Three teaching hours at both levels, outcomes 1.5.1 to 1.5.4. There is no additional higher level content in Structure 1.5.
Guiding question: How does the model of ideal gas behaviour help us to predict the behaviour of real gases?
Structure 1.5 · Ideal gases
1The ideal gas model 1.5.1 SL + HL Repeated
A tyre pumped up on a cold morning reads a higher pressure by the afternoon; a sealed bag of crisps swells at altitude; a syringe of air pushes back harder the further it is compressed. All gases respond to changes of temperature, pressure and volume in almost exactly the same way, whatever they are made of. That uniformity is what makes a single model — the ideal gas — so useful, and understanding where it fails tells us about the forces between real molecules.
An ideal gas consists of moving particles with negligible volume and no intermolecular forces. All collisions between particles are considered elastic — no kinetic energy is lost in them.
The model is the kinetic molecular theory of Structure 1.1 applied to gases. The particles are in constant, random, straight-line motion, and their average kinetic energy is proportional to the absolute temperature. Pressure is the result of particles colliding with the walls of the container: each collision exerts a tiny force, and the pressure is the total force per unit area. Anything that makes collisions with the walls more frequent or more forceful increases the pressure — more particles in the same space, a smaller space for the same particles, or faster particles at a higher temperature.
Two of the assumptions are what make every gas behave alike. If the particles have no volume of their own, the volume of a gas depends only on how many particles there are, not on their size. If there are no forces between them, their behaviour does not depend on their chemistry. They are also the two assumptions that a real gas breaks.
2Real gases and the limits of the model 1.5.2 SL + HL Strong
Real gases deviate from the ideal gas model, particularly at low temperature and high pressure. No mathematical treatment is required: every mark here is for explanation.
At high pressure the particles are forced close together. The volume of the particles themselves is no longer negligible compared with the volume of the container, so the gas occupies more space than the ideal model predicts: part of the measured volume is molecules, not empty space. At low temperature the particles move slowly enough for intermolecular attractions to pull them together during collisions. The gas then occupies less space, and exerts less pressure, than predicted. As the temperature falls further, these attractions eventually condense the gas to a liquid — a change the ideal model cannot describe at all.
A convenient measure of the deviation is pV/nRT, which equals exactly 1 for an ideal gas. Values below 1 show the effect of attractions; values above 1 show the effect of particle volume. At everyday conditions — around 100 kPa and room temperature — most gases have pV/nRT within a few tenths of a per cent of 1, which is why the ideal gas equation works so well in the laboratory.
Different gases deviate by different amounts under the same conditions. The stronger the intermolecular forces, the larger the deviation: polar molecules and large molecules with many electrons (and so strong London forces) deviate more than small non-polar ones. Hydrogen and helium are the most nearly ideal gases; ammonia, with hydrogen bonding, and carbon dioxide deviate much more. The reasoning is the chain of Structure 2.2: structure → polarity → type of intermolecular force → strength → extent of deviation.
How far does a real gas depart from ideal?
Pick a gas, a temperature and a pressure. The curve shows pV/nRT, which is exactly 1 for an ideal gas. Below 1, attractions between molecules are winning; above 1, the volume of the molecules themselves is. The other gases are drawn faintly for comparison.
Computed with the van der Waals model and published constants for each gas — enrichment: the syllabus asks only for the explanation, not the equation.
Exam focus · what the published papers show
A published question asks: Outline why the volume occupied by propane (g) at very high pressure is higher than the value calculated using PV = nRT. The scheme wants ideal gas molecules have no volume or volume of «propane» molecules is not negligible.
Intermolecular forces score nothing, and a report says why: candidates talked about the presence of intermolecular forces which would not cause the volume to be higher than that predicted by the ideal gas equation. The mean mark on that question was 0.78 out of 2. If a question states the direction of the deviation, it has already told you which assumption has failed.
On a one-mark version — state one reason why gases such as carbon dioxide and ethenone become less ideal at higher pressures — a report records that only about 20 % of candidates could explain it and 14 % did not attempt it; another report calls the reasons for deviation not a strong area of understanding. Here the scheme accepts either assumption — intermolecular forces no longer negligible or volume occupied by molecules no longer negligible — because no direction is specified.
Question. In which set of conditions does a gas behave most like an ideal gas? The options pair H2 or NH3 with 100 or 50 kPa and 273 or 473 K.
