What is the Kinetic Theory of Gases and how does it relate to ideal gases?

The Kinetic Theory of Gases is a model that explains the behavior of gases in terms of the motion of their molecules. It assumes that gas molecules are in constant random motion and that they collide elastically with each other and the walls of their container. This theory is fundamental to understanding ideal gases, which are hypothetical gases that perfectly follow the assumptions of the kinetic theory. In the context of Edexcel International A Levels Physics, this theory helps explain concepts like pressure, temperature, and volume relationships in gases.

How do the assumptions of the Kinetic Theory apply to real gases?

The Kinetic Theory assumes that gas molecules have negligible volume, experience no intermolecular forces, and undergo perfectly elastic collisions. While these assumptions hold true for ideal gases, real gases deviate from this behavior under high pressure and low temperature, where intermolecular forces and molecular volume become significant. In Edexcel International A Levels, understanding these deviations is crucial for studying real-world applications and the limitations of the ideal gas law.

What is the Ideal Gas Law and how is it derived from the Kinetic Theory?

The Ideal Gas Law is an equation of state for an ideal gas, expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature in Kelvin. It is derived from the Kinetic Theory by relating the average kinetic energy of gas molecules to temperature and using the assumptions of the theory to establish relationships between pressure, volume, and temperature. This law is a key part of the Edexcel International A Levels Physics curriculum, providing a foundation for understanding gas behavior.

Why is temperature important in the Kinetic Theory of Gases?

In the Kinetic Theory of Gases, temperature is directly related to the average kinetic energy of gas molecules. As temperature increases, the kinetic energy of the molecules increases, leading to faster movement and more frequent collisions, which in turn affects pressure and volume. Understanding this relationship is essential for students studying Edexcel International A Levels Physics, as it explains how temperature influences gas behavior and is crucial for solving thermodynamics problems.

How does the concept of pressure arise from the Kinetic Theory of Gases?

Pressure in the Kinetic Theory of Gases is the result of gas molecules colliding with the walls of their container. Each collision exerts a force on the wall, and the cumulative effect of countless collisions over a given area results in pressure. This concept is vital for Edexcel International A Levels Physics students, as it links microscopic molecular behavior to macroscopic properties like pressure, which is a key parameter in thermodynamics and gas laws.

Why is "Using temperature in degrees Celsius instead of kelvin in gas-law calculations" a common mistake in Kinetic Theory and Ideal Gases?

Why it happens: Temperatures are usually quoted in the question in °C, and students substitute the Celsius number straight into pV=NkT or a ratio without converting. How to avoid it: ALWAYS convert to kelvin first: T/K = /^C + 273. The gas laws use absolute temperature — p and V are proportional to T only in kelvin. Using 27 instead of 300 K, for example, makes a ratio ten times wrong. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Forgetting the final square root when finding the r.m.s. speed (quoting ⟨c²⟩ as the speed)" a common mistake in Kinetic Theory and Ideal Gases?

Why it happens: Rearranging \tfrac12 m c^2 = \tfrac32 kT gives c^2, and students stop there, forgetting that this is a mean-*square* speed, not a speed. How to avoid it: c^2 has units m^2s^-2 — a speed squared. Take the square root: c_rms = √( c^2) = √(3kT/m). If your 'speed' looks like \sim10^5 instead of \sim10^2–10^3, you forgot the root. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Thinking the mean kinetic energy depends on the type or mass of the gas molecule" a common mistake in Kinetic Theory and Ideal Gases?

Why it happens: Students associate 'more massive' with 'more energy' and assume a heavier molecule at the same temperature has a greater mean kinetic energy. How to avoid it: The mean translational kinetic energy is \tfrac12 m c^2 = \tfrac32 kT — it depends **only on the absolute temperature**. At the same T, all molecules have the *same* mean KE; the heavier ones simply move more slowly (smaller r.m.s. speed). Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Applying $p \propto 1/V$ (Boyle's law) when the temperature is not constant" a common mistake in Kinetic Theory and Ideal Gases?

Why it happens: Boyle's law is memorised as 'pressure and volume are inversely proportional', dropping the essential condition that the temperature (and mass of gas) must be constant. How to avoid it: p 1/V holds **only at constant temperature** for a fixed mass of gas. If T changes too, use the full relation p_1V_1/T_1 = p_2V_2/T_2. In CP14 the gas is compressed *slowly* precisely to keep T constant. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Giving vague or wrong assumptions of an ideal gas (e.g. 'molecules are small', 'no collisions')" a common mistake in Kinetic Theory and Ideal Gases?

Why it happens: Students half-remember the kinetic-theory assumptions and state them imprecisely, or confuse 'negligible volume' with 'no collisions'. How to avoid it: Be precise: molecular volume is **negligible compared with the container**; there are **no intermolecular forces except during collisions**; collisions are **perfectly elastic**; the **collision time is negligible** compared with the time between collisions; a **large number** of molecules in **random motion**. Molecules DO collide (that is what causes pressure). Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Confusing the number of molecules N with the number of moles n in pV = NkT" a common mistake in Kinetic Theory and Ideal Gases?

Why it happens: The Edexcel form uses N (molecules) with k, whereas the more familiar pV=nRT uses n (moles) with R; students mix the two. How to avoid it: In pV=NkT, N is the **number of molecules** and k is the Boltzmann constant. If given moles n, first convert: N = nN_A. Pairing n with k (or N with R) gives an answer wrong by a factor of N_A 6×10^23. Source: Edexcel IAL Physics Unit 5 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Kinetic Theory and Ideal Gases

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Short Notes - Kinetic Theory and Ideal Gases

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Detailed Study Notes

Detailed notes on Thermodynamics, Radiation, Oscillations and Cosmology for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Kinetic Theory and Ideal Gases — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 5)

The ideal gas equation pV = NkT with temperature in kelvin; Boyle's law and the p–V and p–(1/V) graphs; the kinetic theory model and the assumptions of an ideal gas; how temperature is linked to the mean molecular kinetic energy (½m⟨c²⟩ = 3/2 kT) and the r.m.s. speed; absolute zero; and Core Practical 14 — investigating the pressure–volume relationship of a gas at constant temperature.

At a glance

Ideal gas equation: $pV = NkT$ , where $N$ is the number of molecules, $k=1.38\times10^{-23}\ \text{J K}^{-1}$ and $T$ is in kelvin. Convert °C → K with $T=\theta+273$ .
Boyle's law: at constant temperature for a fixed mass of gas, $pV=\text{constant}$ , so $p_1V_1=p_2V_2$ . A graph of $p$ against $1/V$ is a straight line through the origin.
Kinetic theory model: a large number of molecules in random motion; negligible molecular volume; no forces except in collisions; elastic collisions; collision time ≪ time between collisions.
Pressure arises from molecules colliding with the walls and changing momentum — pressure is the resulting force per unit area.
Temperature ↔ energy: the mean translational KE per molecule is $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ — it depends only on $T$ , not on the type of gas.
R.m.s. speed: $c_{\text{rms}}=\sqrt{\langle c^2\rangle}=\sqrt{3kT/m}$ , so $c_{\text{rms}}\propto\sqrt{T}$ . Lighter molecules move faster at the same temperature.
Absolute zero: $0\ \text{K}=-273\ ^{\circ}\text{C}$ — the temperature of minimum internal energy.
Core Practical 14: trap a fixed mass of gas, vary $V$ at constant $T$ , plot $p$ vs $1/V$ → straight line through the origin confirms Boyle's law.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Use the ideal gas equation $pV = NkT$ , converting temperatures to kelvin, and relate it to $pV=nRT$ via $N=nN_A$ .
State Boyle's law and use $p_1V_1=p_2V_2$ ; interpret $p$ – $V$ and $p$ – $(1/V)$ graphs at constant temperature.
State the assumptions of the kinetic theory model of an ideal gas and use them to explain the origin of gas pressure.
Relate the absolute temperature of an ideal gas to the mean translational kinetic energy of its molecules, $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ , and calculate r.m.s. speeds.
Explain the meaning of absolute zero as the temperature of minimum internal energy ( $0\ \text{K}=-273\ ^{\circ}\text{C}$ ).
CP14 — investigate the $p$ – $V$ relationship for a gas at constant temperature and evaluate the experiment.

