The Boltzmann Constant

Spec mapping: OCR H556 Module 5.2 — Ideal gases (the Boltzmann constant $k$k as a fundamental molecular-scale constant; the relationship $k = R/N_A$k=R/NA​; the molecular form $pV = NkT$pV=NkT of the equation of state; the Avogadro constant $N_A$NA​). Refer to the official OCR H556 specification document for exact wording.

Lesson 6 wrote the equation of state of an ideal gas as $pV = nRT$pV=nRT, working in terms of the number of moles $n$n and the molar gas constant $R = 8.31$R=8.31 J mol$^{-1}$−1 K$^{-1}$−1. The form is excellent for bulk-scale problems where mass and moles are the natural quantities. As soon as you want to think about individual molecules, however — their speeds, their kinetic energies, their bombardment of container walls — counting in moles becomes inconvenient. The mole is, after all, just a counting unit ($1$1 mol = $6.022 \times 10^{23}$6.022×1023 entities, by the modern SI definition), so dividing through by Avogadro's number converts macroscopic moles into microscopic counts of individual particles.

That conversion is what the Boltzmann constant accomplishes. Defined as

$k = \frac{R}{N_A}$k=NA​R​

the Boltzmann constant has the value $1.38 \times 10^{-23}$1.38×10−23 J K$^{-1}$−1 and lets us rewrite the ideal gas equation in its molecular form:

$pV = NkT$pV=NkT

with $N$N the total number of molecules. The Boltzmann constant is, in a deep sense, the per-molecule analogue of the molar gas constant. Where $R$R tells you how much $pV$pV you get per mole per kelvin, $k$k tells you how much $pV$pV you get per molecule per kelvin. The number $kT$kT itself — about $4 \times 10^{-21}$4×10−21 J at room temperature — is the natural energy unit of thermal physics, the energy scale of typical molecular motion.

This lesson develops the Boltzmann constant carefully, derives its value from $R$R and $N_A$NA​, works seven numerical examples spanning rooms, balloons, atmospheres and partial pressures, and ends with the misconceptions H556 examiners watch for in mole-vs-molecule confusion.

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From Moles to Molecules

Recall that one mole contains N_A = 6.02 × 10²³ molecules — Avogadro's number. If a gas contains n moles, then it contains

$$N = n N_A$$N=nNA​

molecules in total. Substituting into pV = nRT:

$$\begin{aligned} pV &= (N/N_A) R T \\ &= N \times (R/N_A) \times T \end{aligned}$$pV​=(N/NA​)RT=N×(R/NA​)×T​

The combination R/N_A appears so often in physics that it has been given its own symbol: k (or sometimes k_B), the Boltzmann constant.

$$k = R / N_A$$k=R/NA​

With this definition, the ideal gas equation becomes

$$pV = NkT$$pV=NkT

which is a form of the same law, expressed in terms of individual molecules rather than moles.

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The Numerical Value of k

From R = 8.31 J mol⁻¹ K⁻¹ and N_A = 6.02 × 10²³ mol⁻¹:

$$\begin{aligned} k &= R / N_A \\ &= 8.31 / (6.02 \times 10^{23}) \\ &\approx 1.38 \times 10^{-23}\,\text{J K}^{-1} \end{aligned}$$k​=R/NA​=8.31/(6.02×1023)≈1.38×10−23J K−1​

So the Boltzmann constant is

$$k = 1.38 \times 10^{-23}\,\text{J K}^{-1} \quad \text{(OCR data sheet value)}$$k=1.38×10−23J K−1(OCR data sheet value)

Notice the units: joules per kelvin — no "per mole". The Boltzmann constant converts a temperature directly into an energy per molecule, without any mention of the macroscopic quantity "mole".

Exam Tip: It is easy to confuse k (Boltzmann's constant) with K (kelvin). In your exam script, use lowercase k for Boltzmann's constant and uppercase K for temperature units. The context almost always makes the meaning clear.

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The Avogadro Constant

The Avogadro constant N_A = 6.02 × 10²³ mol⁻¹ gives the number of elementary entities in one mole. Originally defined in terms of carbon-12 (specifically, so that 12 g of carbon-12 contains exactly N_A atoms), it has since 2019 been defined as an exact number by SI: N_A = 6.022 140 76 × 10²³ mol⁻¹ exactly.