Two tests. Closest to ideal means the lowest pressure (particles far apart, their volume negligible) and the highest temperature (fast enough to ignore attractions): 50 kPa and 473 K. Then the gas: H2 is small and non-polar, with only weak London forces; NH3 forms hydrogen bonds. H2 at 50 kPa and 473 K. A parallel item pairs phosphine and ammonia; PH3 deviates less because it has no hydrogen bonding, although it is the heavier molecule — molar mass is not the criterion.
Asked why NO deviates more than N2, the scheme has two halves: M1 NO polar AND N2 non-polar OR NO has dipole-dipole «and LDF» AND N2 has London/dispersion «forces between molecules»; M2 stronger intermolecular forces in NO.
A report records that the majority of students struggled with this part … Most of the answers were not based on the type of intermolecular forces and their strength. Name the forces in both gases, then compare their strengths.
3Molar volume, and the relationships between p, V and T 1.5.3 SL + HL Strong
The molar volume of an ideal gas is a constant at a specific temperature and pressure. At STP (273 K and 100 kPa) it is 22.7 dm3 mol−1 (data booklet). It is the same for every ideal gas — Avogadro’s law restated.
At other conditions the molar volume changes: at 298 K and 100 kPa it is 24.8 dm3 mol−1, and at 373 K and 100 kPa it is 31.0 dm3 mol−1. So 22.7 dm3 mol−1 may only be used when the conditions are STP.
For a fixed amount of an ideal gas, three relationships follow from the particle model and can be investigated experimentally. The names of the individual gas laws are not assessed, and every one can be read from the ideal gas equation in section 4.
- Pressure and volume, at constant temperature: p ∝ 1/V, so pV is constant. Halving the volume doubles the frequency with which particles hit each unit area of the wall, so the pressure doubles.
- Volume and temperature, at constant pressure: V ∝ T (in kelvin). Faster particles hit the walls harder and more often; for the pressure to stay the same, the gas must expand.
- Pressure and temperature, at constant volume: p ∝ T (in kelvin). In a rigid container the faster particles simply exert a greater pressure.
A straight line through the origin is the signature of direct proportionality, so plotting p against 1/V, or V against T in kelvin, turns the data into a straight line whose gradient can be measured. On a Celsius axis the V–T line is still straight but does not pass through the origin: extrapolated back, it meets zero volume at −273 °C, which is how absolute zero was first estimated. Graphs may be sketched, showing only the shape, or plotted accurately from data; a sketch shows the relationship, a plot allows the constant to be determined and the fit to be judged.
The gas relationships, read from pV = nRT
Choose a relationship, fix the amount of gas and the variable held constant, then move along the curve. Every value is computed from pV = nRT, so the last row stays at R = 8.31 J K⁻¹ mol⁻¹ whatever you change.
Calculated for an ideal gas in kPa, dm³ and K (kPa × dm³ = J). On the θ/°C plot the dashed extension shows where the line would meet zero volume.
Exam focus · what the published papers show
A report records candidates who used the molar volume at STP of 22.7 which meant M2 was lost. If the question says “at STP”, use Vm; if it gives a temperature and pressure, use pV = nRT.
Question. The volume of a fixed mass of an ideal gas was measured at constant temperature at different pressures. Which graph shows the relationship between pV and p?
pV = nRT, and n, R and T are all constant, so pV does not depend on p: a horizontal line. The examiners, answering a teacher objection, wrote that this was considered a fair question as the students can use the ideal gas equation to deduce that pV must be constant — the intended route is the equation, not a memorized shape.
Question. Which graph shows the relationship between the volume and temperature of an ideal gas at constant pressure? Two options plot V against T in kelvin and two against °C.
V = (nR/p)T: a straight line through the origin, but only when T is in kelvin. Against °C the line is shifted and crosses the temperature axis at −273 °C. A report says the item required awareness of how the graph would be when different units of temperature are used. The candidates got confused with the axes. Read the axis label before the line.
4The ideal gas equation and the combined gas law 1.5.4 SL + HL Strong
pV = nRT
p pressure in Pa · V volume in m3 · n amount in mol · R = 8.31 J K−1 mol−1 · T temperature in K
p1V1 ÷ T1 = p2V2 ÷ T2 fixed amount of gas; no R needed
Both equations and the value of R are in the data booklet. The syllabus requires SI units: pressure in pascals and volume in cubic metres. Because 1 kPa × 1 dm3 = 1 Pa × 1 m3 × (103 × 10−3) = 1 J, kilopascals with cubic decimetres give the same answer, and published schemes accept either consistent set.