The ideal gas equation and temperature in kelvin

pV = NkT ties pressure, volume, number of molecules and absolute temperature together — always work in kelvin.

A gas has three easily measured properties — pressure $p$ , volume $V$ and temperature $T$ — plus the amount of gas present. For an ideal gas these are linked by a single equation of state, which Edexcel writes as

$pV = NkT$

$p$ = pressure in pascals (Pa),
$V$ = volume in cubic metres ( $\text{m}^3$ ),
$N$ = the number of molecules (just a number, no unit),
$k$ = the Boltzmann constant $= 1.38\times10^{-23}\ \text{J K}^{-1}$ ,
$T$ = the absolute temperature in kelvin (K).

Temperature must be in kelvin. Convert a Celsius temperature by adding 273: $T/\text{K} = \theta/^{\circ}\text{C} + 273$ So $27\ ^{\circ}\text{C} = 300\ \text{K}$ . This matters because the gas laws say $p$ and $V$ are proportional to $T$ — and proportionality only works from an absolute zero, which the kelvin scale provides.

Moles or molecules? You may meet the chemistry form $pV = nRT$ , where $n$ is the number of moles and $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ is the molar gas constant. The two are identical because the number of molecules is $N = nN_A$ (with $N_A = 6.02\times10^{23}\ \text{mol}^{-1}$ ) and $R = N_A k$ . Edexcel's booklet lists $pV = NkT$ , so if a question gives moles, convert to molecules first with $N = nN_A$ .

The ideal gas equation pV = NkT. Every calculation uses the absolute temperature in kelvin; convert from Celsius by adding 273. If given moles n, convert to molecules with N = n N_A before substituting.

$pV = NkT$ : $N$ = number of molecules, $k = 1.38\times10^{-23}\ \text{J K}^{-1}$ , $T$ in kelvin.
Convert temperature: $T/\text{K} = \theta/^{\circ}\text{C} + 273$ .
Given moles? Convert first: $N = nN_A$ (equivalently use $pV=nRT$ , $R=N_A k$ ).
Keep volume in $\text{m}^3$ and pressure in Pa for absolute calculations.

See the full worked example for kinetic theory and ideal gases →

Boyle's law and the p–V relationship

At constant temperature pV is constant; p against 1/V is a straight line through the origin.

Hold the temperature constant and keep a fixed mass of gas, and $pV = NkT$ tells us that $pV$ itself is constant (everything on the right-hand side is fixed). This is Boyle's law:

$pV = \text{constant} \qquad\Rightarrow\qquad p_1V_1 = p_2V_2$

So pressure is inversely proportional to volume: halve the volume and you double the pressure.

Two graphs, one law. A graph of $p$ against $V$ is a curve (a rectangular hyperbola) — awkward to test by eye. Rearranging to $p = (\text{constant})\times\dfrac{1}{V}$ and plotting $p$ against $\dfrac{1}{V}$ gives a straight line through the origin, which is far easier to confirm. This linearisation is the heart of Core Practical 14.

At constant temperature, p against V is a curve (left) while p against 1/V is a straight line through the origin (right). The straight-line graph is used to confirm Boyle's law; its gradient equals the constant pV.

Boyle's law is a special case of the more general combined relation. Because $\dfrac{pV}{T} = Nk$ is constant for a fixed mass of gas, if all three of $p$ , $V$ , $T$ change then $\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}.$ Use this when the temperature also changes; use $p_1V_1=p_2V_2$ only when it does not.

Boyle's law: fixed mass, constant $T$ ⇒ $pV=\text{constant}$ , so $p_1V_1=p_2V_2$ .
$p$ vs $V$ is a curve; $p$ vs $1/V$ is a straight line through the origin.
Only ratios matter for Boyle's law, so volume units cancel (no need to convert).
If $T$ also changes, use $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$ .

See the full worked example for kinetic theory and ideal gases →

The kinetic theory model and the origin of pressure

A gas is countless tiny molecules in random motion; their collisions with the walls are what we feel as pressure.

The kinetic theory explains the large-scale behaviour of a gas from the motion of its molecules. The model rests on a set of assumptions of an ideal gas:

A gas contains a very large number of molecules in continuous, random motion.
The volume of the molecules themselves is negligible compared with the volume of the container (they are treated as points).
There are no intermolecular forces except during collisions, so between collisions each molecule travels in a straight line at constant velocity.
All collisions (with each other and with the walls) are perfectly elastic — kinetic energy is conserved.
The time of each collision is negligible compared with the time between collisions.

Where pressure comes from. A molecule heading for a wall collides with it and rebounds, so its momentum reverses. A change of momentum requires a force (Newton's second law: force = rate of change of momentum), and by Newton's third law the molecule pushes back on the wall with an equal and opposite force. Averaged over the countless molecules hitting every part of the wall each second, this gives a steady force; the pressure is that total force per unit area.

Molecules move randomly and collide elastically with the walls. Each collision reverses a molecule's momentum, exerting a force on the wall; the pressure is the total force per unit area from all such collisions.

This picture also explains the gas laws qualitatively. Heating at constant volume makes molecules move faster, so they hit the walls more often and with a bigger momentum change each time — the pressure rises. Compressing at constant temperature (Boyle's law) leaves the molecular speeds unchanged but packs the molecules into a smaller space, so they strike the walls more frequently — again the pressure rises.

Assumptions: many molecules, random motion; negligible molecular volume; no forces except in collisions; elastic collisions; collision time ≪ time between collisions.
Pressure = molecules colliding with walls, reversing momentum → force per unit area.
Heating (constant V): faster molecules → more frequent, harder hits → higher $p$ .
Compressing (constant T): same speeds, but more frequent hits → higher $p$ .

See the full worked example for kinetic theory and ideal gases →

Temperature, molecular energy and r.m.s. speed

Absolute temperature is a direct measure of the mean kinetic energy of a molecule: ½m⟨c²⟩ = 3/2 kT.

The kinetic theory delivers one of the most important results in the whole topic. Combining the microscopic model with $pV = NkT$ gives the link between temperature and the mean kinetic energy of a molecule:

$\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$

$m$ = mass of one molecule (kg),
$\langle c^2\rangle$ = the mean-square speed (the average of the squares of the molecular speeds), units $\text{m}^2\text{s}^{-2}$ ,
$T$ = absolute temperature (K).

Read this carefully: the mean translational kinetic energy of a molecule depends only on the absolute temperature. It does not depend on the type of gas or the mass of the molecule. So at the same temperature, a light helium atom and a heavy oxygen molecule have the same mean kinetic energy — the heavy one simply moves more slowly.

Root-mean-square (r.m.s.) speed. To find a representative molecular speed, rearrange for $\langle c^2\rangle$ and take the square root:

$\langle c^2\rangle = \frac{3kT}{m} \qquad\Rightarrow\qquad c_{\text{rms}} = \sqrt{\langle c^2\rangle} = \sqrt{\frac{3kT}{m}}$

The final square root is essential — $\langle c^2\rangle$ is a speed squared. Because $c_{\text{rms}} \propto \sqrt{T}$ , you have to quadruple the absolute temperature to double the r.m.s. speed.

Because ½m⟨c²⟩ = 3/2 kT, the mean-square speed is proportional to absolute temperature and the r.m.s. speed is proportional to √T. Doubling the r.m.s. speed requires a four-fold increase in the kelvin temperature.

Absolute zero. As $T$ falls, the mean kinetic energy $\tfrac32 kT$ falls with it. At absolute zero, $T = 0\ \text{K} = -273\ ^{\circ}\text{C}$ , the molecules have their minimum possible internal energy — the coldest anything can be. Extrapolating a pressure–temperature graph for a gas back to zero pressure reaches this same point, $-273\ ^{\circ}\text{C}$ , which is how the kelvin scale is anchored.

