Avogadro Constant, Molar Gas Constant, Boltzmann Constant, and Molar/Molecular Mass Revision Notes | A Level AQA Physics

1. Why These Constants Matter

Gases contain an almost incomprehensible number of particles — a single breath holds roughly $10^{22}$ molecules. Tracking each one individually is impossible, so physicists developed a set of constants that bridge the microscopic world of individual particles and the macroscopic world of pressure gauges and thermometers.

The Avogadro constant $N_A$, the molar gas constant $R$, and the Boltzmann constant $k$ are the three pillars of thermal physics at A Level. Master them and the ideal gas equation, kinetic theory, and internal energy calculations all fall into place.

2. The Avogadro Constant

Think of a mole the same way you think of a dozen — it is a counting unit. One dozen always means 12 items; one mole always means $6.02 \times 10^{23}$ particles. The Avogadro constant $N_A$ is that fixed number.

The unit mol$^{-1}$ tells you that $N_A$ is a count per mole — divide the total number of particles $N$ by $N_A$ and you get the number of moles $n$. Rearranging gives $N = n N_A$, which you will use constantly in kinetic theory problems.

To put the scale in perspective: if you stacked $6.02 \times 10^{23}$ grains of sand, the pile would cover the entire surface of the Earth to a depth of several metres. The Avogadro constant is genuinely enormous, which is why we need it — atoms are genuinely tiny.

3. Molar Mass and Molecular Mass

Now that we have a way to count particles in bulk, we need a way to connect that count to mass — because in the lab, you weigh substances, you do not count molecules. That is where molar mass and molecular mass come in.

Molecular mass (also called the mass of one molecule) is the mass of a single particle. You obtain it by dividing the molar mass by the Avogadro constant: $m = M / N_A$. The unit is kg (or sometimes atomic mass units, u, in chemistry, but for A Level physics calculations always convert to kg).

The connecting equation is $m = M / N_A$. Notice that $M$ and $N_A$ both refer to the mole scale, so dividing them gives you the per-particle scale. This equation appears directly in AQA mark schemes for kinetic theory questions.

Worked example: Calculating the Mass of a Single Nitrogen Molecule

Nitrogen gas consists of diatomic molecules N$_2$. The molar mass of N$_2$ is $28$ g mol$^{-1}$. Calculate the mass of one nitrogen molecule in kg.

1. $M = 28 \text{ g mol}^{-1} = 28 \times 10^{-3} \text{ kg mol}^{-1} = 0.028 \text{ kg mol}^{-1}$ — AQA calculations require SI units throughout. Always convert g mol$^{-1}$ to kg mol$^{-1}$ before substituting. [M1]
2. $m = \dfrac{M}{N_A}$ — This divides the mass of one mole by the number of particles in one mole, giving the mass of one particle. [M1]
3. $m = \dfrac{0.028}{6.02 \times 10^{23}}$ — Numerator is in kg mol$^{-1}$; denominator is in mol$^{-1}$. Units: kg mol$^{-1}$ ÷ mol$^{-1}$ = kg. [M1]
4. $m = \dfrac{0.028}{6.02 \times 10^{23}} = \dfrac{2.8 \times 10^{-2}}{6.02 \times 10^{23}} = 4.65 \times 10^{-26} \text{ kg}$ — Divide the coefficients: $2.8 \div 6.02 \approx 0.465$. Then $10^{-2} \div 10^{23} = 10^{-25}$. So $0.465 \times 10^{-25} = 4.65 \times 10^{-26}$ kg. [A1]
5. Atomic masses are of order $10^{-27}$ kg per nucleon. N$_2$ has 28 nucleons, so expect $\approx 28 \times 10^{-27} = 2.8 \times 10^{-26}$ kg. Our answer $4.65 \times 10^{-26}$ kg is the same order — consistent. — Always check your power of ten against a rough estimate. A factor-of-two difference here is acceptable given the rough estimate used.

Answer: $m = 4.65 \times 10^{-26}$ kg

4. The Molar Gas Constant and the Boltzmann Constant

With molar mass and the Avogadro constant established, we can now look at the two constants that govern how gases behave with temperature. They describe the same physical reality — just at different scales.

The molar gas constant $R$ operates at the mole scale — it tells you how much energy (in joules) one mole of gas gains for every kelvin rise in temperature. But what about a single molecule? That is where the Boltzmann constant enters.

Here is the key relationship that ties everything together: $k = R / N_A$. Dividing the molar gas constant by the Avogadro constant scales it down from one mole to one particle. This is not a coincidence — it is the same logic as dividing molar mass by $N_A$ to get molecular mass.