The ideal gas equation combines the three proportionalities of section 3 with Avogadro’s law (V ∝ n). It applies to any ideal gas, or mixture of gases, and connects the measurable conditions to the amount in moles — and so, through n = m/M, to mass. Use it when a question involves an amount, a mass or a molar mass. Use the combined gas law when a fixed sample moves from one set of conditions to another; only the temperature needs converting, because the units of p and V cancel provided each pair matches.
| Quantity | SI unit | Conversions |
|---|---|---|
| Pressure | Pa | 1 kPa = 103 Pa; 1 atm = 101 325 Pa |
| Volume | m3 | 1 dm3 = 10−3 m3; 1 cm3 = 10−6 m3 |
| Temperature | K | T/K = θ/°C + 273 |
Molar mass of a gas or volatile liquid
Combining pV = nRT with n = m/M gives M = mRT ÷ pV. A known mass of a volatile liquid is injected into a heated gas syringe, where it evaporates completely; the volume of vapour, its temperature and the pressure are measured, and the molar mass follows. The method assumes the vapour behaves ideally, so it gives better results well above the boiling point and at low pressure.
Question. 0.363 g of organic liquid Y was vaporized completely at 95.0 °C and 100.0 kPa. The gas volume was 81.0 cm3. Determine the molar mass of Y. [3]
| Convert first | T = 368 K; V = 0.0810 dm3 (8.10 × 10−5 m3); p = 100.0 kPa (1.000 × 105 Pa) |
| Amount | n = pV/RT = (100.0 × 0.0810) ÷ (8.31 × 368) = 0.00265 mol |
| Molar mass | M = m ÷ n = 0.363 ÷ 0.00265 = 137 g mol−1 |
Check. In pure SI, (1.000 × 105 × 8.10 × 10−5) ÷ (8.31 × 368) = 0.00265 mol — the same. Mixed units would disagree by exactly a factor of 1000. A report notes candidates struggled with rearranging the ideal gas and density equations, often giving up without obtaining a value.
Question. Calculate the volume, in dm3, occupied by 6.45 g of propane at 100 kPa and 15 °C. [2]
n = 6.45 ÷ 44.11 = 0.146 mol; V = nRT/p = 0.146 × 8.31 × 288 ÷ 100 = 3.49 dm3
The scheme accepts answers in the range 3.49 – 3.59 dm3, the spread produced by rounding at different points. Leaving the temperature as 15 would give 0.182 dm3 — a fifth of a litre for six grams of gas, which should look wrong before you check it.
Exam focus · what the published papers show
Published schemes award a mark for the converted values alone, before any arithmetic. One gives it for either consistent set: T = 368 K AND P = 100.0 kPa AND V = 0.0810 dm3 OR T = 368 K AND P = 100 000 Pa AND V = 8.1 × 10−5 m3. Another opens with T = 200 + 273 K as its first marking point. Write that line out: it is worth a mark and it is where the errors happen.
A multiple-choice item: a fixed volume of gas at 30 °C and pressure P is heated to 60 °C; what is the new pressure? In kelvin the ratio is 333/303 = 1.10 — the pressure rises by about a tenth, not doubles. One report calls it one of the worst answered questions, with less than 25 % of students remembering to convert to Kelvin; the report on the other level’s paper in the same session: The majority did not convert temperature to Kelvin.
A companion item states that a gas volume is increased by 100 % at constant temperature. That is a doubling, so the pressure halves — a 50 % decrease, not a “100 % decrease”, which would leave no pressure at all.
One report states that candidates were forgetting that the value of R is tied to its units; another that more than half of the candidates were confused about the appropriate unit of volume to be used in the ideal gas equation to calculate pressure in Pascals, and selected dm³. R is a constant: one item offers “the gas constant, R” among factors that depend on temperature, and it does not. A report also lists calculation using the ideal gas equation among areas of good preparation while listing rearranging the ideal gas equation to calculate molar mass as difficult — practise solving for each of p, V, n and T in turn.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.
Why do gases deviate from the ideal gas law at high pressures?
Which factors affect the molar volume of an ideal gas?
I. Pressure II. Temperature III. Empirical formula
The volume of a sample of gas measured at 27 °C is 10.0 dm3. What is the temperature when the volume is reduced to 9.0 dm3 at the same pressure?
What is the volume of gas when the pressure on 100 cm3 of gas is changed from 400 kPa to 200 kPa at constant temperature?
Two containers are connected by a valve. Container 1: volume 4.0 dm3, pressure 2.0 × 105 Pa. Container 2: volume 2.0 dm3, pressure 3.0 × 105 Pa. What is the total pressure after the valve is opened and the two gas samples are allowed to mix at constant temperature?