Mean translational KE per molecule: $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ — depends only on $T$ .
Same temperature ⇒ same mean KE for every gas; lighter molecules move faster.
$c_{\text{rms}} = \sqrt{\langle c^2\rangle} = \sqrt{3kT/m}$ — remember the final square root.
$c_{\text{rms}} \propto \sqrt{T}$ : quadruple $T$ (in K) to double the r.m.s. speed.
Absolute zero: $0\ \text{K} = -273\ ^{\circ}\text{C}$ = minimum internal energy.

See the full worked example for kinetic theory and ideal gases →

Core Practical 14 — the p–V relationship at constant temperature

Trap a fixed mass of gas, change its volume slowly, and plot p against 1/V to confirm Boyle's law.

Core Practical 14 investigates how the pressure of a fixed mass of gas depends on its volume at constant temperature — a direct test of Boyle's law.

Apparatus. A common set-up traps a fixed mass of air in a gas syringe whose outlet is connected to a pressure sensor (or Bourdon gauge). The syringe's own scale reads the volume of the trapped gas, and the sensor reads the pressure. (An alternative uses a column of gas trapped above oil in a sealed tube, with a foot-pump varying the pressure.)

A fixed mass of gas is trapped in a syringe connected to a pressure sensor. Moving the plunger changes the volume (read from the scale) while the sensor reads the pressure. Compress slowly so the temperature stays constant.

Method.

Record the initial volume $V$ and pressure $p$ .
Move the plunger to a new volume, slowly, and wait for the reading to settle before recording $p$ and $V$ .
Repeat over a good range (say 6–8 volumes).

Analysis. Boyle's law predicts $pV = \text{constant}$ , i.e. $p = (\text{constant})\times\dfrac{1}{V}$ . Calculate $\dfrac{1}{V}$ for each reading and plot $p$ (y-axis) against $\dfrac{1}{V}$ (x-axis). A straight line through the origin confirms $p \propto \dfrac{1}{V}$ — Boyle's law holds and the gas is behaving ideally over the range tested. (Equivalently, check that $pV$ is the same for every reading.)

Evaluating the experiment.

Compress slowly. Compressing a gas warms it; if you change the volume quickly the temperature rises and Boyle's law (constant $T$ ) fails. Pausing before each reading lets the gas return to room temperature.
Avoid leaks. The mass of gas must be fixed — a leaking seal makes $pV$ drift.
Read a good range and use a best-fit line so a single anomalous reading does not distort the conclusion.

Trap a fixed mass of gas in a syringe + pressure sensor; read $p$ and $V$ .
Vary the volume in steps over a good range; record $p$ at each.
Plot $p$ against $1/V$ : straight line through the origin confirms Boyle's law.
Compress SLOWLY (constant $T$ ); no leaks (fixed mass); best-fit line.

See the full worked example for kinetic theory and ideal gases →

Quick recap

Ideal gas equation $pV = NkT$ : $N$ = number of molecules, $k = 1.38\times10^{-23}\ \text{J K}^{-1}$ , temperature in kelvin ( $T = \theta + 273$ ).
Boyle's law: fixed mass, constant $T$ ⇒ $pV = \text{constant}$ ; $p$ vs $1/V$ is a straight line through the origin; if $T$ changes use $pV/T = \text{constant}$ .
Kinetic theory assumptions: many molecules in random motion; negligible molecular volume; no forces except in collisions; elastic collisions; short collision time.
Pressure comes from molecules colliding with the walls and reversing their momentum — force per unit area.
$\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ : mean molecular KE depends only on absolute temperature, not on the gas.
R.m.s. speed $c_{\text{rms}} = \sqrt{3kT/m} \propto \sqrt{T}$ ; remember the square root; quadruple $T$ to double the speed.
Absolute zero $= 0\ \text{K} = -273\ ^{\circ}\text{C}$ , the temperature of minimum internal energy.
CP14: trap a fixed mass of gas, vary $V$ slowly at constant $T$ , plot $p$ vs $1/V$ → straight line through origin confirms Boyle's law.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$pV = NkT$ — $N$ is the number of molecules, $k = 1.38\times10^{-23}\ \text{J K}^{-1}$ , $T$ in kelvin.
Convert temperature: $T/\text{K} = \theta/^{\circ}\text{C} + 273$ .
Boyle's law: $pV = \text{constant}$ at constant $T$ ; $p_1V_1 = p_2V_2$ ; $p$ vs $1/V$ straight line through origin.
Combined: $\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}$ (fixed mass of gas).
$\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ — mean molecular KE depends only on absolute temperature.
$c_{\text{rms}} = \sqrt{\langle c^2\rangle} = \sqrt{3kT/m}$ ; $c_{\text{rms}} \propto \sqrt{T}$ .
$N = nN_A$ with $N_A = 6.02\times10^{23}\ \text{mol}^{-1}$ (and $R = N_A k = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ ).
Absolute zero $= 0\ \text{K} = -273\ ^{\circ}\text{C}$ = minimum internal energy.

How it’s examined

Kinetic theory and ideal gases is Unit 5 content, examined on the Unit 5 paper (WPH15, Thermodynamics, Radiation, Oscillations and Cosmology), a structured IA2 paper with short calculations, 'show that' items, graph work and extended-answer questions. Typical formats: convert a temperature to kelvin and substitute into $pV=NkT$ (1–3 marks, AO2); a Boyle's-law calculation with $p_1V_1=p_2V_2$ or a combined-law calculation where you must spot which quantities change (2–4 marks); 'state the assumptions of an ideal gas' (up to 4 marks, AO1 — precise wording matters); a kinetic-theory explanation of how a gas exerts pressure or why heating/compressing changes the pressure (3–4 marks, AO2); a mean-KE or r.m.s.-speed calculation using $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ (2–4 marks — do not forget the square root); and a CP14 data/experiment question (plot $p$ vs $1/V$ , precautions, constant temperature). ALWAYS work in kelvin, show the equation, substitute with units, and quote answers to 3 s.f. For r.m.s. speed remember the final square root; for assumptions be precise ('negligible molecular volume', not just 'small').

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 5 (WPH15) examiner reports; Edexcel IAL Physics Unit 5 sample assessment materials and mark schemes. Last reviewed 2026-07-06.

Take this whole topic with you

Step-by-step worked examples — Kinetic Theory and Ideal Gases

Step-by-step solutions to past-paper-style questions on kinetic theory and ideal gases, written exactly the way a tutor would explain them at the board.

Question type:

