Molecular nature of matter and the behaviour of gases
The smell of a perfume spreads across a room because its molecules keep moving. Link to earlier classes : particles of matter of class 9 and the gas laws of chapter 10.
Matter is made of atoms and molecules in constant motion. A gas is mostly empty space; its molecules are far apart and move freely. One mole of any substance contains Avogadro's number N_A = 6.022 × 10²³ particles.
Rule 1: Ideal gas in terms of molecules — P V = N k_B T (N = number of molecules, k_B = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹). Also k_B = R ÷ N_A. A real gas behaves like an ideal gas at low pressure and high temperature.
- Easy : 1 mol contains 6.022 × 10²³ molecules.
- Medium : The number density at 1 atm and 300 K is n = P ÷ (k_B T) = 1.013 × 10⁵ ÷ (1.38 × 10⁻²³ × 300) = 2.4 × 10²⁵ m⁻³.
- Exam : The number of molecules in 1 litre at 273 K and 1 atm = 1.013 × 10⁵ × 10⁻³ ÷ (1.38 × 10⁻²³ × 273) = 2.7 × 10²² (Avogadro's law : equal volumes, same T and P, same number of molecules).
P V = N k_B T = n R T; k_B = R ÷ N_A.
Mixing up k_B (per molecule) and R (per mole).
Memory clue : "Big N with small k, big n with big R."
Spot the difference
| Constant | Value | Per |
|---|---|---|
| R | 8.314 J mol⁻¹ K⁻¹ | mole |
| k_B | 1.38 × 10⁻²³ J K⁻¹ | molecule |
Kinetic theory and the pressure of an ideal gas
A balloon stays inflated because countless molecules keep hitting its walls. Link to earlier classes : momentum and impulse of chapter 4 are used for the collisions of molecules with the wall.
In the kinetic theory, an ideal gas is a large number of tiny, identical, randomly moving molecules that collide elastically with each other and with the walls and exert no force except during collisions. The pressure of the gas is the average force per area from these collisions.
Rule 1: Pressure of an ideal gas — P = (1 ÷ 3) n m v̄² (n = number of molecules per unit volume, m = mass of one molecule, v̄² = mean of the squares of the speeds). Since n m = ρ, P = (1 ÷ 3) ρ v̄².
- Easy : Air of density 1.16 kg m⁻³ with v_rms = 517 m s⁻¹ gives P = (1 ÷ 3) × 1.16 × 517² = 1.03 × 10⁵ Pa, close to 1 atm.
- Medium : If the rms speed doubles, the pressure becomes 4 times.
- Exam : Doubling the number of molecules in the same volume at the same speed doubles the pressure.
P = (1 ÷ 3) n m v̄² = (1 ÷ 3) ρ v̄²; the factor ⅓ comes from three directions.
Using the average speed instead of the rms speed (the mean of the squares).
Memory clue : "One-Third Nice Mass V-square."
Spot the difference
| Quantity | Meaning |
|---|---|
| Average speed | mean of the speeds |
| rms speed | square root of the mean of the squares |
Kinetic interpretation of temperature
Hot air feels different from cool air because its molecules move faster. Link to earlier classes : kinetic energy ½mv² of chapter 5.
The absolute temperature of an ideal gas is a measure of the average translational kinetic energy of its molecules. It does not depend on the kind of gas.
Rule 1: Temperature and energy — ½ m v̄² = (3 ÷ 2) k_B T, so v_rms = √(3 k_B T ÷ m). At the same temperature a lighter gas has a larger rms speed (v_rms ∝ 1 ÷ √m).
- Easy : At 300 K the average translational energy of a molecule is 1.5 × 1.38 × 10⁻²³ × 300 = 6.2 × 10⁻²¹ J (0.039 eV).
- Medium : For nitrogen (m = 4.65 × 10⁻²⁶ kg) at 300 K, v_rms = 517 m s⁻¹; for oxygen (m = 5.3 × 10⁻²⁶ kg) it is 484 m s⁻¹.
- Exam : The temperature at which argon atoms (m = 6.64 × 10⁻²⁶ kg) have v_rms = 2.0 km s⁻¹ is T = m v² ÷ (3 k_B) = 6.64 × 10⁻²⁶ × 4 × 10⁶ ÷ (3 × 1.38 × 10⁻²³) = 6400 K. Hydrogen molecules move 4 times as fast as oxygen molecules at the same temperature.
Average kinetic energy = (3 ÷ 2) k_B T; v_rms ∝ √(T ÷ m); the energy depends only on T.
