Chapter 12: Kinetic Theory
Unit IX – Behaviour of Perfect Gases and Kinetic Theory of Gases
12.1 Molecular Nature of Matter
- All matter is made of atoms and molecules.
- Ideal gas: point masses with no intermolecular forces except during elastic collisions.
- Brownian motion provides direct evidence for molecular motion.
12.2 Ideal Gas Equation
Combining Boyle’s Law, Charles’ Law, and Avogadro’s Law:
\[ PV = nRT \]
where:
- \(P\) = pressure (Pa)
- \(V\) = volume (m³)
- \(n\) = number of moles
- \(R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}\) (universal gas constant)
- \(T\) = temperature (K)
Also:
\[ PV = NkT \]
where \(N\) = number of molecules and \(k_B = \frac{R}{N_A} = 1.38 \times 10^{-23} \text{ J/K}\) (Boltzmann constant).
12.3 Kinetic Theory of Gases — Assumptions
- Gas consists of a large number of small, identical, point masses.
- Molecules move randomly in all directions.
- Collisions between molecules and container walls are perfectly elastic.
- The volume of molecules is negligible compared to the volume of the container.
- There are no intermolecular forces (except during collisions).
- Time of collision is negligible.
12.4 Kinetic Interpretation of Pressure and Temperature
Pressure from kinetic theory:
\[ P = \frac{1}{3}\frac{mNv_{rms}^2}{V} = \frac{1}{3}\rho v_{rms}^2 \]
Average kinetic energy per molecule:
\[ \langle KE \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{3}{2}k_B T \]
Thus temperature is proportional to average KE of molecules.
12.5 RMS, Mean, and Most Probable Speeds
For a gas of molar mass \(M\) at temperature \(T\):
\[ v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_BT}{m}} \]
\[ \bar{v} = \sqrt{\frac{8RT}{\pi M}} \quad \text{(mean speed)} \]
\[ v_{mp} = \sqrt{\frac{2RT}{M}} \quad \text{(most probable speed)} \]
Ratio: \(v_{mp} : \bar{v} : v_{rms} = 1 : 1.128 : 1.225\)
12.6 Maxwell Speed Distribution
12.7 Law of Equipartition of Energy
Each degree of freedom of a molecule has average energy \(\dfrac{1}{2}k_BT\).
Degrees of freedom:
| Molecule type | Translational | Rotational | Total f |
|---|---|---|---|
| Monoatomic | 3 | 0 | 3 |
| Diatomic (rigid) | 3 | 2 | 5 |
| Diatomic (non-rigid) | 3 | 2 | 7 |
| Polyatomic (non-linear) | 3 | 3 | 6 |
Total energy per mole:
\[ U = \frac{f}{2}RT \]
Molar specific heats for ideal gas:
\[ C_V = \frac{f}{2}R, \quad C_P = C_V + R = \frac{f+2}{2}R, \quad \gamma = \frac{C_P}{C_V} = \frac{f+2}{f} \]
| Gas | f | \(C_V\) | \(C_P\) | \(\gamma\) |
|---|---|---|---|---|
| Monoatomic | 3 | 3R/2 | 5R/2 | 5/3 |
| Diatomic | 5 | 5R/2 | 7R/2 | 7/5 |
12.8 Mean Free Path
The average distance a molecule travels between successive collisions:
\[ \ell = \frac{1}{\sqrt{2},\pi d^2 n} \]
where \(d\) = diameter of molecule, \(n\) = number density (molecules/m³).
As pressure decreases (or temperature increases), mean free path increases.
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(PV = nRT\) | Ideal gas equation |
| \(\langle KE \rangle = \frac{3}{2}k_BT\) | Mean KE per molecule |
| \(v_{rms} = \sqrt{\frac{3RT}{M}}\) | RMS speed |
| \(U = \frac{f}{2}RT\) | Internal energy per mole |
| \(\gamma = \frac{f+2}{f}\) | Ratio of specific heats |
| \(\ell = \frac{1}{\sqrt{2}\pi d^2 n}\) | Mean free path |
Practice Questions
Section A – MCQ (1 mark each)
Q1. The value of \(\gamma\) for a diatomic gas is:
(a) 5/3 (b) 7/5 (c) 4/3 (d) 3/2
Answer
(b) 7/5 = 1.4 — diatomic gas has \(f = 5\), so \(\gamma = (5+2)/5 = 7/5\).
Q2. The rms speed of nitrogen molecules at 27°C is approximately:
(Given: M = 28 g/mol, R = 8.314 J/mol·K)
(a) 316 m/s (b) 517 m/s (c) 658 m/s (d) 710 m/s
Answer
(b) 517 m/s
\(v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}} = \sqrt{267214} \approx 517 \text{ m/s}\)
Section B – Short Answer (2–3 marks)
Q3. State the law of equipartition of energy and use it to find \(C_V\) and \(C_P\) for a monoatomic ideal gas.
Answer
Law: Each degree of freedom of a molecule (in translational, rotational, or vibrational motion) contributes \(\frac{1}{2}k_BT\) of energy on average.
Monoatomic gas: \(f = 3\) (only translational)
\[ U = \frac{3}{2}RT \text{ per mole} \Rightarrow C_V = \frac{dU}{dT} = \frac{3R}{2} = 12.47 \text{ J/mol·K} \]
\[ C_P = C_V + R = \frac{5R}{2} = 20.78 \text{ J/mol·K} \]
Section D – Competency-Based Questions
Q4. (Case Study) Hydrogen gas (\(M = 2 \times 10^{-3}\) kg/mol) is at 300 K.
(i) Find the rms speed of hydrogen molecules.
(ii) At what temperature will the rms speed of hydrogen be double that at 300 K?
(iii) Compare the rms speed of hydrogen and oxygen at the same temperature. (\(M_O = 32 \times 10^{-3}\) kg/mol)
(iv) Why do lighter gases escape from planetary atmospheres more easily?
Answer
(i) \(v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{2 \times 10^{-3}}} = \sqrt{3.74 \times 10^6} \approx \mathbf{1934 \text{ m/s}}\)
(ii) \(v_{rms} \propto \sqrt{T} \Rightarrow 2v_{rms} \propto \sqrt{T’}\)
\(\frac{v’}{v} = 2 = \sqrt{\frac{T’}{300}} \Rightarrow T’ = 4 \times 300 = \mathbf{1200 \text{ K}}\)
(iii) \(\frac{v_H}{v_O} = \sqrt{\frac{M_O}{M_H}} = \sqrt{\frac{32}{2}} = \sqrt{16} = \mathbf{4}\)
Hydrogen moves 4 times faster than oxygen at the same temperature.
(iv) Lighter gases have higher rms speeds at any given temperature. If the rms speed exceeds the escape velocity of the planet, the gas molecules escape into space. This is why planets with low gravity (like the Moon) cannot hold light gases like hydrogen or helium.