Kinetic Theory of Gases

The gas laws describe what gases do — how pressure, volume, and temperature are related. Kinetic theory explains why they behave this way, by modelling a gas as a collection of tiny particles (molecules) moving randomly and colliding with each other and the walls of their container.

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Assumptions of the Kinetic Theory Model

The kinetic theory of gases is built on a set of simplifying assumptions. These are crucial to understand and are frequently examined:

1. A gas contains a very large number of identical molecules. The sheer number (on the order of 10²³ or more) ensures that statistical averages are meaningful.

2. The molecules are in constant, random motion. They move in all directions with a range of speeds. There is no preferred direction of motion.

3. Collisions between molecules, and between molecules and the container walls, are perfectly elastic. This means kinetic energy is conserved in every collision. No energy is lost to deformation, heat, or sound.

4. The volume of the molecules themselves is negligible compared to the volume of the container. The molecules are treated as point particles — almost all the space in the container is empty.

5. There are no intermolecular forces between molecules except during collisions. Between collisions, molecules travel in straight lines at constant speed (Newton's first law). There are no attractive or repulsive forces acting at a distance.

6. The duration of a collision is negligible compared to the time between collisions. Molecules spend almost all their time in free flight, not in contact with other molecules.

These assumptions define an ideal gas. Real gases deviate from these assumptions, particularly at high pressures (where molecular volume matters) and low temperatures (where intermolecular forces become significant).

graph TD
    A["Kinetic Theory\nAssumptions"] --> B["1. Very large number\nof identical molecules"]
    A --> C["2. Constant random\nmotion"]
    A --> D["3. Elastic\ncollisions"]
    A --> E["4. Negligible\nmolecular volume"]
    A --> F["5. No intermolecular\nforces"]
    A --> G["6. Negligible\ncollision duration"]
    B --> H["Statistical averages\nare meaningful"]
    C --> I["No preferred\ndirection"]
    D --> J["Total KE\nconserved"]
    E --> K["Molecules are\npoint particles"]
    F --> L["Straight-line motion\nbetween collisions"]
    G --> M["Most time spent\nin free flight"]

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Deriving pV = (1/3)Nm<c²> — Qualitative Understanding

The Edexcel specification requires a qualitative understanding of how kinetic theory leads to the expression for pressure. Here is the argument:

Step 1: A single molecule bouncing between walls.

Consider a single molecule of mass m moving with velocity c in the x-direction inside a cubic box of side length L. It bounces back and forth between two opposite walls.

Step 2: Momentum change at the wall.

Each time the molecule hits a wall, its momentum changes from +mc to −mc (or vice versa). The change in momentum per collision is:

Δp = 2mc

Step 3: Time between collisions with the same wall.

The molecule must travel a distance of 2L (to the opposite wall and back) before hitting the same wall again. The time between collisions with one wall is:

t = 2L/c

Step 4: Force from one molecule.

Force = rate of change of momentum = Δp/t = 2mc/(2L/c) = mc²/L

Step 5: Pressure from one molecule.

Pressure = Force/Area. The area of the wall is L². So:

p = mc²/(L × L²) = mc²/L³ = mc²/V (since L³ = V, the volume of the box)

Step 6: Extending to N molecules moving in three dimensions.

Not all molecules move in the x-direction. In three dimensions, on average, only one-third of the molecular kinetic energy is associated with motion in any one direction (x, y, or z). Also, different molecules have different speeds, so we use the mean square speed <c²>.

For N molecules:

pV = (1/3)Nm<c²>

where:

• p = pressure (Pa)
• V = volume (m³)
• N = total number of molecules
• m = mass of one molecule (kg)
• <c²> = mean square speed (m² s⁻²)

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Why the Factor of 1/3?

This is frequently examined. The factor arises because molecular motion is equally distributed among three perpendicular directions (x, y, z). If we define the speed c of a molecule as:

c² = v_x² + v_y² + v_z²

Then for random motion, on average:

<v_x²> = <v_y²> = <v_z²> = (1/3)<c²>

Only the x-component of velocity contributes to pressure on walls perpendicular to the x-axis. Since this is one-third of the total, the factor of 1/3 appears.

