Pressure from Molecular Bombardment

Spec mapping: OCR H556 Module 5.2 — Kinetic theory of gases (derivation of $pV = \tfrac{1}{3} N m \langle c^{2} \rangle$pV=31​Nm⟨c2⟩ from a cubic-box geometry with elastic molecular collisions and random isotropic motion; the factor of $\tfrac{1}{3}$31​ as a consequence of isotropy in three dimensions; the root-mean-square speed $c_{\text{rms}}$crms​). Refer to the official OCR H556 specification document for exact wording.

Lesson 8 stated the six assumptions of the kinetic theory of an ideal gas. We now put them to work to derive, from Newton's second law applied to a single molecule, the celebrated pressure equation:

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

This formula relates the macroscopic quantities $p$p and $V$V (which you can read off a pressure gauge and a measuring cylinder) to the microscopic quantities $N$N (number of molecules), $m$m (mass of one molecule), and $\langle c^{2} \rangle$⟨c2⟩ (the mean-square speed of the molecules — see below). It is the microscopic equation of state, in contrast with the macroscopic ideal gas equation $pV = NkT$pV=NkT from Lesson 7; combining the two in Lesson 10 reveals the deep connection between temperature and molecular kinetic energy.

OCR H556 examiners expect students to be able to derive this formula. The derivation is a classic piece of physics that bundles together Newton's second law, momentum change in an elastic collision, kinematics on a closed trajectory, isotropic random motion in three dimensions, and statistical averaging. The factor of $\tfrac{1}{3}$31​ in particular — sitting at the heart of the formula — comes from the isotropy of the motion (the mean-square components $\langle u^{2} \rangle$⟨u2⟩, $\langle v^{2} \rangle$⟨v2⟩, $\langle w^{2} \rangle$⟨w2⟩ are equal, so each is one third of $\langle c^{2} \rangle$⟨c2⟩). Lose that step and the derivation collapses; OCR mark schemes routinely test whether candidates can explain it.

This lesson walks through the derivation in six numbered steps, with a worked-example check at the end, and ends with the contract sections (synoptic links, specimen question, mark-loss patterns, going further, A-Level misconceptions, summary) that we use across the H556 platform.

---

The Strategy

We will consider a single molecule in a cubic box, compute the momentum change per unit time it imposes on one wall (which by Newton's second law gives the force), and then multiply up to get the total force due to all molecules. Dividing by wall area gives pressure. Finally, we multiply by volume to get pV on the left-hand side.

The key assumptions we rely on, from Lesson 8:

• Random motion in all directions (assumption 2).
• Elastic collisions with the walls (assumption 4).
• Short collision time compared to the time between collisions (assumption 5).
• No intermolecular forces between collisions (assumption 6).
• Many molecules, so averages are meaningful (assumption 1).

---

Setting Up the Problem

Imagine a cubic container of side length L, so the volume is V = L³. Inside are N identical molecules, each of mass m. One of these molecules moves with velocity (u₁, v₁, w₁) — i.e. with velocity components u₁ in the x-direction, v₁ in the y-direction, w₁ in the z-direction.

We will focus on one particular wall — let us say the right-hand wall, perpendicular to the x-axis. This wall has area A = L².

graph TD
    M[Molecule with velocity u along x]
    L1[Left wall] -- u --- M
    M -- u --- R["Right wall<br/>area L²"]
    M -. bounces elastically .- R

---

Step 1: Momentum Change per Collision

The molecule travels across the box in the x-direction. When it hits the right-hand wall, the x-component of its velocity reverses (because the collision is elastic and the wall is much more massive than the molecule). Its new velocity is (-u₁, v₁, w₁): the y and z components are unchanged, but u₁ → -u₁.

The change in the molecule's x-momentum is therefore

$$\Delta p_x = m(-u_{1}) - m(u_{1}) = -2 m u_{1}$$Δpx​=m(−u1​)−m(u1​)=−2mu1​

By Newton's third law, the wall receives an equal and opposite impulse: the wall's x-momentum increases by +2 m u₁ for each collision with this molecule.

---

Step 2: Time Between Successive Collisions

How often does this molecule hit the right-hand wall?

After bouncing off the right wall, the molecule travels at speed u₁ across the box, hits the left wall at distance L away, bounces back, and returns to the right wall after travelling a total distance 2L. (We ignore collisions with the top, bottom, front and back walls, because those affect only the y- and z-components of velocity, not the x-component.)

The time between successive collisions of this molecule with the right-hand wall is therefore

$$\Delta t = 2L / u_{1}$$Δt=2L/u1​

And the number of collisions per unit time is u₁/(2L).