1Converting a temperature to kelvin (2 marks)
Getting startedDirect calculation• temperature, kelvin, Unit 5
Question
A gas in a container is at a temperature of 27 °C. State the thermodynamic (absolute) temperature of the gas in kelvin, and explain why this conversion matters for the ideal gas equation. (2 marks)
Step-by-step solution
1. Step 1
  Convert a Celsius temperature to kelvin by adding 273: $T/\text{K} = \theta/^{\circ}\text{C} + 273$ .
  $T = 27 + 273 = 300\ \text{K}$
2. Step 2
  The ideal gas equation $pV = NkT$ uses the absolute (kelvin) temperature. Kelvin measures the internal energy from absolute zero, so $p$ and $V$ are proportional to $T$ only when $T$ is in kelvin. Substituting a Celsius value would be physically wrong.
  $pV = NkT \quad (T\ \text{in kelvin})$
Answer
$T = 300\ \text{K}$ .
Examiner tip
Mark scheme: (1) add 273 (M); (2) $300\ \text{K}$ with the reason that gas laws require absolute temperature (A). Every gas calculation on this paper needs kelvin — forgetting to convert is the single most common error and it makes ratios (e.g. Charles' law) meaningless.
2Pressure of a gas from pV = NkT (2 marks)
Getting startedDirect calculation• ideal gas, pV=NkT, AO2, Unit 5
Question
A sealed container of volume 2.0 × 10⁻³ m³ holds 3.0 × 10²³ molecules of an ideal gas at a temperature of 300 K. Calculate the pressure of the gas. (Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹.) (2 marks)
Step-by-step solution
1. Step 1
  Use the ideal gas equation in the Edexcel form $pV = NkT$ and rearrange for pressure: $p = \dfrac{NkT}{V}$ .
  $p = \frac{NkT}{V}$
2. Step 2
  Substitute $N = 3.0\times10^{23}$ , $k = 1.38\times10^{-23}\ \text{J K}^{-1}$ , $T = 300\ \text{K}$ and $V = 2.0\times10^{-3}\ \text{m}^3$ .
  $p = \frac{3.0\times10^{23} \times 1.38\times10^{-23} \times 300}{2.0\times10^{-3}}$
3. Step 3
  Evaluate the numerator ( $3.0\times10^{23}\times1.38\times10^{-23} = 4.14$ ; $\times300 = 1242$ ), then divide.
  $p = \frac{1242}{2.0\times10^{-3}} = 6.21\times10^{5}\ \text{Pa}$
Answer
$p = 6.21\times10^{5}\ \text{Pa}$ .
Examiner tip
Mark scheme: (1) use of $pV=NkT$ (M); (2) $6.21\times10^{5}\ \text{Pa}$ (A). The temperature is already in kelvin here. Keep the volume in $\text{m}^3$ — using $\text{cm}^3$ or litres without converting is a frequent slip.
3Boyle's law at constant temperature (2 marks)
Getting startedDirect calculation• Boyle's law, pV=const, Unit 5
Question
A fixed mass of gas has a volume of 500 cm³ at a pressure of 1.0 × 10⁵ Pa. It is compressed slowly at constant temperature until its volume is 200 cm³. Calculate the new pressure. (2 marks)
Step-by-step solution
1. Step 1
  At constant temperature for a fixed mass of gas, Boyle's law gives $pV = \text{constant}$ , so $p_1V_1 = p_2V_2$ .
  $p_1 V_1 = p_2 V_2$
2. Step 2
  Rearrange for $p_2$ and substitute. The volumes can be left in $\text{cm}^3$ because only their ratio matters.
  $p_2 = p_1\frac{V_1}{V_2} = 1.0\times10^{5} \times \frac{500}{200}$
3. Step 3
  Evaluate ( $500/200 = 2.5$ ).
  $p_2 = 1.0\times10^{5} \times 2.5 = 2.5\times10^{5}\ \text{Pa}$
Answer
$p_2 = 2.5\times10^{5}\ \text{Pa}$ .
Examiner tip
Mark scheme: (1) use of $p_1V_1=p_2V_2$ (M); (2) $2.5\times10^{5}\ \text{Pa}$ (A). Because it is a ratio the volume units cancel, so converting to $\text{m}^3$ is not required here. Boyle's law only holds because the temperature is constant — the phrase 'compressed slowly' signals that the gas stays at constant temperature.
4Pressure from a number of moles (4 marks)
Building confidenceMulti-step problem• ideal gas, moles, Avogadro, Unit 5
Question
A container of volume 0.025 m³ holds 0.80 mol of an ideal gas at 17 °C. Calculate the pressure of the gas. (Avogadro constant N_A = 6.02 × 10²³ mol⁻¹; k = 1.38 × 10⁻²³ J K⁻¹.) (4 marks)
Step-by-step solution
1. Step 1
  Find the number of molecules from the number of moles: $N = n N_A$ .
  $N = n N_A = 0.80 \times 6.02\times10^{23} = 4.82\times10^{23}$
2. Step 2
  Convert the temperature to kelvin.
  $T = 17 + 273 = 290\ \text{K}$
3. Step 3
  Use $pV = NkT$ , rearranged for pressure.
  $p = \frac{NkT}{V}$
4. Step 4
  Substitute and evaluate.
  $p = \frac{4.82\times10^{23} \times 1.38\times10^{-23} \times 290}{0.025} = 7.71\times10^{4}\ \text{Pa}$
Answer
$p = 7.71\times10^{4}\ \text{Pa}$ .
Examiner tip
Mark scheme: (1) $N=nN_A$ (M); (2) $T=290\ \text{K}$ ; (3) use of $pV=NkT$ (M); (4) $7.71\times10^{4}\ \text{Pa}$ (A). Equivalently $pV=nRT$ with $R=N_A k=8.31\ \text{J mol}^{-1}\text{K}^{-1}$ gives the same answer — but Edexcel's booklet lists $pV=NkT$ , so convert moles to molecules first.
5Mean kinetic energy of a molecule (3 marks)
Building confidenceDirect calculation• kinetic energy, temperature, AO2, Unit 5
Question
Calculate the mean translational kinetic energy of a single molecule of an ideal gas at a temperature of 350 K. (k = 1.38 × 10⁻²³ J K⁻¹.) (3 marks)
Step-by-step solution
1. Step 1
  The mean translational kinetic energy per molecule depends only on the absolute temperature: $\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$ .
  $E_k = \tfrac{3}{2}kT$
2. Step 2
  Substitute $k = 1.38\times10^{-23}\ \text{J K}^{-1}$ and $T = 350\ \text{K}$ .
  $E_k = \tfrac{3}{2} \times 1.38\times10^{-23} \times 350$
3. Step 3
  Evaluate ( $1.38\times10^{-23}\times350 = 4.83\times10^{-21}$ ; $\times1.5$ ).
  $E_k = 7.25\times10^{-21}\ \text{J}$
Answer
$E_k = 7.25\times10^{-21}\ \text{J}$ .
Examiner tip
Mark scheme: (1) use of $E_k=\tfrac32 kT$ (M); (2) substitution; (3) $7.25\times10^{-21}\ \text{J}$ (A). This is the mean KE of ONE molecule and is independent of the gas — a helium atom and a much heavier molecule have the same mean KE at the same temperature (the heavier one just moves more slowly).
6Root-mean-square speed of a molecule (4 marks)
Building confidenceMulti-step problem• rms speed, kinetic energy, AO2, Unit 5
Question
Nitrogen gas is at a temperature of 300 K. One nitrogen molecule has a mass of 4.65 × 10⁻²⁶ kg. Calculate the root-mean-square (r.m.s.) speed of a nitrogen molecule. (k = 1.38 × 10⁻²³ J K⁻¹.) (4 marks)
Step-by-step solution
1. Step 1
  Equate the mean kinetic energy expression: $\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$ , then make the mean-square speed the subject.
  $\langle c^2\rangle = \frac{3kT}{m}$
2. Step 2
  Substitute the values.
  $\langle c^2\rangle = \frac{3 \times 1.38\times10^{-23} \times 300}{4.65\times10^{-26}}$
3. Step 3
  Evaluate the mean-square speed ( $3\times1.38\times10^{-23}\times300 = 1.242\times10^{-20}$ ).
  $\langle c^2\rangle = \frac{1.242\times10^{-20}}{4.65\times10^{-26}} = 2.671\times10^{5}\ \text{m}^2\,\text{s}^{-2}$
4. Step 4
  The r.m.s. speed is the square root of the mean-square speed: $c_{\text{rms}} = \sqrt{\langle c^2\rangle}$ .
  $c_{\text{rms}} = \sqrt{2.671\times10^{5}} = 517\ \text{m s}^{-1}$
Answer
$c_{\text{rms}} = 517\ \text{m s}^{-1}$ .
Examiner tip
Mark scheme: (1) equate $\tfrac12 m\langle c^2\rangle=\tfrac32 kT$ (M); (2)/(3) $\langle c^2\rangle = 3kT/m$ (M); (4) take the square root, $517\ \text{m s}^{-1}$ (A). Do NOT forget the final square root — $\langle c^2\rangle$ is a mean-square speed with units $\text{m}^2\text{s}^{-2}$ ; the r.m.s. speed is its square root.
7Finding the temperature to double the r.m.s. speed (5 marks)
StretchMulti-step problem• rms speed, proportionality, AO2, Unit 5
Question
The r.m.s. speed of the molecules of an ideal gas is 480 m s⁻¹ at 300 K. To what temperature must the gas be heated so that the r.m.s. speed of the molecules is doubled to 960 m s⁻¹? (5 marks)
Step-by-step solution
1. Step 1
  From $\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$ , the mean-square speed is proportional to absolute temperature: $\langle c^2\rangle \propto T$ , so $c_{\text{rms}} \propto \sqrt{T}$ .
  $c_{\text{rms}} \propto \sqrt{T}$
2. Step 2
  Write the ratio of the two states. Doubling $c_{\text{rms}}$ means the ratio of speeds is 2.
  $\frac{c_2}{c_1} = \sqrt{\frac{T_2}{T_1}} \;\Rightarrow\; \frac{T_2}{T_1} = \left(\frac{c_2}{c_1}\right)^2$
3. Step 3
  The speed ratio is $c_2/c_1 = 960/480 = 2$ , so the temperature ratio is $2^2 = 4$ .
  $\frac{T_2}{T_1} = 2^2 = 4$
4. Step 4
  Solve for $T_2$ with $T_1 = 300\ \text{K}$ .
  $T_2 = 4 \times 300 = 1200\ \text{K}$
5. Step 5
  Check: at $1200\ \text{K}$ , $\langle c^2\rangle$ is four times larger, so $c_{\text{rms}}$ is $\sqrt{4}=2$ times larger — as required.
  $c_{\text{rms}}(1200\ \text{K}) = 2 \times 480 = 960\ \text{m s}^{-1}\ \checkmark$
Answer
$T_2 = 1200\ \text{K}$ (about 927 °C).
Examiner tip
Mark scheme: (1) $c_{\text{rms}}\propto\sqrt{T}$ (M); (2)/(3) $T_2/T_1=(c_2/c_1)^2=4$ (M); (4) $1200\ \text{K}$ (A). The trap is thinking that doubling the speed doubles the temperature — because the KE (and hence $T$ ) depends on $c^2$ , doubling the speed quadruples the absolute temperature.
8Combined gas behaviour with changing p, V and T (6 marks)
StretchMulti-step problem• ideal gas, combined, AO2, Unit 5
Question
A weather balloon contains 25 m³ of helium at ground level, where the pressure is 1.0 × 10⁵ Pa and the temperature is 290 K. The balloon rises until the pressure is 4.0 × 10⁴ Pa and the temperature is 250 K. Assuming no gas escapes and the helium behaves ideally, calculate the new volume of the balloon. (6 marks)
Step-by-step solution
1. Step 1
  For a fixed number of molecules $N$ , $pV = NkT$ gives $\dfrac{pV}{T} = Nk = \text{constant}$ , so $\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}$ .
  $\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$
2. Step 2
  Rearrange for the new volume $V_2$ .
  $V_2 = V_1 \times \frac{p_1}{p_2} \times \frac{T_2}{T_1}$
3. Step 3
  Substitute the pressure ratio $p_1/p_2 = 1.0\times10^{5} / 4.0\times10^{4} = 2.5$ .
  $\frac{p_1}{p_2} = \frac{1.0\times10^{5}}{4.0\times10^{4}} = 2.5$
4. Step 4
  Substitute the temperature ratio $T_2/T_1 = 250/290 = 0.862$ .
  $\frac{T_2}{T_1} = \frac{250}{290} = 0.862$
5. Step 5
  Combine the factors with $V_1 = 25\ \text{m}^3$ .
  $V_2 = 25 \times 2.5 \times 0.862$
6. Step 6
  Evaluate.
  $V_2 = 53.9\ \text{m}^3$
Answer
$V_2 = 53.9\ \text{m}^3$ .
Examiner tip
Mark scheme: (1) $pV/T=$ constant (M); (2) rearrange (M); (3)/(4) both ratios (M); (5)/(6) $53.9\ \text{m}^3$ (A). Both temperatures are in kelvin (given), so no conversion is needed here — but always check. The balloon expands because the fall in pressure (×2.5) outweighs the modest cooling (×0.86).
9Core Practical 14 — verifying Boyle's law from a graph (6 marks)
StretchGraph or diagram• Adapted from Edexcel IAL Physics Unit 5 (CP14)• CP14, Boyle's law, graph, AO3, Unit 5
Question
In Core Practical 14 a student traps a fixed mass of air in a gas syringe connected to a pressure sensor and slowly reduces the volume, recording pressure p and volume V at constant temperature. Explain how plotting a graph of p against 1/V confirms Boyle's law, and describe how the graph would be used to test whether the gas behaves ideally. State one precaution needed for a valid result. (6 marks)
Step-by-step solution
1. Step 1
  Boyle's law states $pV = \text{constant}$ at constant temperature for a fixed mass of gas. Rearranging gives $p = \text{constant} \times \dfrac{1}{V}$ .
  $pV = \text{constant} \;\Rightarrow\; p = (\text{const})\frac{1}{V}$
2. Step 2
  Comparing $p = (\text{const})\dfrac{1}{V}$ with $y = mx$ : plotting $p$ (y-axis) against $1/V$ (x-axis) should give a straight line through the origin, whose gradient equals the constant $pV$ .
  $y = mx, \quad m = pV = \text{constant}$
3. Step 3
  A straight line passing through the origin confirms that $p$ is inversely proportional to $V$ — i.e. Boyle's law holds and the gas is behaving (close to) ideally over the range tested. Any curvature or a non-zero intercept indicates a departure from ideal behaviour.
  $p \propto \frac{1}{V} \iff \text{straight line through origin}$
4. Step 4
  Precaution: compress the gas SLOWLY so the temperature stays constant (compression warms a gas; the heat must have time to leave so $T$ returns to room temperature before each reading). Boyle's law is only valid at constant temperature.
  $T = \text{constant (compress slowly)}$
Answer
A graph of $p$ against $1/V$ that is a straight line through the origin confirms $p \propto 1/V$ (Boyle's law); its gradient is the constant $pV$ . Departures (curvature/non-zero intercept) show non-ideal behaviour. Precaution: change the volume slowly so the temperature stays constant.
Examiner tip
Mark scheme: (1) $pV=$ const rearranged to $p=\text{const}/V$ (M); (2) plot $p$ vs $1/V$ , expect a straight line through the origin (M); (3) straight line through origin ⇒ $p\propto 1/V$ / ideal (A); (4) named valid precaution: compress slowly to keep $T$ constant (or allow thermal equilibrium before each reading) (M+A). 'Plot $p$ against $V$ ' scores nothing — that gives a curve; only $p$ against $1/V$ linearises the relationship.