You should be comfortable with these three equivalent relationships:

Relation	What it tells you
N = n N_A	Number of molecules from number of moles
n = m/M	Number of moles from mass and molar mass
N = (m/M) N_A	Number of molecules from mass and molar mass

These let you move freely among mass (in kg), amount (in moles), and number (dimensionless).

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The Two Forms Side by Side

Form	Best when	Notes
pV = nRT	You know the amount in moles or mass	Use molar gas constant R
pV = NkT	You know the number of molecules or need molecular-scale results	Use Boltzmann constant k

Both forms give exactly the same answer for any given physical situation. Choose whichever is most convenient for the numbers you have.

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Worked Example 1: Molecules in a Room

How many molecules of air are there in a typical classroom of dimensions 10 m × 8 m × 3 m, at atmospheric pressure p = 1.01 × 10⁵ Pa and temperature T = 293 K?

$$\begin{aligned} V &= 10 \times 8 \times 3 = 240\,\text{m}^{3} \\ N &= pV / (kT) \\ &= (1.01 \times 10^{5})(240) / ((1.38 \times 10^{-23})(293)) \\ &= 2.424 \times 10^{7} / (4.04 \times 10^{-21}) \\ &\approx 6.0 \times 10^{27} \end{aligned}$$VN​=10×8×3=240m3=pV/(kT)=(1.01×105)(240)/((1.38×10−23)(293))=2.424×107/(4.04×10−21)≈6.0×1027​

Six billion billion billion molecules. This gives you a sense of just how crowded even "empty" rooms are with gas molecules.

Equivalently, in moles:

$$\begin{aligned} n &= pV / (RT) \\ &= (1.01 \times 10^{5})(240) / ((8.31)(293)) \\ &\approx 9.96 \times 10^{3}\,\text{mol} \end{aligned}$$n​=pV/(RT)=(1.01×105)(240)/((8.31)(293))≈9.96×103mol​

And indeed N = nN_A = (9.96 × 10³)(6.02 × 10²³) ≈ 6.0 × 10²⁷. The two methods agree perfectly.

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Worked Example 2: Molecules per Cubic Metre at STP

How many molecules of an ideal gas are there per cubic metre at standard temperature and pressure (STP: T = 273 K, p = 1.01 × 10⁵ Pa)?

$$\begin{aligned} N/V &= p / (kT) \\ &= (1.01 \times 10^{5}) / ((1.38 \times 10^{-23})(273)) \\ &= 1.01 \times 10^{5} / (3.77 \times 10^{-21}) \\ &\approx 2.68 \times 10^{25}\,\text{molecules m}^{-3} \end{aligned}$$N/V​=p/(kT)=(1.01×105)/((1.38×10−23)(273))=1.01×105/(3.77×10−21)≈2.68×1025molecules m−3​

This number — called Loschmidt's number — is one of the most important in atomic physics. It says that a cubic metre of air at STP contains about 2.7 × 10²⁵ molecules, and a cubic centimetre contains about 2.7 × 10¹⁹. These are numbers worth committing to memory for a rough sense of scale.

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Worked Example 3: Molecules in a Balloon

A balloon of volume 2.5 × 10⁻³ m³ contains helium at 1.2 × 10⁵ Pa and 295 K. How many helium atoms are inside?

$$\begin{aligned} N &= pV / (kT) \\ &= (1.2 \times 10^{5})(2.5 \times 10^{-3}) / ((1.38 \times 10^{-23})(295)) \\ &= 300 / (4.07 \times 10^{-21}) \\ &\approx 7.4 \times 10^{22}\,\text{atoms} \end{aligned}$$N​=pV/(kT)=(1.2×105)(2.5×10−3)/((1.38×10−23)(295))=300/(4.07×10−21)≈7.4×1022atoms​

We say "atoms" rather than "molecules" because helium is monatomic. Physically it makes no difference for the ideal gas law: N is just the number of independent particles, whatever they are.

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The Mass of a Single Molecule

If you know the molar mass M (kg mol⁻¹) of a substance, the mass of one molecule is

$$m = M / N_A$$m=M/NA​

For example, nitrogen N₂ has M = 0.028 kg mol⁻¹, so

$$m = 0.028 / (6.02 \times 10^{23}) \approx 4.65 \times 10^{-26}\,\text{kg}$$m=0.028/(6.02×1023)≈4.65×10−26kg

And hydrogen H₂ has M = 0.002 kg mol⁻¹, giving m ≈ 3.32 × 10⁻²⁷ kg per molecule — about twice the mass of a proton, as expected.