---

Step 3: Force on the Wall from One Molecule

The average force on the right-hand wall due to this single molecule is the momentum transfer per collision times the number of collisions per unit time:

$$\begin{aligned} F_{1} &= (2 m u_{1}) \times (u_{1} / (2L)) \\ &= m u_{1}^{2} / L \end{aligned}$$F1​​=(2mu1​)×(u1​/(2L))=mu12​/L​

So a single molecule with x-velocity u₁ exerts a (time-averaged) force of m u₁² / L on the right-hand wall.

---

Step 4: Total Force from All Molecules

Now sum over all N molecules. Each molecule i has its own x-velocity component u_i, and contributes a force m u_i² / L to the right wall. The total force is

$$\begin{aligned} F &= \sum_{i=1}^{N} (m u_i^{2} / L) \\ &= (m/L) \sum_{i=1}^{N} u_i^{2} \end{aligned}$$F​=i=1∑N​(mui2​/L)=(m/L)i=1∑N​ui2​​

We now introduce the mean-square velocity component in the x-direction, $\langle u^{2} \rangle$⟨u2⟩, defined as

$$\langle u^{2} \rangle = (1/N) \sum_{i=1}^{N} u_i^{2}$$⟨u2⟩=(1/N)i=1∑N​ui2​

In words: $\langle u^{2} \rangle$⟨u2⟩ is the average value of $u^{2}$u2 over all $N$N molecules. Then $\sum_i u_i^{2} = N \langle u^{2} \rangle$∑i​ui2​=N⟨u2⟩, and

$$F = (m/L) \times N \langle u^{2} \rangle = N m \langle u^{2} \rangle / L$$F=(m/L)×N⟨u2⟩=Nm⟨u2⟩/L

---

Step 5: Pressure from the Force

The pressure on the right-hand wall is the force per unit area:

$$\begin{aligned} p &= F/A = F/L^{2} \\ &= (N m \langle u^{2} \rangle / L) / L^{2} \\ &= N m \langle u^{2} \rangle / L^{3} \end{aligned}$$p​=F/A=F/L2=(Nm⟨u2⟩/L)/L2=Nm⟨u2⟩/L3​

Since $L^{3} = V$L3=V (the volume of the cube), we can write

$$p V = N m \langle u^{2} \rangle$$pV=Nm⟨u2⟩

That is our result so far: the pressure on one wall, expressed in terms of the mean square of the x-component of velocity.

---

Step 6: From $\langle u^{2} \rangle$⟨u2⟩ to $\langle c^{2} \rangle$⟨c2⟩

We now invoke the randomness of the motion. Each molecule has a speed $c_i$ci​ with components $(u_i, v_i, w_i)$(ui​,vi​,wi​), so by Pythagoras,

$$c_i^{2} = u_i^{2} + v_i^{2} + w_i^{2}$$ci2​=ui2​+vi2​+wi2​

Averaging over all molecules,

$$\langle c^{2} \rangle = \langle u^{2} \rangle + \langle v^{2} \rangle + \langle w^{2} \rangle$$⟨c2⟩=⟨u2⟩+⟨v2⟩+⟨w2⟩

where $\langle c^{2} \rangle$⟨c2⟩ is the mean-square speed, $\langle v^{2} \rangle$⟨v2⟩ is the mean-square y-velocity component, and so on.

Now comes the crucial assumption of isotropy: because the motion is random, there is no preferred direction. On average, the x, y, and z components of velocity-squared are the same:

$$\langle u^{2} \rangle = \langle v^{2} \rangle = \langle w^{2} \rangle$$⟨u2⟩=⟨v2⟩=⟨w2⟩

Therefore

$$\langle c^{2} \rangle = 3 \langle u^{2} \rangle$$⟨c2⟩=3⟨u2⟩

or equivalently

$$\langle u^{2} \rangle = \tfrac{1}{3} \langle c^{2} \rangle$$⟨u2⟩=31​⟨c2⟩

Substituting back into $pV = N m \langle u^{2} \rangle$pV=Nm⟨u2⟩:

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

This is the pressure equation of the kinetic theory of gases. It holds for any ideal gas in any container — the cubical shape was just a convenient choice for the derivation.