Model Answers — Kinetic Theory and Ideal Gases

High-scoring sample answers for kinetic theory and ideal gases on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State the ideal gas equation in the form pV = NkT, defining each symbol, and state the SI unit of temperature that must be used.
Model answer
The ideal gas equation is

$pV = NkT$

where $p$ is the pressure (Pa), $V$ is the volume ( $\text{m}^3$ ), $N$ is the number of molecules, $k$ is the Boltzmann constant ( $1.38\times10^{-23}\ \text{J K}^{-1}$ ) and $T$ is the absolute (thermodynamic) temperature.

The temperature must be in kelvin (K).
Why this scores
Why this scores 2/2. (1) the equation with symbols defined; (2) temperature in kelvin. Every symbol must be named. Stating $N$ is the number of molecules (not moles) is essential in this Edexcel $pV=NkT$ form.
Question 2
3 marks
Q (3 marks). State what is meant by absolute zero, give its value in degrees Celsius, and explain, in terms of molecular motion, what happens to the gas molecules as this temperature is approached.
Model answer
Absolute zero is the lowest possible temperature, 0 K, which is equal to −273 °C (more precisely −273.15 °C).

At absolute zero the molecules have their minimum possible internal (kinetic) energy. As the temperature falls towards absolute zero, the mean kinetic energy of the molecules falls (since $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ ), so the molecules move more and more slowly; at $0\ \text{K}$ their translational kinetic energy would be zero (a minimum).