These per-molecule masses are needed in the kinetic theory, where the mean kinetic energy per molecule is (1/2) m <c²>.

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Why Does the Molecular Form Exist at All?

You might wonder why we bother with two different forms of the ideal gas equation. The answer is that each is natural in a different context:

• Chemists typically measure gases in grams and moles. They usually want pV = nRT, which connects directly to moles and molar masses.
• Physicists doing statistical mechanics or kinetic theory work with individual molecules. They want pV = NkT, which connects to the number of molecules and the Boltzmann constant.

The Boltzmann constant is the more "fundamental" of the two in a sense: it tells you that every molecule carries (3/2)kT of mean kinetic energy (at least for a monatomic ideal gas; see Lesson 10). It is a universal relationship at the molecular level, without any reference to the arbitrary unit of "mole".

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The Mole as a Convenient Bookkeeping Device

It is worth stepping back and appreciating the role of the mole. The mole is not a physical unit in the sense that kelvin or joule is: it is a counting unit, like "dozen" (= 12) or "gross" (= 144). "One mole of atoms" simply means "Avogadro's number of atoms", which is a lot — about 6 × 10²³. The mole was invented because bulk chemistry deals with amounts so large that counting molecules individually is absurd, but dividing by a fixed large number (N_A) gives workable small numbers.

When you move from chemistry to physics, you often want to undo this redefinition and go back to counting molecules. The Boltzmann constant is exactly the factor that does this:

$$k = R / N_A \quad \text{(per-molecule version of } R \text{)}$$k=R/NA​(per-molecule version of R)

In this sense, writing pV = NkT is "more fundamental" than writing pV = nRT — the former doesn't mention the anthropic choice of Avogadro's number at all.

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Worked Example 4: Density from Pressure and Temperature

Find the density of air at atmospheric pressure (1.01 × 10⁵ Pa) and 20 °C (293 K), given that the molar mass of air is about 0.029 kg mol⁻¹.

From the ideal gas equation:

$$\begin{aligned} n &= pV / (RT) \\ m &= nM = (pV/RT) \cdot M \quad \text{(mass of gas)} \\ \rho &= m/V = pM/(RT) \quad \text{(density)} \\ &= (1.01 \times 10^{5})(0.029) / ((8.31)(293)) \\ &= 2929 / 2434.8 \\ &\approx 1.20\,\text{kg m}^{-3} \end{aligned}$$nmρ​=pV/(RT)=nM=(pV/RT)⋅M(mass of gas)=m/V=pM/(RT)(density)=(1.01×105)(0.029)/((8.31)(293))=2929/2434.8≈1.20kg m−3​

This matches the familiar "density of air ≈ 1.2 kg m⁻³" commonly quoted in fluid dynamics problems.

Note the useful general formula

$$\rho = pM / (RT)$$ρ=pM/(RT)

for the density of an ideal gas in terms of its pressure, temperature, and molar mass. Equivalently, ρ = pm/(kT) where m is the mass of a single molecule — which follows by dividing numerator and denominator by N_A.

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Worked Example 5: Avogadro's Principle

Equal volumes of any two ideal gases, at the same temperature and pressure, contain the same number of molecules. Show this from pV = NkT.

Fix V, T, p to be the same for two different gases, labelled with subscripts $1$1 and $2$2. Then

$$\begin{aligned} p V &= N_1 k T \quad \text{(gas 1)} \\ p V &= N_2 k T \quad \text{(gas 2)} \end{aligned}$$pVpV​=N1​kT(gas 1)=N2​kT(gas 2)​

Dividing the two equations and cancelling $pV$pV and $kT$kT on the left and right respectively,

$$N_1 = N_2$$N1​=N2​

regardless of what the gases are made of. This is Avogadro's principle, originally an empirical guess ($1811$1811) that turned out to be exactly what the ideal gas law predicts.

It does not, however, mean that the two gases weigh the same — they have different molecular masses. But the number of molecules is the same.

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Worked Example 6: Comparing Two Samples

A sample of oxygen contains 1.0 × 10²⁴ molecules. A sample of hydrogen contains 4.0 × 10²³ molecules. Both are at 300 K and 1.0 × 10⁵ Pa. Which sample has a greater volume, and by what factor?

The ideal gas equation pV = NkT gives V = NkT/p. At the same p and T, V ∝ N. So the oxygen sample has 1.0 × 10²⁴ / 4.0 × 10²³ = 2.5 times the volume of the hydrogen sample. It does not matter that the molecules are of different species.