---

The Result Unpacked

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

Let us be clear about every symbol:

Symbol	Meaning	Units
$p$p	Pressure of the gas	Pa ($=$= N m$^{-2}$−2)
$V$V	Volume of the container	m$^{3}$3
$N$N	Total number of molecules	(dimensionless)
$m$m	Mass of one molecule	kg
$\langle c^{2} \rangle$⟨c2⟩	Mean-square speed of the molecules	m$^{2}$2 s$^{-2}$−2

Note the factor of $\tfrac{1}{3}$31​. It comes from the randomness of motion in three dimensions, specifically from the step $\langle u^{2} \rangle = \tfrac{1}{3} \langle c^{2} \rangle$⟨u2⟩=31​⟨c2⟩. In two dimensions (a gas confined to a plane) the factor would be $\tfrac{1}{2}$21​. In one dimension it would be $1$1. The factor of $\tfrac{1}{3}$31​ is the signature of three-dimensional isotropic motion — and OCR mark schemes specifically test whether candidates can identify this origin.

---

Alternative Forms of the Pressure Equation

Several equivalent forms of the same equation appear in textbooks and on data sheets:

Form 1 (with molecule count and per-molecule mass):

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

This is the form given on the OCR data sheet.

Form 2 (with density):

$$p = \tfrac{1}{3} \rho \langle c^{2} \rangle$$p=31​ρ⟨c2⟩

where $\rho = Nm/V$ρ=Nm/V is the mass density of the gas. This is useful because density is often more directly measurable than $N$N.

Form 3 (with total gas mass):

$$pV = \tfrac{1}{3} M_{\text{total}} \langle c^{2} \rangle$$pV=31​Mtotal​⟨c2⟩

where M_total = Nm is the total mass of the gas in the container.

All three are the same equation in different guises. You should be able to convert between them fluently.

---

Root-Mean-Square Speed

The quantity $\langle c^{2} \rangle$⟨c2⟩ is called the mean-square speed. Its square root is the root-mean-square speed:

$$c_{\text{rms}} = \sqrt{\langle c^{2} \rangle}$$crms​=⟨c2⟩​

We can then write the pressure equation as

$$pV = \tfrac{1}{3} N m (c_{\text{rms}})^{2}$$pV=31​Nm(crms​)2

The root-mean-square speed is not the same as the mean speed $\langle c \rangle$⟨c⟩. It is slightly larger, because the averaging is weighted toward the higher speeds: $\langle c^{2} \rangle \geq \langle c \rangle^{2}$⟨c2⟩≥⟨c⟩2 always (by the Cauchy-Schwarz inequality), with equality only if all molecules have exactly the same speed. For a Maxwell-Boltzmann distribution at thermal equilibrium, $c_{\text{rms}}$crms​ is about $8.5\%$8.5% larger than $\langle c \rangle$⟨c⟩. We use $c_{\text{rms}}$crms​ rather than $\langle c \rangle$⟨c⟩ in the pressure equation because it is directly related to the per-molecule kinetic energy: $\tfrac{1}{2} m \langle c^{2} \rangle = \tfrac{1}{2} m (c_{\text{rms}})^{2}$21​m⟨c2⟩=21​m(crms​)2.

---

Worked Example 1: Using the Pressure Equation

A gas has density $\rho = 1.2$ρ=1.2 kg m$^{-3}$−3 and is at atmospheric pressure $p = 1.01 \times 10^{5}$p=1.01×105 Pa. Find the rms speed of its molecules.

Using $p = \tfrac{1}{3} \rho \langle c^{2} \rangle$p=31​ρ⟨c2⟩:

$$\begin{aligned} \langle c^{2} \rangle &= 3 p / \rho \\ &= (3)(1.01 \times 10^{5}) / 1.2 \\ &= 2.525 \times 10^{5}\,\text{m}^{2}\,\text{s}^{-2} \\ c_{\text{rms}} &= \sqrt{\langle c^{2} \rangle} \approx 503\,\text{m s}^{-1} \end{aligned}$$⟨c2⟩crms​​=3p/ρ=(3)(1.01×105)/1.2=2.525×105m2s−2=⟨c2⟩​≈503m s−1​

So air molecules move at about 500 m s⁻¹ on average (root-mean-square). This is a staggering speed — faster than the speed of sound, in fact.

---

Worked Example 2: Mass of an Individual Molecule

A gas consists of $1.5 \times 10^{24}$1.5×1024 molecules in a container of volume $V = 0.050$V=0.050 m$^{3}$3 at pressure $p = 1.0 \times 10^{5}$p=1.0×105 Pa. The rms speed of the molecules is $c_{\text{rms}} = 500$crms​=500 m s$^{-1}$−1. What is the mass of one molecule?