Because pressure arises from molecular collisions with the walls, the pressure of an ideal gas would also fall to zero at absolute zero.
Why this scores
Why this scores 3/3. (1) $0\ \text{K} = -273\ ^{\circ}\text{C}$ ; (2) minimum internal energy / molecules have least KE; (3) link to $\tfrac32 kT$ so mean KE → minimum as $T\to0$ . Saying molecules 'stop completely' is acceptable at this level as minimum KE; the key mark is the value $-273\ ^{\circ}\text{C}$ and 'minimum energy'.
Question 3
3 marks
Q (3 marks). Calculate the mean translational kinetic energy of a molecule of an ideal gas at 400 K, and state whether a heavier molecule at the same temperature would have a larger, smaller or equal mean kinetic energy. (k = 1.38 × 10⁻²³ J K⁻¹.)
Model answer
The mean translational kinetic energy per molecule is

$E_k = \tfrac{3}{2}kT = \tfrac{3}{2}\times 1.38\times10^{-23}\times 400 = 8.28\times10^{-21}\ \text{J}$

A heavier molecule at the same temperature would have the same mean kinetic energy, because the mean kinetic energy $\tfrac32 kT$ depends only on the absolute temperature, not on the mass. The heavier molecule simply moves more slowly (smaller r.m.s. speed) so that $\tfrac12 m\langle c^2\rangle$ is unchanged.
Why this scores
Why this scores 3/3. (1) use of $E_k=\tfrac32 kT$ ; (2) $8.28\times10^{-21}\ \text{J}$ ; (3) equal mean KE, because it depends only on $T$ . The common wrong answer is 'the heavier molecule has more KE' — mean KE is set by temperature alone.
Question 4
4 marks
Q (4 marks). State four assumptions of the kinetic theory model of an ideal gas.
Model answer
Any four of the following assumptions of an ideal gas:

The gas contains a very large number of molecules in continuous random motion.

The volume of the molecules is negligible compared with the volume of the container (the molecules are treated as point particles).

There are no intermolecular forces between molecules except during collisions (so molecules move in straight lines between collisions).

Collisions between molecules, and with the walls, are perfectly elastic (kinetic energy is conserved).

The duration of a collision is negligible compared with the time between collisions.

Together these mean the internal energy of an ideal gas is entirely the translational kinetic energy of its molecules.
Why this scores
Why this scores 4/4. One mark for each valid assumption (max 4): large number in random motion; negligible molecular volume; negligible intermolecular forces except in collisions; elastic collisions; collision time ≪ time between collisions. Vague statements ('molecules are small') are not enough — say negligible compared with the container volume.
Question 5
4 marks
Q (4 marks). Using the kinetic theory, explain how a gas exerts a pressure on the walls of its container, and hence explain why the pressure increases when the gas is heated at constant volume.
Model answer
Origin of pressure. The gas molecules are in continuous random motion and repeatedly collide with the walls of the container. In each collision a molecule's momentum reverses (changes), so by Newton's second law the wall exerts a force on the molecule and, by Newton's third law, the molecule exerts an equal and opposite force on the wall. The pressure is the total force from all these collisions per unit area of wall.

Heating at constant volume. Raising the temperature increases the mean kinetic energy of the molecules ( $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ ), so they move faster. Faster molecules:

hit the walls more frequently, and

undergo a larger momentum change at each collision.

Both effects increase the rate of change of momentum at the walls, so the force per unit area — the pressure — increases. (With $V$ fixed, $pV=NkT$ confirms $p\propto T$ .)
Why this scores
Why this scores 4/4. (1) molecules collide with walls and change momentum; (2) force on wall = rate of change of momentum, pressure = force/area; (3) heating raises mean KE / speed; (4) more frequent collisions + greater momentum change ⇒ higher pressure. Linking to $p\propto T$ from $pV=NkT$ is a good confirming detail.
Question 6
4 marks
Q (4 marks). A fixed mass of gas is compressed to half its volume at constant temperature. State what happens to its pressure, and explain this using the kinetic theory.
Model answer
By Boyle's law, at constant temperature $pV = \text{constant}$ , so halving the volume doubles the pressure.

Kinetic-theory explanation. At constant temperature the molecules keep the same mean kinetic energy and therefore the same mean speed — each collision still delivers the same momentum change. However, when the volume is halved the molecules are in a smaller space, so on average each molecule travels a shorter distance between hitting a wall.

This means the molecules collide with the walls more frequently (there are more collisions per second per unit area). More collisions per second, each with the same momentum change, means a greater total force per unit area, so the pressure doubles.
Why this scores
Why this scores 4/4. (1) pressure doubles (from $pV=$ const); (2) constant $T$ ⇒ same molecular speed/KE; (3) smaller volume ⇒ molecules hit walls more often (shorter distance between collisions); (4) more frequent collisions ⇒ higher pressure. The key is that the speed is unchanged (constant $T$ ) — only the collision frequency increases.
Question 7
4 marks
Q (4 marks). A container holds a mixture of helium (molecular mass 6.64 × 10⁻²⁷ kg) and oxygen (molecular mass 5.31 × 10⁻²⁶ kg) at the same temperature. Compare (i) the mean kinetic energies and (ii) the r.m.s. speeds of the two types of molecule, justifying each answer.
Model answer
(i) Mean kinetic energies. Both gases are at the same temperature, and the mean translational kinetic energy of a molecule is $\tfrac32 kT$ , which depends only on $T$ . So the helium and oxygen molecules have the same mean kinetic energy.

(ii) R.m.s. speeds. Since $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ is the same for both, but the helium molecule has a much smaller mass $m$ , its mean-square speed $\langle c^2\rangle = 3kT/m$ must be larger. Therefore the helium molecules have the greater r.m.s. speed.

Quantitatively $c_{\text{rms}} \propto \sqrt{1/m}$ , so the ratio is

$\frac{c_{\text{He}}}{c_{\text{O}_2}} = \sqrt{\frac{m_{\text{O}_2}}{m_{\text{He}}}} = \sqrt{\frac{5.31\times10^{-26}}{6.64\times10^{-27}}} = \sqrt{8.0} = 2.83$

so helium molecules move about 2.8 times faster (r.m.s.) than oxygen molecules.
Why this scores
Why this scores 4/4. (i) equal mean KE because it depends only on $T$ (2 marks: statement + reason); (ii) helium faster because $c_{\text{rms}}\propto\sqrt{1/m}$ and He is lighter (2 marks: statement + $\sqrt{}$ reasoning or ratio 2.8). The classic error is assuming the lighter gas has less KE — at the same $T$ the KE is identical; only the speeds differ.
Question 8
6 marks
Q (6 marks). A gas syringe contains 60 cm³ of air at a pressure of 1.0 × 10⁵ Pa and a temperature of 300 K. (a) The plunger is pushed in slowly, at constant temperature, until the volume is 24 cm³. Calculate the new pressure. (b) The syringe (now at 24 cm³) is then sealed and heated to 360 K. Calculate the pressure at 360 K. (c) State one assumption you made.
Model answer
(a) At constant temperature, Boyle's law gives $p_1V_1 = p_2V_2$ :

$p_2 = p_1\frac{V_1}{V_2} = 1.0\times10^{5}\times\frac{60}{24} = 2.5\times10^{5}\ \text{Pa}$

(b) Now the volume is fixed at $24\ \text{cm}^3$ and the gas is heated from $300\ \text{K}$ to $360\ \text{K}$ . At constant volume $\dfrac{p}{T} = \text{constant}$ (from $pV=NkT$ ), so

$p_3 = p_2\frac{T_3}{T_2} = 2.5\times10^{5}\times\frac{360}{300} = 3.0\times10^{5}\ \text{Pa}$

(c) Assumption: the mass of gas is fixed (the syringe is sealed and no air leaks), and the gas behaves as an ideal gas so that $pV=NkT$ applies. (Also that the temperature was truly constant in part (a).)
Why this scores
Why this scores 6/6. (a) $p_1V_1=p_2V_2$ ⇒ $2.5\times10^{5}\ \text{Pa}$ (M+A); (b) $p/T=$ const, $360/300$ , ⇒ $3.0\times10^{5}\ \text{Pa}$ (M+A); (c) valid assumption — fixed mass of gas / ideal gas / constant temperature in (a). Note the two stages use different gas laws: Boyle (constant $T$ ) then the pressure law (constant $V$ ). Temperatures already in kelvin.
Question 9
Adapted from Edexcel IAL Physics Unit 5 (CP14)6 marks
Q (6 marks). Describe how Core Practical 14 can be carried out to investigate the relationship between the pressure and volume of a fixed mass of gas at constant temperature. Include the apparatus, the measurements taken, the graph plotted and how it confirms Boyle's law, and one precaution.
Model answer
Apparatus. Trap a fixed mass of air in a gas syringe (or a sealed tube of gas above a column of oil). Connect the trapped gas to a pressure sensor (or Bourdon gauge) so its pressure can be read, and use the syringe scale to read the volume of the gas.

