recall that the Boltzmann constant k is given by k = R / NA

Equation of State – Cambridge A‑Level Physics (9702)

Learning Objective

Recall that the Boltzmann constant k links the macroscopic gas constant R to the number of particles per mole N_A:

\$k = \frac{R}{N_{\mathrm A}}\$

1. The Mole

• Definition: One mole of a substance contains exactly N_A = 6.022 140 76 × 10²³ elementary entities (atoms, molecules, ions, …).
• SI unit of amount of substance: mol.
• Molar mass (M) is the mass of one mole of a substance (units g mol⁻¹). Numerically it equals the average mass of a single particle expressed in atomic mass units (u).
• Conversion between amount of substance and number of particles:

  \$n = \frac{N}{N{\mathrm A}}\qquad\text{or}\qquad N = n\,N{\mathrm A}\$

  where n is the amount in moles and N the total number of particles.

Worked examples

1. From mass to particles (CO₂):

   The molar mass of carbon dioxide is 44.01 g mol⁻¹. A 22.0 g sample contains

   \$n = \frac{22.0\;\text{g}}{44.01\;\text{g mol}^{-1}} = 0.500\;\text{mol}\$

   and therefore

   \$N = 0.500\;\text{mol}\times 6.022\times10^{23}\;\text{mol}^{-1}=3.01\times10^{23}\;\text{molecules}.\$

2. From particles to mass (H₂):

   How many grams of hydrogen correspond to \(2.0\times10^{23}\) molecules?

   First find the number of moles:

   \$n = \frac{N}{N_{\mathrm A}} = \frac{2.0\times10^{23}}{6.022\times10^{23}} = 0.332\;\text{mol}.\$

   The molar mass of H₂ is 2.016 g mol⁻¹, so the mass is

   \$m = nM = 0.332\;\text{mol}\times 2.016\;\text{g mol}^{-1}=0.670\;\text{g}.\$

2. Ideal‑Gas Law (Macroscopic Form)

The equation of state for an ideal gas is

\$pV = nRT\$

where

• p – pressure (Pa)
• V – volume (m³)
• n – amount of substance (mol)
• R – universal gas constant (8.314 462 618 J mol⁻¹ K⁻¹)
• T – absolute temperature (K)

3. The Three Simple Gas Laws

Bridge paragraph: Combining Boyle’s, Charles’s and Avogadro’s laws eliminates the three separate constants and yields the single macroscopic ideal‑gas equation \(pV = nRT\). This unified form can be rewritten in microscopic terms by substituting \(n = N/N_{\mathrm A}\), where \(N\) is the total number of particles.

4. Microscopic Form of the Equation of State

Replace the amount of substance in the macroscopic law:

\$pV = \frac{N}{N_{\mathrm A}}RT\$

Define the Boltzmann constant \(k = R/N_{\mathrm A}\) to obtain

\$pV = NkT\$

5. Deriving the Boltzmann Constant

From the two equivalent expressions

• Macroscopic: \(pV = nRT\)
• Microscopic: \(pV = NkT\)

and using \(n = N/N_{\mathrm A}\):

\$\frac{N}{N{\mathrm A}}RT = NkT \;\;\Longrightarrow\;\; k = \frac{R}{N{\mathrm A}}.\$

6. Kinetic Theory of Gases – Key Assumptions

• The gas consists of a very large number of tiny particles (atoms or molecules) in constant random motion.
• The volume of the particles themselves is negligible compared with the container volume.
• Collisions between particles, and between particles and the container walls, are perfectly elastic (no loss of kinetic energy).
• No intermolecular forces act except during collisions.
• The average translational kinetic energy of the particles depends only on temperature.

7. From Kinetic Theory to the Microscopic Equation

Consider a cubic container of side \(L\) (so \(V = L^{3}\)) and a single molecule of mass \(m\) moving with speed component \(c_{x}\) perpendicular to a wall.

1. Change in momentum on a collision: \(\Delta p = 2mc_{x}\).
2. Time between successive collisions with the same wall: \(\Delta t = 2L/c_{x}\).
3. Force exerted on the wall by this molecule:

   \$F = \frac{\Delta p}{\Delta t}= \frac{2mc{x}}{2L/c{x}}= \frac{mc_{x}^{2}}{L}.\$

4. Pressure contributed by this molecule:

   \$p = \frac{F}{A}= \frac{mc{x}^{2}}{L^{3}} = \frac{mc{x}^{2}}{V}.\$

5. For many molecules the motion is isotropic, so

   \$\langle c{x}^{2}\rangle = \langle c{y}^{2}\rangle = \langle c_{z}^{2}\rangle = \frac{1}{3}\langle c^{2}\rangle.\$

6. Summing over all \(N\) particles gives the kinetic‑theory equation of state:

   \$pV = \frac{1}{3}\,N\,m\,\langle c^{2}\rangle.\$

8. Connecting Kinetic Theory to Temperature

The average translational kinetic energy of a molecule is

\$\langle E_{\text{kin}}\rangle = \frac{1}{2}m\langle c^{2}\rangle.\$

Comparing this with the microscopic ideal‑gas law (\(pV = NkT\)) leads to the fundamental result

\$\boxed{\langle E_{\text{kin}}\rangle = \frac{3}{2}\,kT}\$

and the root‑mean‑square speed

\$v_{\text{rms}} = \sqrt{\langle c^{2}\rangle}= \sqrt{\frac{3kT}{m}}.\$

9. Important Physical Constants

10. Sample Application

Problem: Calculate the rms speed of nitrogen (N₂) molecules at 300 K. (Molar mass of N₂ = 28.02 g mol⁻¹.)

1. Mass of one N₂ molecule:

   \$m = \frac{M}{N_{\mathrm A}} = \frac{28.02\times10^{-3}\;\text{kg mol}^{-1}}{6.022\times10^{23}\;\text{mol}^{-1}} = 4.65\times10^{-26}\;\text{kg}.\$

2. Insert into the rms‑speed formula:

   \$v_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3(1.380649\times10^{-23}\;\text{J K}^{-1})(300\;\text{K})}{4.65\times10^{-26}\;\text{kg}}} \approx 5.2\times10^{2}\;\text{m s}^{-1}.\$

11. Quick Revision Checklist

• Can you write the ideal‑gas law in both macroscopic (pV = nRT) and microscopic (pV = NkT) forms?
• Do you know the definition of a mole, its SI unit (mol), and how to convert between mass, moles, and number of particles?
• Can you list the three simple gas laws (Boyle, Charles, Avogadro) and explain how they combine to give the macroscopic equation of state?
• Are you able to derive k = R/N_A and the kinetic‑theory result pV = (1/3)Nm⟨c²⟩?
• Can you use k to calculate average kinetic energy (3/2 kT) and rms speed (√(3kT/m))?
• Do you remember the numerical values and units of R, N_A and k?