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Mean Square Speed and RMS Speed

The mean square speed <c²> is the average of the squares of the speeds of all molecules:

<c²> = (c₁² + c₂² + c₃² + ... + cₙ²) / N

This is different from the square of the mean speed. Squaring first and then averaging gives greater weight to faster molecules.

The root mean square (rms) speed is:

c_rms = √<c²>

This is the single speed value most commonly used to represent the "typical" speed of molecules in kinetic theory calculations.

Worked calculation of rms speed from a data set:

Five molecules have speeds of 200, 400, 500, 600, and 800 m s⁻¹.

Mean speed = (200 + 400 + 500 + 600 + 800)/5 = 2500/5 = 500 m s⁻¹

<c²> = (200² + 400² + 500² + 600² + 800²)/5 = (40 000 + 160 000 + 250 000 + 360 000 + 640 000)/5 = 1 450 000/5 = 290 000 m² s⁻²

c_rms = √290 000 = 539 m s⁻¹

Notice that c_rms (539 m s⁻¹) is greater than the mean speed (500 m s⁻¹). This is always the case because squaring gives extra weight to the faster molecules.

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Connecting to the Ideal Gas Equation

We now have two expressions involving pV:

From kinetic theory: pV = (1/3)Nm<c²> From the ideal gas equation: pV = NkT

Setting them equal:

(1/3)Nm<c²> = NkT

The N cancels:

(1/3)m<c²> = kT

This is the bridge between the microscopic (molecular) world and the macroscopic (measurable) world. It shows that temperature is directly related to the mean kinetic energy of the molecules — a profound result.

Rearranging:

<c²> = 3kT/m

This means: at a given temperature, lighter molecules have greater mean square speed (and hence greater rms speed). Heavier molecules move slower.

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Worked Example 1: RMS Speed of Nitrogen

Calculate the rms speed of nitrogen molecules (N₂) at 300 K. (Molar mass of N₂ = 0.028 kg mol⁻¹, Nₐ = 6.02 × 10²³, k = 1.38 × 10⁻²³ J K⁻¹)

First, find the mass of one molecule: m = M/Nₐ = 0.028/(6.02 × 10²³) = 4.65 × 10⁻²⁶ kg

Using <c²> = 3kT/m:

<c²> = (3 × 1.38 × 10⁻²³ × 300) / (4.65 × 10⁻²⁶)

<c²> = (1.242 × 10⁻²⁰) / (4.65 × 10⁻²⁶) = 2.67 × 10⁵ m² s⁻²

c_rms = √(2.67 × 10⁵) = 517 m s⁻¹

Nitrogen molecules at room temperature move at roughly 500 m s⁻¹ — faster than the speed of sound in air (about 340 m s⁻¹).

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Worked Example 2: Using the Molar Form

Calculate the rms speed of helium atoms at 20 °C. (M = 0.004 kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹)

Using the molar form: c_rms = √(3RT/M)

T = 20 + 273 = 293 K

c_rms = √(3 × 8.31 × 293 / 0.004) = √(7304.5 / 0.004) = √(1 826 130)

c_rms = 1352 m s⁻¹

Helium atoms at room temperature move at about 1350 m s⁻¹ — much faster than nitrogen because helium is much lighter.

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Worked Example 3: Pressure from Molecular Data

A container of volume 0.050 m³ holds 4.0 × 10²⁴ oxygen molecules (mass of one molecule = 5.3 × 10⁻²⁶ kg). The rms speed of the molecules is 450 m s⁻¹. Calculate the pressure.

pV = (1/3)Nm<c²>

p = (1/3) × N × m × <c²> / V

<c²> = c_rms² = 450² = 202 500 m² s⁻²

p = (1/3) × 4.0 × 10²⁴ × 5.3 × 10⁻²⁶ × 202 500 / 0.050

Numerator: (1/3) × 4.0 × 10²⁴ × 5.3 × 10⁻²⁶ × 202 500 = (1/3) × 42.93 = 14.31

p = 14.31 / 0.050 = 286 000 Pa ≈ 2.9 × 10⁵ Pa

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RMS Speeds of Common Gases at 20 °C (293 K)

Gas	M / kg mol⁻¹	c_rms / m s⁻¹
H₂	0.002	1905
He	0.004	1352
H₂O	0.018	637
N₂	0.028	511
O₂	0.032	478
Ar	0.040	428
CO₂	0.044	408

Notice the trend: lighter molecules have significantly higher rms speeds. This has practical consequences — hydrogen escapes from containers more easily, and helium balloons deflate faster than air-filled ones.