Method. Change the volume of the trapped gas in steps by moving the plunger (or by using a pump/foot-pump on the oil reservoir). At each setting record the volume $V$ from the syringe scale and the pressure $p$ from the sensor. Take a good range of values (e.g. 6–8 readings).

Analysis. Boyle's law predicts $pV = \text{constant}$ at constant temperature, i.e. $p = (\text{const})\times\dfrac{1}{V}$ . Calculate $\dfrac{1}{V}$ for each reading and plot $p$ (y-axis) against $\dfrac{1}{V}$ (x-axis). A straight line through the origin confirms that $p \propto \dfrac{1}{V}$ , i.e. Boyle's law holds. (Alternatively, showing $pV$ is constant for every reading also confirms it.)

Precaution. Change the volume slowly and pause before each reading, so that any heating caused by compression dissipates and the gas returns to room temperature — Boyle's law is only valid at constant temperature.
Why this scores
Why this scores 6/6. (1) gas syringe with fixed mass of gas + pressure sensor; (2) vary volume, record $p$ and $V$ over a range; (3) rearrange to $p=\text{const}/V$ ; (4) plot $p$ against $1/V$ ; (5) straight line through origin ⇒ $p\propto1/V$ / Boyle's law; (6) precaution — compress slowly to keep $T$ constant (with reason). Naming the apparatus and the $p$ -versus- $1/V$ graph are the precision marks; 'plot $p$ against $V$ ' would lose the analysis marks.
Question 10
6 marks
Q (6 marks). An ideal gas of 0.50 mol is contained in a volume of 0.012 m³ at a pressure of 1.1 × 10⁵ Pa. (a) Calculate the temperature of the gas. (b) Calculate the r.m.s. speed of the molecules, given that the mass of one molecule is 5.3 × 10⁻²⁶ kg. (k = 1.38 × 10⁻²³ J K⁻¹; N_A = 6.02 × 10²³ mol⁻¹.)
Model answer
(a) First find the number of molecules: $N = nN_A = 0.50\times6.02\times10^{23} = 3.01\times10^{23}$ .

Rearranging $pV = NkT$ for temperature:

$T = \frac{pV}{Nk} = \frac{1.1\times10^{5}\times 0.012}{3.01\times10^{23}\times 1.38\times10^{-23}}$

$T = \frac{1320}{4.154} = 318\ \text{K}$

(b) The r.m.s. speed comes from $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ , so $\langle c^2\rangle = \dfrac{3kT}{m}$ :

$\langle c^2\rangle = \frac{3\times 1.38\times10^{-23}\times 318}{5.3\times10^{-26}} = \frac{1.317\times10^{-20}}{5.3\times10^{-26}} = 2.484\times10^{5}\ \text{m}^2\text{s}^{-2}$

$c_{\text{rms}} = \sqrt{2.484\times10^{5}} = 498\ \text{m s}^{-1}$

So the temperature is about 318 K and the r.m.s. speed is about 498 m s⁻¹.
Why this scores
Why this scores 6/6. (a) $N=nN_A$ (M), $T=pV/(Nk)=318\ \text{K}$ (M+A); (b) $\langle c^2\rangle=3kT/m$ (M), take the square root ⇒ $498\ \text{m s}^{-1}$ (A). Carry the temperature from part (a) at full precision into part (b). The final square root is essential — dropping it leaves the mean-square speed, not the r.m.s. speed.

Key Formulae — Kinetic Theory and Ideal Gases

The formulae you need to memorise for kinetic theory and ideal gases on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Ideal gas equation
$pV = NkT$
$p$
pressure of the gas / Pa
$V$
volume of the gas / $\text{m}^3$
$N$
number of molecules (dimensionless)
$k$
Boltzmann constant $= 1.38\times10^{-23}\ \text{J K}^{-1}$
$T$
absolute (thermodynamic) temperature / K
When to use
Relating pressure, volume, number of molecules and temperature for an ideal gas. Temperature MUST be in kelvin. In the Edexcel formulae booklet. Equivalent to $pV=nRT$ with $N=nN_A$ and $R=N_A k$ .
Example
$N=3.0\times10^{23}$ , $V=2.0\times10^{-3}\ \text{m}^3$ , $T=300\ \text{K}$ : $p=NkT/V=6.21\times10^{5}\ \text{Pa}$ .
Boyle's law
$pV = \text{constant} \quad (\text{constant } T)$
$p$
pressure / Pa
$V$
volume / $\text{m}^3$
$T$
temperature (held constant) / K
When to use
A fixed mass of gas at constant temperature: $p_1V_1 = p_2V_2$ . Plotting $p$ against $1/V$ gives a straight line through the origin. Only valid at constant temperature.
Example
$500\ \text{cm}^3$ at $1.0\times10^{5}\ \text{Pa}$ compressed to $200\ \text{cm}^3$ : $p_2=1.0\times10^{5}\times500/200=2.5\times10^{5}\ \text{Pa}$ .
Mean translational kinetic energy of a molecule
$\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$
$m$
mass of one molecule / kg
$\langle c^2\rangle$
mean-square speed of the molecules / $\text{m}^2\text{s}^{-2}$
$k$
Boltzmann constant / $\text{J K}^{-1}$
$T$
absolute temperature / K
When to use
Linking absolute temperature to the mean kinetic energy of a single molecule. In the Edexcel formulae booklet. The mean KE depends only on $T$ — not on the type of gas.
Example
At $T=350\ \text{K}$ : mean KE $=\tfrac32 kT=\tfrac32\times1.38\times10^{-23}\times350=7.25\times10^{-21}\ \text{J}$ .
Root-mean-square speed
$c_{\text{rms}} = \sqrt{\langle c^2\rangle} = \sqrt{\frac{3kT}{m}}$
$c_{\text{rms}}$
root-mean-square speed / $\text{m s}^{-1}$
$\langle c^2\rangle$
mean-square speed / $\text{m}^2\text{s}^{-2}$
$m$
mass of one molecule / kg
$k$
Boltzmann constant / $\text{J K}^{-1}$
$T$
absolute temperature / K
When to use
Finding the r.m.s. molecular speed by rearranging $\tfrac12 m\langle c^2\rangle=\tfrac32 kT$ to $\langle c^2\rangle=3kT/m$ , then taking the square root. Because $c_{\text{rms}}\propto\sqrt{T}$ , quadrupling $T$ doubles the r.m.s. speed.
Example
Nitrogen ( $m=4.65\times10^{-26}\ \text{kg}$ ) at $300\ \text{K}$ : $c_{\text{rms}}=\sqrt{3kT/m}=517\ \text{m s}^{-1}$ .
Molecules from moles
$N = n N_A$
$N$
number of molecules
$n$
amount of substance / mol
$N_A$
Avogadro constant $= 6.02\times10^{23}\ \text{mol}^{-1}$
When to use
Converting a number of moles to a number of molecules so that $pV=NkT$ can be used. Also links the two forms of the gas law: $R = N_A k = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ .
Example
$n=0.80\ \text{mol}$ : $N=0.80\times6.02\times10^{23}=4.82\times10^{23}$ molecules.