A 0.150 g sample of menthol, when vaporized, had a volume of 0.0337 dm3 at 150 °C and 100.2 kPa. Calculate its molar mass showing your working. [2]
2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g). A 0.200 g piece of lithium was placed in 500.0 cm3 of water.
(ii) Calculate the volume of hydrogen gas produced, in cm3, if the temperature was 22.5 °C and the pressure was 103 kPa. Use sections 1 and 2 of the data booklet. [2]
(iii) Suggest a reason why the volume of hydrogen gas collected was smaller than predicted. [1]
Determine the concentration, in mol dm−3, of the solution formed when 900.0 dm3 of NH3(g) at 300.0 K and 100.0 kPa is dissolved in water to form 2.00 dm3 of solution. Use sections 1 and 2 of the data booklet. [2]
Solutions and mark-scheme guidance · Set 1I
1I.1 A
At high pressure the molecules are close together and their own volume is no longer negligible. Attractive (cohesive) forces would decrease the volume, so B is wrong in direction; D is true but is not a failure of the model.
1I.2 A
Vm = RT/p depends only on temperature and pressure; the identity of an ideal gas is irrelevant (Avogadro’s law).
1I.3 A
V ∝ T in kelvin: T2 = 300 × 9.0/10.0 = 270 K = −3 °C. B scales the Celsius temperature (27 × 0.9 = 24.3), the trap the item is built on.
1I.4 C
pV constant: V2 = 400 × 100 ÷ 200 = 200 cm3. Halving the pressure doubles the volume.
1I.5 B
At constant T the total pV is conserved: (2.0 × 105 × 4.0 + 3.0 × 105 × 2.0) ÷ 6.0 = 2.3 × 105 Pa. C is the unweighted average; D simply adds the pressures.
1I.6 A
M = mRT/pV = 40.0 × 8.31 × 290 ÷ (98 × 0.220): temperature converted to kelvin (290 K) and volume to dm3 to match kPa. C leaves the temperature in °C; B and D invert the expression.
1I.7 D
n(H2) = 10 000 g ÷ 2 g mol−1 = 5 × 103 mol; p = nRT/V = 5 × 103 × 8.31 × 298 ÷ 1.0. The kilograms must become grams, H2 has M = 2, and T must be in kelvin.
1I.8 [2]
T = 423 K OR M = mRT/pV ✔; M = 0.150 × 8.31 × 423 ÷ (100.2 × 0.0337) = 156 g mol−1 ✔. Award [1] for correct answer with no working shown. (Menthol, C10H20O, M = 156.3.)
1I.9 [3]
(ii) n(H2) = ½ × 0.0288 = 0.0144 mol; V = nRT/p = 0.0144 × 8.31 × 295.5 ÷ 103 ✔ = 0.343 dm3 = 343 cm3 ✔. Accept 334–344 cm3. Award [1 max] for 0.343 «cm3», for 26.1 cm3 (using 22.5 K) or for 687 cm3 (using 0.0288 mol).
(iii) Lithium impure/partially oxidized OR gas leaked/ignited ✔ (accept “gas dissolved”).
1I.10 [2]
n = pV/RT = 100.0 × 900.0 ÷ (8.31 × 300.0) = 36.1 mol ✔; c = 36.1 ÷ 2.00 = 18.1 mol dm−3 ✔. Structure 1.5 and 1.4 in one question: gas to amount, amount to concentration.
Review · Structure 1.5
5Misconceptions, the examiner’s view, and the question types
- “Real gases occupy more volume than predicted because of intermolecular forces.” Why it is wrong: attractions pull molecules together and reduce the volume. Correct model: a larger volume comes from the molecules’ own volume at high pressure. Consequence: zero marks on the most quoted question in the topic.
- “Heavier gases deviate more.” Intermolecular force strength, not molar mass, decides it.
- “Doubling from 30 °C to 60 °C doubles the pressure.” Only kelvin ratios count: 333/303 = 1.10.
- “22.7 dm3 mol−1 is the volume of any mole of gas.” Only at STP.
- “R changes with temperature.” R is a constant; its numerical value depends only on the units used.
- “V against T is always a line through the origin.” Only with T in kelvin.
Evidence base: the IB Diploma chemistry subject reports quoted in the exam-focus sections of this page.
Answered well: straightforward calculations using the ideal gas equation, and converting the temperature correctly in structured questions.
