption of no intermolecular forces (ammonia liquefies at −33 °C). Helium and neon are single, non-polar atoms with only very weak instantaneous dipole-induced dipole forces, so they are among the most nearly ideal gases. Hydrogen molecules are small and non-polar with very few electrons, so their attractions are also very weak.
M = mRT/(pV) with T = 100 + 273 = 373 K, p = 1.01 × 105 Pa and V = 80.0 × 10−6 m3: M = (0.200 × 8.31 × 373) ÷ (1.01 × 105 × 8.00 × 10−5) = 620 ÷ 8.08 = 76.7 g mol−1, so Mr = 76.7. 7.67 × 10−5 leaves the volume in cm3. 20.6 uses T = 100 instead of 373 K. 7.67 × 104 uses the pressure in kPa instead of Pa.
1.25 g dm−3 = 1250 g m−3 (1 m3 = 1000 dm3), T = 273 K and p = 1.01 × 105 Pa. M = ρRT/p = (1250 × 8.31 × 273) ÷ 1.01 × 105 = 28.1 g mol−1, consistent with N2 or CO. 0.0281 uses 1.25 g m−3, forgetting to convert the density. 2.81 × 104 leaves the pressure in kPa. 0.0356 inverts the expression, calculating p/(ρRT) instead of ρRT/p.
At very high pressure the molecules occupy a large fraction of the container, so the space they can actually move in is less than V and the gas cannot be compressed as much as an ideal gas; the measured volume is larger than nRT/p and pV/nRT rises above 1. Intermolecular attractions pull molecules together and reduce the pressure they exert, which lowers pV/nRT; that is the cause of the dip below 1 in region 1, not the rise in region 2. The model of elastic collisions is not what fails here, and inelastic collisions would lose energy, not raise the pressure. The graph is drawn at constant temperature, so the average speed of the molecules does not change.
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.