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Worked Example 7: Gases in Equilibrium

A container of volume V = 0.10 m³ is separated into two halves by a fixed partition. The left half contains N_1 = 2.0 × 10²⁴ nitrogen molecules; the right half contains N_2 = 3.0 × 10²⁴ oxygen molecules. Both halves are at T = 300 K. What is the pressure in each half?

Each half has volume V/2 = 0.050 m³. Apply pV = NkT:

$$\begin{aligned} &\text{Left half:} \\ p_1 &= N_1 k T / (V/2) \\ &= (2.0 \times 10^{24})(1.38 \times 10^{-23})(300) / 0.050 \\ &= 8280 / 0.050 \\ &\approx 1.66 \times 10^{5}\,\text{Pa} \\ &\text{Right half:} \\ p_2 &= N_2 k T / (V/2) \\ &= (3.0 \times 10^{24})(1.38 \times 10^{-23})(300) / 0.050 \\ &= 12420 / 0.050 \\ &\approx 2.48 \times 10^{5}\,\text{Pa} \end{aligned}$$p1​p2​​Left half:=N1​kT/(V/2)=(2.0×1024)(1.38×10−23)(300)/0.050=8280/0.050≈1.66×105PaRight half:=N2​kT/(V/2)=(3.0×1024)(1.38×10−23)(300)/0.050=12420/0.050≈2.48×105Pa​

If the partition were removed and the two gases allowed to mix, they would eventually fill the whole container. The total pressure would then be

$$\begin{aligned} p_{\text{total}} &= (N_1 + N_2) k T / V \\ &= (5.0 \times 10^{24})(1.38 \times 10^{-23})(300) / 0.10 \\ &= 20\,700 / 0.10 \\ &\approx 2.07 \times 10^{5}\,\text{Pa} \end{aligned}$$ptotal​​=(N1​+N2​)kT/V=(5.0×1024)(1.38×10−23)(300)/0.10=20700/0.10≈2.07×105Pa​

which is just the sum of the partial pressures each gas would produce if it occupied the whole container alone (Dalton's law of partial pressures).

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Relating R, k and N_A

Memorise these three relationships; you will use them constantly:

Equation	In words
R = N_A k	Molar gas constant = Avogadro × Boltzmann
k = R/N_A	Boltzmann = molar / Avogadro
pV = nRT = NkT	Two forms of the ideal gas equation

The first two are equivalent; either can be checked using R = 8.31 J mol⁻¹ K⁻¹ and N_A = 6.02 × 10²³ mol⁻¹.

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Synoptic Links

• Ideal gas equation (Lesson 6) — $pV = NkT$pV=NkT is the molecule-counting twin of $pV = nRT$pV=nRT. The bridge $N = n N_A$N=nNA​ and the definition $k = R/N_A$k=R/NA​ make them interchangeable. Choose by what is given: mass/moles ⇒ $R$R; molecule count or microscale variable ⇒ $k$k.
• Kinetic theory pressure equation (Lesson 9) — the derivation gives $pV = (1/3) N m \langle c^{2} \rangle$pV=(1/3)Nm⟨c2⟩, with $N$N and $m$m both molecule-scale quantities. Equating with $pV = NkT$pV=NkT delivers $\tfrac{1}{2} m \langle c^{2}\rangle = \tfrac{3}{2} k T$21​m⟨c2⟩=23​kT (Lesson 10) — the central result of statistical mechanics, in which the Boltzmann constant is the natural conversion between temperature and microscopic kinetic energy.
• Internal energy of an ideal monatomic gas (Module 5.1) — $U = \tfrac{3}{2} N k T = \tfrac{3}{2} n R T$U=23​NkT=23​nRT. Either form works; the Boltzmann form makes the per-molecule energy $\tfrac{3}{2} kT$23​kT explicit.
• Black-body radiation and Wien's displacement (Module 5.5) — the spectral peak of a black-body satisfies $\lambda_{\max} T = 2.898 \times 10^{-3}$λmax​T=2.898×10−3 m K, a Wien constant proportional to $hc/k$hc/k. The Boltzmann constant therefore sits at the heart of stellar thermometry.
• Statistical mechanics (post-A-Level) — Boltzmann factors of the form $\exp(-E/(kT))$exp(−E/(kT)) are the universal weighting for energy levels in thermal equilibrium. The Boltzmann constant is what gives the exponent its physical meaning.

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Specimen question modelled on the OCR H556 paper format