Using $pV = \tfrac{1}{3} N m (c_{\text{rms}})^{2}$pV=31​Nm(crms​)2:

$$\begin{aligned} m &= 3 p V / (N (c_{\text{rms}})^{2}) \\ &= (3)(1.0 \times 10^{5})(0.050) / ((1.5 \times 10^{24})(500^{2})) \\ &= 1.5 \times 10^{4} / (1.5 \times 10^{24} \times 2.5 \times 10^{5}) \\ &= 1.5 \times 10^{4} / (3.75 \times 10^{29}) \\ &\approx 4.0 \times 10^{-26}\,\text{kg} \end{aligned}$$m​=3pV/(N(crms​)2)=(3)(1.0×105)(0.050)/((1.5×1024)(5002))=1.5×104/(1.5×1024×2.5×105)=1.5×104/(3.75×1029)≈4.0×10−26kg​

This is close to the mass of a nitrogen molecule (m ≈ 4.65 × 10⁻²⁶ kg), which is consistent with the gas being nitrogen (or possibly air, which is about 78% nitrogen).

---

Worked Example 3: Comparing Two Gases

A container holds oxygen at pressure p_1 and rms speed c_1. At the same temperature, what is the rms speed of hydrogen molecules in a separate identical container? Given that the molar mass of oxygen is 32 g mol⁻¹ and of hydrogen is 2 g mol⁻¹.

In Lesson 10 we will show that the average kinetic energy per molecule depends only on temperature:

$$\tfrac{1}{2} m \langle c^{2} \rangle = \tfrac{3}{2} k T$$21​m⟨c2⟩=23​kT

Since the temperature is the same for both gases,

$$\begin{aligned} \tfrac{1}{2} m_{O_2} \langle c_{O_2}^{2} \rangle &= \tfrac{1}{2} m_{H_2} \langle c_{H_2}^{2} \rangle \\ m_{O_2} \langle c_{O_2}^{2} \rangle &= m_{H_2} \langle c_{H_2}^{2} \rangle \\ \frac{\langle c_{H_2}^{2} \rangle}{\langle c_{O_2}^{2} \rangle} &= \frac{m_{O_2}}{m_{H_2}} = \frac{32}{2} = 16 \\ \frac{c_{H_2,\text{rms}}}{c_{O_2,\text{rms}}} &= \sqrt{16} = 4 \end{aligned}$$21​mO2​​⟨cO2​2​⟩mO2​​⟨cO2​2​⟩⟨cO2​2​⟩⟨cH2​2​⟩​cO2​,rms​cH2​,rms​​​=21​mH2​​⟨cH2​2​⟩=mH2​​⟨cH2​2​⟩=mH2​​mO2​​​=232​=16=16​=4​

So hydrogen molecules at the same temperature move about four times faster than oxygen molecules, in direct consequence of the 1:16 ratio of their masses.

---

Worked Example 4: Pressure Check at Room Conditions

Let us check that $pV = \tfrac{1}{3} N m \langle c^{2} \rangle$pV=31​Nm⟨c2⟩ gives a sensible answer at room conditions. Consider air at $T = 293$T=293 K, $p = 1.01 \times 10^{5}$p=1.01×105 Pa.

Take the average molar mass of air as $M = 0.029$M=0.029 kg mol$^{-1}$−1, so the mass of one "average" air molecule is $m = M/N_A = 0.029 / (6.02 \times 10^{23}) \approx 4.82 \times 10^{-26}$m=M/NA​=0.029/(6.02×1023)≈4.82×10−26 kg.

Number density $N/V = p/(kT)$N/V=p/(kT):

$$\begin{aligned} N/V &= (1.01 \times 10^{5}) / ((1.38 \times 10^{-23})(293)) \\ &\approx 2.50 \times 10^{25}\,\text{m}^{-3} \end{aligned}$$N/V​=(1.01×105)/((1.38×10−23)(293))≈2.50×1025m−3​

Now use $p = \tfrac{1}{3}(N/V) m \langle c^{2} \rangle$p=31​(N/V)m⟨c2⟩:

$$\begin{aligned} \langle c^{2} \rangle &= 3 p / ((N/V) m) \\ &= (3)(1.01 \times 10^{5}) / ((2.50 \times 10^{25})(4.82 \times 10^{-26})) \\ &= 3.03 \times 10^{5} / 1.21 \\ &\approx 2.51 \times 10^{5}\,\text{m}^{2}\,\text{s}^{-2} \\ c_{\text{rms}} &\approx 501\,\text{m s}^{-1} \end{aligned}$$⟨c2⟩crms​​=3p/((N/V)m)=(3)(1.01×105)/((2.50×1025)(4.82×10−26))=3.03×105/1.21≈2.51×105m2s−2≈501m s−1​

The same answer as Worked Example 1. The pressure equation is internally consistent with $pV = NkT$pV=NkT and with the lemma $\tfrac{1}{2} m \langle c^{2} \rangle = \tfrac{3}{2} k T$21​m⟨c2⟩=23​kT that we will derive in Lesson 10.

---

Why This Is Important