The Boltzmann Constant

In Lesson 6 we derived the ideal gas equation pV = nRT in terms of moles of gas. But for many purposes — especially the kinetic theory of gases in Lessons 9 and 10 — it is more natural to work with the number of molecules rather than the number of moles. This leads to the molecular form of the gas equation,

pV = NkT

where N is the number of molecules and k is the Boltzmann constant. This is the form that connects the macroscopic gas laws to the microscopic world of individual molecules, and it will be central to everything we do in the rest of this course.

Module 5.1.3 of the OCR A-Level Physics A specification (H556) requires you to know the Boltzmann constant, its relationship to the Avogadro and molar gas constants, and how to use the molecular form of the ideal gas equation in calculations.

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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

molecules in total. Substituting into pV = nRT:

pV = (N/N_A) R T
   = N × (R/N_A) × 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

With this definition, the ideal gas equation becomes

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:

k = R / N_A
  = 8.31 / (6.02 × 10²³)
  ≈ 1.38 × 10⁻²³ J K⁻¹

So the Boltzmann constant is

k = 1.38 × 10⁻²³ J 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?

V = 10 × 8 × 3 = 240 m³

N = 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:

n = pV / (RT)
  = (1.01 × 10⁵)(240) / ((8.31)(293))
  ≈ 9.96 × 10³ mol

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)?

N/V = p / (kT)
    = (1.01 × 10⁵) / ((1.38 × 10⁻²³)(273))
    = 1.01 × 10⁵ / (3.77 × 10⁻²¹)
    ≈ 2.68 × 10²⁵ molecules 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?

N = pV / (kT)
  = (1.2 × 10⁵)(2.5 × 10⁻³) / ((1.38 × 10⁻²³)(295))
  = 300 / (4.07 × 10⁻²¹)
  ≈ 7.4 × 10²² atoms

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

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

m = 0.028 / (6.02 × 10²³) ≈ 4.65 × 10⁻²⁶ kg

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.