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Summary

• Kinetic theory models a gas as many tiny, fast-moving particles colliding elastically.
• The six key assumptions define an ideal gas.
• The derivation of pV = (1/3)Nm<c²> links molecular motion to macroscopic pressure.
• The 1/3 factor comes from three-dimensional random motion being split equally among x, y, z.
• Mean square speed <c²> and rms speed c_rms = √<c²> are key quantities.
• c_rms is always greater than the mean speed.
• Combining with pV = NkT gives (1/3)m<c²> = kT, linking temperature to molecular kinetic energy.
• Lighter molecules move faster at the same temperature: c_rms = √(3kT/m) = √(3RT/M).

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A-Level Deep Dive: Kinetic Theory of Gases

Spec mapping

Edexcel 9PH0 Topic 9 — Thermodynamics requires candidates to understand and apply the kinetic theory model of an ideal gas, including the six standard assumptions, the derivation of $pV = \tfrac{1}{3}Nm\langle c^2\rangle$pV=31​Nm⟨c2⟩ from first principles, the use of root-mean-square speed $c_{\text{rms}}$crms​, and the relationship $\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$21​m⟨c2⟩=23​kT linking microscopic kinetic energy to absolute temperature (refer to the official specification document for exact wording). Although this material is concentrated in Topic 9, it is examined synoptically throughout 9PH0-02. Kinetic theory underpins the ideal gas equation $pV = NkT = nRT$pV=NkT=nRT, energy concepts in Topic 6 (Further Mechanics — momentum, impulse and elastic collisions), the equipartition ideas hinted at in Topic 13 (Oscillations), and the molecular interpretation of internal energy used throughout the thermodynamics chapter. The Edexcel data and formulae booklet provides $pV = \tfrac{1}{3}Nm\langle c^2\rangle$pV=31​Nm⟨c2⟩ and $\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$21​m⟨c2⟩=23​kT — but candidates must still be able to derive the former from a one-molecule-in-a-box argument when "show that" or "explain how" appears in the question stem.

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Worked example with full mark scheme

Question (8 marks):

(a) A cubical container of side $L$L contains $N$N identical molecules each of mass $m$m. By considering a single molecule's elastic collisions with one wall and then generalising, derive the result $pV = \tfrac{1}{3}Nm\langle c^2\rangle$pV=31​Nm⟨c2⟩, stating clearly any assumptions used. (6)

(b) Hence calculate the root-mean-square speed of nitrogen molecules ($m = 4.65 \times 10^{-26}\ \text{kg}$m=4.65×10−26 kg) at $T = 300\ \text{K}$T=300 K. Take $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$k=1.38×10−23 J K−1. (2)

Solution with mark scheme:

(a) Step 1 — single molecule, single direction. Consider one molecule of velocity components $(u, v, w)$(u,v,w). On hitting the wall perpendicular to the $x$x-axis it elastically reverses $u \to -u$u→−u. Change of momentum: $\Delta p = 2mu$Δp=2mu.

M1 — correct change-of-momentum statement (factor of 2 essential; "elastic" must be stated or used).

Step 2 — frequency of collisions. Time between successive collisions with the same wall: $\Delta t = 2L/u$Δt=2L/u. Force on the wall from this molecule: $F_1 = \Delta p / \Delta t = 2mu / (2L/u) = mu^2 / L$F1​=Δp/Δt=2mu/(2L/u)=mu2/L.

M1 — correct expression for force per molecule. Common mistake: writing $\Delta t = L/u$Δt=L/u (forgetting the round trip), which doubles the answer.

Step 3 — sum over $N$N molecules. Total force in the $x$x-direction: $F_x = \dfrac{m}{L}\sum_{i=1}^{N} u_i^2 = \dfrac{Nm\langle u^2\rangle}{L}$Fx​=Lm​∑i=1N​ui2​=LNm⟨u2⟩​, where $\langle u^2\rangle$⟨u2⟩ is the mean of $u_i^2$ui2​.

M1 — recognising the sum becomes $N\langle u^2\rangle$N⟨u2⟩ by definition of the mean.