You can verify this numerically: $R / N_A = 8.31 / (6.02 \times 10^{23}) = 1.38 \times 10^{-23}$ J K$^{-1}$, which matches the accepted value of $k$ exactly. AQA examiners sometimes ask you to show this derivation, so it is worth being comfortable with the arithmetic.

5. Connecting the Two Forms of the Ideal Gas Equation

The relationship $k = R / N_A$ is not just a curiosity — it explains why the ideal gas equation has two equivalent forms. Depending on whether you are working with moles or with individual molecules, you choose the appropriate version.

Both forms of the ideal gas equation are equivalent because substituting $n = N / N_A$ and $R = k N_A$ into $pV = nRT$ gives $pV = (N / N_A)(k N_A)T = NkT$. The two forms are the same equation viewed at different scales.

Worked example: Verifying the Relationship $k = R / N_A$ Numerically

Show that dividing $R = 8.31$ J mol$^{-1}$ K$^{-1}$ by $N_A = 6.02 \times 10^{23}$ mol$^{-1}$ gives the Boltzmann constant $k = 1.38 \times 10^{-23}$ J K$^{-1}$.

1. $k = \dfrac{R}{N_A} = \dfrac{8.31}{6.02 \times 10^{23}}$ — This is the fundamental link between the mole-scale and particle-scale constants. [M1]
2. $k = \dfrac{8.31}{6.02} \times \dfrac{1}{10^{23}}$ — Splitting the calculation makes the arithmetic cleaner and reduces errors with powers of ten.
3. $8.31 \div 6.02 = 1.380...$ — Verify: $6.02 \times 1.38 = 6.02 \times 1 + 6.02 \times 0.38 = 6.02 + 2.2876 = 8.3076 \approx 8.31$. Confirmed. [M1]
4. $\dfrac{1}{10^{23}} = 10^{-23}$ — Dividing by $10^{23}$ is the same as multiplying by $10^{-23}$.
5. $k = 1.38 \times 10^{-23} \text{ J K}^{-1}$ — Units check: J mol$^{-1}$ K$^{-1}$ divided by mol$^{-1}$ gives J K$^{-1}$. This matches the accepted value of $k$. [A1]

Answer: $k = 1.38 \times 10^{-23}$ J K$^{-1}$, consistent with the accepted value.

6. Worked Application: Finding the Number of Molecules

Exam questions often give you a mass of gas and ask you to find the number of molecules — or vice versa. The route always passes through moles, molar mass, and the Avogadro constant. Here is a full worked example that chains all three constants together.

Worked example: Finding the Number of Molecules in a Sample of Gas

A container holds $3.2$ g of oxygen gas O$_2$. The molar mass of O$_2$ is $32$ g mol$^{-1}$. Calculate the number of oxygen molecules in the container. ($N_A = 6.02 \times 10^{23}$ mol$^{-1}$)

1. $n = \dfrac{\text{mass}}{M} = \dfrac{3.2 \text{ g}}{32 \text{ g mol}^{-1}} = 0.10 \text{ mol}$ — Dividing mass by molar mass gives moles. Units: g ÷ g mol$^{-1}$ = mol. [M1]
2. $N = n N_A$ — Multiplying moles by the Avogadro constant scales up from the mole to the particle level. [M1]
3. $N = 0.10 \times 6.02 \times 10^{23}$ — Multiply the coefficient: $0.10 \times 6.02 = 0.602$. Then $0.602 \times 10^{23} = 6.02 \times 10^{22}$.
4. $N = 6.02 \times 10^{22}$ molecules — No unit for $N$ itself (it is a pure count), but always specify what is being counted. [A1]
5. $3.2$ g is one tenth of a mole, so we expect one tenth of $N_A$: $6.02 \times 10^{23} \div 10 = 6.02 \times 10^{22}$. Confirmed. — Fraction-of-a-mole checks are fast and reliable. If your answer is not a simple fraction of $N_A$, re-examine your mole calculation.

Answer: $N = 6.02 \times 10^{22}$ molecules

7. Exam Technique and Common Mistakes

These constants appear in almost every thermal physics question at A Level, so the marks available for using them correctly add up quickly. The mistakes that cost students marks are almost always the same ones — and they are all avoidable.

Taken together, $N_A$, $R$, $k$, molar mass, and molecular mass form a complete toolkit for moving between the macroscopic and microscopic descriptions of a gas — which is precisely what AQA thermal physics questions demand.