Key Definitions and Keywords — Kinetic Theory and Ideal Gases

Definitions to memorise and the exact keywords mark schemes credit for kinetic theory and ideal gases answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Ideal gas
Examiner keyword
A gas that obeys the equation $pV = NkT$ exactly at all pressures and temperatures. It is a model gas whose molecules have negligible volume and no intermolecular forces except during collisions. Real gases approximate this at low pressure and high temperature.
Ideal gas equation (pV = NkT)
Examiner keyword
The equation of state of an ideal gas: $pV = NkT$ , where $p$ is pressure (Pa), $V$ is volume ( $\text{m}^3$ ), $N$ is the number of molecules, $k$ is the Boltzmann constant and $T$ is the absolute temperature (K). Equivalent to $pV=nRT$ .
Boltzmann constant (k)
Examiner keyword
The constant $k = 1.38\times10^{-23}\ \text{J K}^{-1}$ linking the mean energy of a molecule to temperature. It equals the molar gas constant divided by the Avogadro constant, $k = R/N_A$ .
Boyle's law
Examiner keyword
For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume: $pV = \text{constant}$ , so $p_1V_1 = p_2V_2$ . A graph of $p$ against $1/V$ is a straight line through the origin.
Absolute (thermodynamic) temperature
Examiner keyword
Temperature measured on the kelvin scale, starting from absolute zero. Convert from Celsius by adding 273: $T/\text{K} = \theta/^{\circ}\text{C} + 273$ . All gas-law calculations must use kelvin.
Absolute zero
Examiner keyword
The lowest possible temperature, $0\ \text{K} = -273\ ^{\circ}\text{C}$ , at which the molecules of an ideal gas have their minimum internal (kinetic) energy and the pressure of an ideal gas would be zero.
Kinetic theory of gases
Examiner keyword
A model that explains the macroscopic behaviour of a gas (pressure, temperature) in terms of a large number of small molecules in continuous random motion, colliding elastically with each other and with the container walls.
Assumptions of an ideal gas
Examiner keyword
A large number of molecules in random motion; negligible molecular volume compared with the container; no intermolecular forces except during collisions; perfectly elastic collisions; the time of a collision is negligible compared with the time between collisions.
Mean-square speed (⟨c²⟩)
The average of the squares of the molecular speeds, $\langle c^2\rangle$ , with units $\text{m}^2\text{s}^{-2}$ . It appears in $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ ; taking its square root gives the r.m.s. speed.
Root-mean-square (r.m.s.) speed
Examiner keyword
The square root of the mean-square speed, $c_{\text{rms}} = \sqrt{\langle c^2\rangle} = \sqrt{3kT/m}$ . It is a representative molecular speed and is proportional to $\sqrt{T}$ .
Mean translational kinetic energy
Examiner keyword
The average kinetic energy of a gas molecule due to its motion, $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ . It depends only on the absolute temperature, not on the mass or type of molecule.
Avogadro constant (N_A)
The number of particles in one mole, $N_A = 6.02\times10^{23}\ \text{mol}^{-1}$ . Used to convert moles to molecules via $N = nN_A$ so that $pV=NkT$ can be applied.
Mole (amount of substance)
The SI unit for amount of substance. One mole contains $N_A = 6.02\times10^{23}$ particles. The number of molecules is $N = nN_A$ , where $n$ is the number of moles.
Pressure (of a gas)
The force per unit area exerted by the gas molecules on the container walls, arising from the rate of change of momentum as molecules collide with and rebound from the walls. Measured in pascals (Pa).
Elastic collision
A collision in which the total kinetic energy is conserved. In the kinetic theory, collisions of molecules with the walls (and with each other) are assumed perfectly elastic, so the gas does not lose kinetic energy over time.

Common Mistakes and Misconceptions — Kinetic Theory and Ideal Gases

The traps other students keep falling into on kinetic theory and ideal gases questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using temperature in degrees Celsius instead of kelvin in gas-law calculations
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Temperatures are usually quoted in the question in °C, and students substitute the Celsius number straight into $pV=NkT$ or a ratio without converting.
How to avoid it
ALWAYS convert to kelvin first: $T/\text{K} = \theta/^{\circ}\text{C} + 273$ . The gas laws use absolute temperature — $p$ and $V$ are proportional to $T$ only in kelvin. Using 27 instead of 300 K, for example, makes a ratio ten times wrong.
✕Forgetting the final square root when finding the r.m.s. speed (quoting ⟨c²⟩ as the speed)
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Rearranging $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ gives $\langle c^2\rangle$ , and students stop there, forgetting that this is a mean-square speed, not a speed.
How to avoid it
$\langle c^2\rangle$ has units $\text{m}^2\text{s}^{-2}$ — a speed squared. Take the square root: $c_{\text{rms}} = \sqrt{\langle c^2\rangle} = \sqrt{3kT/m}$ . If your 'speed' looks like $\sim10^5$ instead of $\sim10^2$ – $10^3$ , you forgot the root.
✕Thinking the mean kinetic energy depends on the type or mass of the gas molecule
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Students associate 'more massive' with 'more energy' and assume a heavier molecule at the same temperature has a greater mean kinetic energy.
How to avoid it
The mean translational kinetic energy is $\tfrac12 m\langle c^2\rangle = \tfrac32 kT$ — it depends only on the absolute temperature. At the same $T$ , all molecules have the same mean KE; the heavier ones simply move more slowly (smaller r.m.s. speed).
✕Applying $p \propto 1/V$ (Boyle's law) when the temperature is not constant
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Boyle's law is memorised as 'pressure and volume are inversely proportional', dropping the essential condition that the temperature (and mass of gas) must be constant.
How to avoid it
$p \propto 1/V$ holds only at constant temperature for a fixed mass of gas. If $T$ changes too, use the full relation $\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}$ . In CP14 the gas is compressed slowly precisely to keep $T$ constant.
✕Giving vague or wrong assumptions of an ideal gas (e.g. 'molecules are small', 'no collisions')
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Students half-remember the kinetic-theory assumptions and state them imprecisely, or confuse 'negligible volume' with 'no collisions'.
How to avoid it
Be precise: molecular volume is negligible compared with the container; there are no intermolecular forces except during collisions; collisions are perfectly elastic; the collision time is negligible compared with the time between collisions; a large number of molecules in random motion. Molecules DO collide (that is what causes pressure).
✕Confusing the number of molecules N with the number of moles n in pV = NkT
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
The Edexcel form uses $N$ (molecules) with $k$ , whereas the more familiar $pV=nRT$ uses $n$ (moles) with $R$ ; students mix the two.
How to avoid it
In $pV=NkT$ , $N$ is the number of molecules and $k$ is the Boltzmann constant. If given moles $n$ , first convert: $N = nN_A$ . Pairing $n$ with $k$ (or $N$ with $R$ ) gives an answer wrong by a factor of $N_A \approx 6\times10^{23}$ .

Kinetic Theory and Ideal Gases — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Kinetic Theory and Ideal Gases

Short Notes - Kinetic Theory and Ideal Gases

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Converting a temperature to kelvin (2 marks)

2Pressure of a gas from pV = NkT (2 marks)

3Boyle's law at constant temperature (2 marks)

4Pressure from a number of moles (4 marks)

5Mean kinetic energy of a molecule (3 marks)

6Root-mean-square speed of a molecule (4 marks)

7Finding the temperature to double the r.m.s. speed (5 marks)

8Combined gas behaviour with changing p, V and T (6 marks)

9Core Practical 14 — verifying Boyle's law from a graph (6 marks)

Ideal gas equation

Boyle's law

Mean translational kinetic energy of a molecule

Root-mean-square speed

Molecules from moles

Ideal gas

Ideal gas equation (pV = NkT)

Boltzmann constant (k)

Boyle's law

Absolute (thermodynamic) temperature

Absolute zero

Kinetic theory of gases

Assumptions of an ideal gas

Mean-square speed (⟨c²⟩)

Root-mean-square (r.m.s.) speed

Mean translational kinetic energy

Avogadro constant (N_A)

Mole (amount of substance)

Pressure (of a gas)

Elastic collision

✕Using temperature in degrees Celsius instead of kelvin in gas-law calculations

✕Forgetting the final square root when finding the r.m.s. speed (quoting ⟨c²⟩ as the speed)

✕Thinking the mean kinetic energy depends on the type or mass of the gas molecule

✕Applying p∝1/Vp \propto 1/Vp∝1/V (Boyle's law) when the temperature is not constant

✕Giving vague or wrong assumptions of an ideal gas (e.g. 'molecules are small', 'no collisions')

✕Confusing the number of molecules N with the number of moles n in pV = NkT

Kinetic Theory and Ideal Gases — frequently asked questions

✕Applying $p \propto 1/V$ (Boyle's law) when the temperature is not constant