Found difficult: (1) explaining deviations — about 20 % correct on a one-mark item, a mean of 0.78 out of 2 on another, and a common answer that pointed the wrong way; (2) comparing two gases by the type and strength of their intermolecular forces; (3) converting to kelvin in proportionality items, with fewer than a quarter correct in both levels’ papers of one session; (4) choosing consistent units, including cm3 or dm3 with pascals; (5) using the STP molar volume at other conditions; (6) rearranging for molar mass; (7) reading gas-law graphs when the temperature axis is in °C. One report lists Structure 1.5 among the areas of the programme and examination which appeared difficult for the students.
What successful answers did: wrote the converted values on their own line before substituting; checked whether a deviation question gave a direction; named the intermolecular forces in both gases and compared their strength; and derived every graph shape from pV = nRT.
Six question types cover the topic.
| If the question asks… | …then |
|---|---|
| An assumption of the model | Moving particles · negligible volume · no intermolecular forces · elastic collisions. |
| Why a gas deviates | Direction given? Larger V → particle volume; smaller V or p → attractions. No direction → either. |
| Which gas / which conditions | Low p, high T; weakest intermolecular forces (polarity, hydrogen bonding, electrons). |
| A graph | Read the axis labels; rearrange pV = nRT with two variables fixed. |
| A calculation with an amount or mass | pV = nRT, SI or kPa/dm3 consistently, T in K; M = mRT/pV. |
| A fixed sample changing conditions | p1V1/T1 = p2V2/T2; only T must be converted. |
6Quick check
- State the four assumptions of the ideal gas model and identify the two that real gases break.
- A real gas at very high pressure occupies a larger volume than pV = nRT predicts. Which assumption has failed, and why does the other not explain it?
- What volume does 0.500 mol of any ideal gas occupy at STP?
- Calculate the amount of gas occupying 2.00 dm3 at 101 kPa and 25 °C.
- A fixed mass of gas at 200 K is heated to 400 K at constant pressure. What happens to its volume — and what if it is heated from 27 °C to 54 °C instead?
- Sketch pV against p at constant temperature, and V against T on kelvin and Celsius axes.
- Explain why NO deviates from ideal behaviour more than N2, in the two steps a scheme wants.
- What volume does 4.00 g of helium occupy at STP?
| 1 | Constant random motion; negligible particle volume; no intermolecular forces; elastic collisions. The middle two fail. |
| 2 | Negligible particle volume: part of the measured volume is the molecules. Attractions would make the volume smaller. |
| 3 | 0.500 × 22.7 = 11.4 dm3, the same for every ideal gas. |
| 4 | n = (101 × 2.00) ÷ (8.31 × 298) = 0.0816 mol. |
| 5 | 200 → 400 K doubles the volume; 300 → 327 K increases it by only 9 %. |
| 6 | pV vs p: horizontal line. V vs T/K: straight line through the origin. V vs T/°C: straight line meeting V = 0 at −273 °C. |
| 7 | NO is polar (dipole–dipole and London forces), N2 non-polar (London only); so the intermolecular forces in NO are stronger. |
| 8 | 4.00 ÷ 4.00 = 1.00 mol → 22.7 dm3. |
7Summary and knowledge organiser
Essential knowledge
- An ideal gas: moving particles of negligible volume, no intermolecular forces, elastic collisions; pressure arises from collisions with the walls.
- Real gases deviate most at low temperature (attractions: smaller volume) and high pressure (particle volume: larger volume); stronger intermolecular forces mean larger deviation.
- Vm is constant at fixed T and p: 22.7 dm3 mol−1 at STP (273 K, 100 kPa).
- For a fixed amount: p ∝ 1/V (constant T); V ∝ T (constant p); p ∝ T (constant V), with T in kelvin.
- pV = nRT (SI: Pa, m3, K; R = 8.31 J K−1 mol−1) and p1V1/T1 = p2V2/T2; M = mRT/pV.
Examination checklist
- Write converted values (T in K; consistent p and V) on their own line.
- Deviation questions: check the direction; name the assumption.
- Compare gases by the type and strength of their intermolecular forces.
- Use 22.7 dm3 mol−1 only at STP.
- Read the temperature axis of any graph before choosing its shape.
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| Ideal gas 1.5.1 | Negligible volume, no IMF, elastic collisions, random motion. | Pressure = collisions with walls. |
| Real gases 1.5.2 | Deviate at high p (volume) and low T (attractions); polar/large deviate more. | Larger V is never caused by attractions. |
| Molar volume, graphs 1.5.3 | Vm = 22.7 dm3 mol−1 at STP; p ∝ 1/V; V ∝ T; p ∝ T. | Kelvin axis through origin; °C axis meets −273. |
| Ideal gas equation 1.5.4 | pV = nRT; combined gas law; M = mRT/pV. | T in K; Pa with m3 or kPa with dm3; never mix. |