Statistical physics: why most molecules stay on the ground floor

The problem

A molecule in a gas suffers billions of collisions every second, and each one changes its speed. Predicting one molecule is impossible. But predicting what fraction of a billion billion molecules have a certain energy turns out to be easy.

Think of a building with lifts that cost energy. Most people stay on the ground floor; fewer reach each higher floor. On a busy, energetic day more people go up, but the ground floor stays the most crowded. Molecules behave the same way.

Boltzmann distribution law: N₂/N₁ = e^(−ΔE/k_B T), where ΔE = E₂ − E₁. The chance of finding a molecule in a state falls exponentially with its energy divided by k_B T, so more particles reside in lower energy states.

Key terms

Statistical physics

The branch of physics that uses probability theory to describe very large numbers of molecules. Instead of following one molecule, it predicts how many are in each state.

Energy states

Because molecules have different speeds, the atoms and molecules of a system exist in different energy states, or levels: E₁, E₂, E₃ and so on.

Boltzmann factor

The ratio of the populations of a higher and a lower level is e^(−ΔE/k_B T). N₁ is the population of the lower state and N₂ of the higher state.

Average kinetic energy

Each collision changes a molecule's speed, but the average kinetic energy at temperature T stays (3/2)k_B T. Here k_B is effectively the gas constant per molecule.

Short questions with model answers

1. Q1. Two levels are 0.026 eV apart. At 300 K, what fraction of the lower-level population is in the upper level? (k_B = 8.62 × 10⁻⁵ eV K⁻¹)

   • k_B T = 8.62 × 10⁻⁵ × 300 = 0.0259 eV
   • ΔE ÷ k_B T = 0.026 ÷ 0.0259 = 1.00
   • N₂/N₁ = e^(−1.00) = 0.37

   N₂/N₁ ≈ 0.37

2. Q2. Now make the gap ten times bigger, ΔE = 0.26 eV, still at 300 K. What happens to N₂/N₁?

   • ΔE ÷ k_B T = 0.26 ÷ 0.0259 = 10.0
   • N₂/N₁ = e^(−10) ≈ 4.5 × 10⁻⁵

   Only about 5 in every 100,000 reach the upper level: a big gap is almost empty.

3. Q3. What will the kinetic energy of gas molecules be near absolute zero? Explain.

   • ⟨K.E.⟩ = (3/2)k_B T → 0 as T → 0
   • N₂/N₁ = e^(−ΔE/k_B T) → 0

   It approaches zero: the molecules almost stop, all in the lowest energy state.

Common mistakes

• ✗ Saying that at a high enough temperature the upper level holds more molecules than the lower one.

  ✓ e^(−ΔE/k_B T) is always less than 1 for positive ΔE. Heating brings N₂ closer to N₁, but never above it.

• ✗ Assuming that doubling ΔE simply halves the upper population.

  ✓ The law is exponential. At 300 K, going from 0.026 eV to 0.26 eV takes N₂/N₁ from 0.37 to about 0.000045.

• ✗ Dividing ΔE in electron volts by k_B T in joules.

  ✓ Use the same unit for both: k_B = 8.62 × 10⁻⁵ eV K⁻¹ with eV, or 1.38 × 10⁻²³ J K⁻¹ with joules (1 eV = 1.6 × 10⁻¹⁹ J).

MCQs

1. 1. According to the Boltzmann distribution law, more particles reside in:

   1. (a) higher energy states
   2. (b) lower energy states
   3. (c) all states equally
   4. (d) no state
   Show answer

   (b) Lower states: the population falls exponentially as energy rises.

2. 2. The ratio of the populations of a higher and a lower energy level is:

   1. (a) e^(ΔE/k_B T)
   2. (b) e^(−ΔE/k_B T)
   3. (c) ΔE/k_B T
   4. (d) k_B T/ΔE
   Show answer

   (b) N₂/N₁ = e^(−ΔE/k_B T), the Boltzmann factor. The minus sign makes it less than 1.

3. 3. At room temperature (300 K), k_B T is about:

   1. (a) 0.026 eV
   2. (b) 1.6 eV
   3. (c) 300 eV
   4. (d) 8.62 eV
   Show answer

   (a) 8.62 × 10⁻⁵ eV K⁻¹ × 300 K = 0.026 eV (4.14 × 10⁻²¹ J).

4. 4. Who described the distribution of molecular speeds in 1860?

   1. (a) Boltzmann
   2. (b) Maxwell
   3. (c) Einstein
   4. (d) Perrin
   Show answer

   (b) James Clerk Maxwell; experiments later confirmed his predictions.

5. 5. In ⟨K.E.⟩ = (3/2)k_B T, the Boltzmann constant k_B is effectively:

   1. (a) the gas constant per molecule
   2. (b) the speed of sound
   3. (c) the number of molecules
   4. (d) the charge on an electron
   Show answer

   (a) k_B = R/N_A: the gas constant shared out per molecule.

Quick revision

• Statistical physics uses probability to describe huge numbers of molecules spread over energy states E₁, E₂, …
• Boltzmann distribution: N₂/N₁ = e^(−ΔE/k_B T); lower states always hold more particles, and heating narrows the gap.
• Average kinetic energy is (3/2)k_B T; at 300 K, k_B T = 0.026 eV.