Believing that heavier molecules have a larger kinetic energy at the same temperature (they have the same energy, only a smaller speed).
Memory clue : "Three-halves k T."
Spot the difference
| Same T | Light gas | Heavy gas |
|---|---|---|
| Average energy | same | same |
| Speed | larger | smaller |
The law of equipartition of energy
A hot diatomic gas keeps energy in rotation as well as in motion along a line. Link to earlier classes : rotational kinetic energy of chapter 6.
Degrees of freedom are the independent ways in which a molecule can store energy. The law of equipartition of energy : in thermal equilibrium, every degree of freedom (every quadratic term in the energy) has an average energy of ½ k_B T per molecule.
Rule 1: Degrees of freedom — Monatomic : 3 (translation). Rigid diatomic : 5 (3 translational + 2 rotational). Non-linear polyatomic : 6 (3 + 3). With vibration each vibrational mode adds 2 (kinetic + potential).
- Easy : A helium atom at 300 K has the average energy (3 ÷ 2) k_B T = 6.2 × 10⁻²¹ J.
- Medium : Internal energy of 2 mol of a monatomic gas at 300 K = (3 ÷ 2) × 2 × 8.314 × 300 = 7480 J.
- Exam : 1 mol of a rigid diatomic gas at 300 K has U = (5 ÷ 2) × 8.314 × 300 = 6240 J.
Each degree of freedom ↔ ½ k_B T per molecule; U = (f ÷ 2) n R T.
Counting rotation about the molecular axis for a monatomic gas.
Memory clue : "Half k T for every freedom."
Spot the difference
| Gas | f | U per mole |
|---|---|---|
| Monatomic | 3 | 3RT ÷ 2 |
| Diatomic | 5 | 5RT ÷ 2 |
| Polyatomic | 6 | 3RT |
Specific heat capacities of gases
It takes more heat to warm a diatomic gas such as air than a monatomic gas such as helium, mole for mole. Link to earlier classes : C_v and C_p of chapter 11.
Using equipartition, the molar specific heat at constant volume is C_v = (f ÷ 2) R, so C_p = (f ÷ 2 + 1) R.
Rule 1: Specific heats from the degrees of freedom — C_v = (f ÷ 2) R, C_p = C_v + R, γ = 1 + 2 ÷ f.
- Easy : Monatomic (f = 3) : C_v = 1.5 R, C_p = 2.5 R and γ = 1 + 2 ÷ 3 = 1.67.
- Medium : Rigid diatomic (f = 5) : C_v = 2.5 R and γ = 1.40.
- Exam : Non-linear polyatomic (f = 6) : C_v = 3 R, C_p = 4 R and γ = 1.33.
C_v = (f ÷ 2) R; fewer degrees of freedom give a larger γ.
Using the same C_v for all gases.
Memory clue : "γ = 1 + 2 ÷ f : 3 → 5 ÷ 3, 5 → 7 ÷ 5."
Spot the difference
| Gas | f | γ |
|---|---|---|
| Monatomic | 3 | 1.67 |
| Diatomic | 5 | 1.40 |
| Polyatomic | 6 | 1.33 |
Mean free path
A smell takes a few seconds to cross a room, not a fraction of a second, because molecules collide again and again. Link to earlier classes : collisions of chapter 5.
The mean free path λ is the average distance a molecule travels between two successive collisions.
Rule 1: Mean free path — λ = 1 ÷ (√2 π d² n) (d = diameter of a molecule, n = number density). It is larger at lower pressure and smaller for a bigger molecule.
- Easy : At 1 atm and 300 K, n = 2.4 × 10²⁵ m⁻³ and d = 3 × 10⁻¹⁰ m ⇒ λ = 1 ÷ (1.414 × 3.14 × 9 × 10⁻²⁰ × 2.4 × 10²⁵) = 1.0 × 10⁻⁷ m, about 100 nm.
- Medium : If the pressure is reduced to a hundredth at the same temperature, n falls to a hundredth and λ becomes 100 times larger.
- Exam : The mean free path is about 300 times the diameter of a molecule, so a gas is mostly empty space.
λ ∝ 1 ÷ n; λ ∝ 1 ÷ d²; at ordinary conditions it is about 10⁻⁷ m.
Confusing mean free path with the average speed.
Memory clue : "More crowd, less free path."
Spot the difference
| Condition | Mean free path |
|---|---|
| High pressure | short |
| Low pressure | long |