Step 4 — three-dimensional isotropy. By symmetry $\langle u^2\rangle = \langle v^2\rangle = \langle w^2\rangle$⟨u2⟩=⟨v2⟩=⟨w2⟩, and $\langle c^2\rangle = \langle u^2\rangle + \langle v^2\rangle + \langle w^2\rangle = 3\langle u^2\rangle$⟨c2⟩=⟨u2⟩+⟨v2⟩+⟨w2⟩=3⟨u2⟩, so $\langle u^2\rangle = \tfrac{1}{3}\langle c^2\rangle$⟨u2⟩=31​⟨c2⟩.

M1 — explicit appeal to isotropy and the factor $\tfrac{1}{3}$31​.

Step 5 — convert force to pressure. Pressure $p = F_x / A = F_x / L^2$p=Fx​/A=Fx​/L2, and volume $V = L^3$V=L3:

$p = \dfrac{Nm\langle u^2\rangle}{L \cdot L^2} = \dfrac{Nm \cdot \tfrac{1}{3}\langle c^2\rangle}{V}$p=L⋅L2Nm⟨u2⟩​=VNm⋅31​⟨c2⟩​

so $pV = \tfrac{1}{3}Nm\langle c^2\rangle$pV=31​Nm⟨c2⟩.

A1 — correct printed result with working consistent.

A1 — at least three of the standard assumptions cited (e.g. "molecules are point particles of negligible volume; collisions with walls and each other are elastic; intermolecular forces are negligible except during collisions; large $N$N so statistical averages are well-defined").

(b) Equate $\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$21​m⟨c2⟩=23​kT, so $\langle c^2\rangle = 3kT/m$⟨c2⟩=3kT/m and $c_{\text{rms}} = \sqrt{3kT/m}$crms​=3kT/m​.

M1 — correct rearrangement.

$c_{\text{rms}} = \sqrt{\dfrac{3 \times 1.38 \times 10^{-23} \times 300}{4.65 \times 10^{-26}}} = \sqrt{2.671 \times 10^{5}} \approx 517\ \text{m s}^{-1}$crms​=4.65×10−263×1.38×10−23×300​​=2.671×105​≈517 m s−1

A1 — answer in the range 510–520 m s$^{-1}$−1, units stated.

Total: 8 marks (M5 A3 as shown).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A sealed flask of volume $2.5 \times 10^{-3}\ \text{m}^3$2.5×10−3 m3 contains $5.0 \times 10^{22}$5.0×1022 helium atoms ($m = 6.64 \times 10^{-27}\ \text{kg}$m=6.64×10−27 kg) at $T = 290\ \text{K}$T=290 K.

(a) Calculate the pressure inside the flask using the kinetic theory result. (3)

(b) Explain, in molecular terms, what happens to the pressure if the temperature is doubled at constant volume. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO2) — using $\langle c^2\rangle = 3kT/m = 3(1.38\times10^{-23})(290)/(6.64\times10^{-27}) \approx 1.808 \times 10^{6}\ \text{m}^2\text{s}^{-2}$⟨c2⟩=3kT/m=3(1.38×10−23)(290)/(6.64×10−27)≈1.808×106 m2s−2.
• M1 (AO2) — substituting into $p = \tfrac{1}{3}(N/V)m\langle c^2\rangle$p=31​(N/V)m⟨c2⟩.
• A1 (AO2) — $p \approx 8.0 \times 10^{4}\ \text{Pa}$p≈8.0×104 Pa (allow 7.9–8.1; equivalent route via $pV = NkT$pV=NkT gives the same value to within rounding).

(b)

• B1 (AO1) — molecules gain kinetic energy / move faster on average (since $\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT$21​m⟨c2⟩=23​kT, so $\langle c^2\rangle$⟨c2⟩ doubles).
• B1 (AO1) — collisions with the walls are more frequent and impart greater impulse per collision.
• B1 (AO2) — therefore pressure doubles (consistent with $pV = NkT$pV=NkT at fixed $V, N$V,N).

Total: 6 marks split AO1 = 2, AO2 = 4. Edexcel typically pairs a numerical kinetic-theory part with a molecular-explanation part. The explanation must mention both collision frequency and momentum change per collision — naming only one is a single B1, not the full pair.

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