The kinetic theory of gases explains the macroscopic behaviour of gases (such as pressure, temperature, and volume) by analyzing the microscopic motion of rapidly moving atoms or molecules. This theory bridges the gap between mechanics and thermodynamics, providing a molecular interpretation of thermodynamic variables.
Historical Chronicle: The Atomic Hypothesis
An ideal gas is a theoretical gas model whose molecules have negligible volume and no mutual interactions. Real gases approach ideal gas behaviour at low pressures and high temperatures .
The state of an ideal gas is governed by the relation:
$$PV = \mu R T$$Where:
Alternative forms of the ideal gas equation:
$$PV = k_B N T \quad \text{or} \quad P = k_B n T$$Where $k_B = \frac{R}{N_A} = 1.38 \times 10^{-23}\text{ J K}^{-1}$ is the Boltzmann constant , $N$ is the total number of molecules, and $n = \frac{N}{V}$ is the number density.
Mathematical Problem Statement: Estimate the ratio of the molecular volume to the total volume occupied by water vapour at $100^\circ\text{C}$ and $1\text{ atm}$ pressure. Given: Density of liquid water = $1000\text{ kg m}^{-3}$, Density of water vapour = $0.6\text{ kg m}^{-3}$.
Step-by-Step Algorithmic Solution Trace:
- The molecular volume is approximately equal to the volume occupied by the same mass of water in its liquid state, where molecules are closely packed.
- Let $M$ be a given mass of water. The volume in the liquid state is $V_{\text{liquid}} = \frac{M}{\rho_{\text{water}}}$.
- The volume in the vapour state is $V_{\text{vapour}} = \frac{M}{\rho_{\text{vapour}}}$.
- The ratio of molecular volume to total volume is: $$\frac{V_{\text{liquid}}}{V_{\text{vapour}}} = \frac{\rho_{\text{vapour}}}{\rho_{\text{water}}} = \frac{0.6}{1000} = 6 \times 10^{-4}$$
Final Evaluation Output: The ratio of molecular volume to total volume is $6 \times 10^{-4}$.
Consider a gas enclosed in a cube of side $l$ (volume $V = l^3$). Let a molecule of mass $m$ move with velocity $(v_x, v_y, v_z)$.
The pressure equation can be rewritten as:
$$PV = \frac{1}{3} N m \bar{v^2} = \frac{2}{3} N \left(\frac{1}{2} m \bar{v^2}\right)$$Comparing this with the ideal gas equation $PV = N k_B T$, we get:
$$\frac{1}{2} m \bar{v^2} = \frac{3}{2} k_B T$$This shows that the average translational kinetic energy of a molecule is directly proportional to the absolute temperature of the gas and is independent of its pressure, volume, or nature.
The total internal energy $E$ of a mole of ideal gas (purely translational) is:
$$E = \frac{3}{2} R T$$The root mean square speed $v_{\text{rms}}$ is defined as:
$$v_{\text{rms}} = \sqrt{\bar{v^2}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M_0}}$$Mathematical Problem Statement: A flask contains argon and chlorine in the ratio of 2:1 by mass at $27^\circ\text{C}$. Find the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed $v_{\text{rms}}$ of the molecules of the two gases. (Atomic mass of $\text{Ar} = 39.9\text{ u}$, Molecular mass of $\text{Cl}_2 = 70.9\text{ u}$).
Step-by-Step Algorithmic Solution Trace:
- The average kinetic energy per molecule of any ideal gas depends only on temperature: $E_{\text{avg}} = \frac{3}{2} k_B T$. Since both gases are at the same temperature ($27^\circ\text{C}$), the ratio of their average kinetic energies is $1:1$.
- The rms speed is given by $v_{\text{rms}} = \sqrt{\frac{3 R T}{M_0}}$.
- The ratio of rms speeds is inversely proportional to the square root of their molecular masses: $$\frac{v_{\text{rms, Ar}}}{v_{\text{rms, Cl}}} = \sqrt{\frac{M_{\text{Cl}}}{M_{\text{Ar}}}} = \sqrt{\frac{70.9}{39.9}} \approx 1.33$$
Final Evaluation Output: (i) Ratio of average kinetic energy = $1:1$, (ii) Ratio of rms speeds = $1.33$.
The law of equipartition of energy states that for any dynamic system in thermal equilibrium, the total energy is distributed equally among all its degrees of freedom, and the average energy associated with each quadratic degree of freedom is $\frac{1}{2} k_B T$.
The number of independent coordinates required to completely specify the position and configuration of a system is called its degrees of freedom .
Using the law of equipartition of energy, we can theoretically predict the molar specific heat capacities at constant volume ($C_v$) and constant pressure ($C_p$), and their ratio ($\gamma = \frac{C_p}{C_v}$).
| Gas Type | Degrees of Freedom ($f$) | Internal Energy ($U$) | $C_v$ ($\text{J mol}^{-1}\text{K}^{-1}$) | $C_p$ ($\text{J mol}^{-1}\text{K}^{-1}$) | $\gamma = \frac{C_p}{C_v}$ |
|---|---|---|---|---|---|
| Monatomic | 3 | $\frac{3}{2}RT$ | $\frac{3}{2}R \approx 12.5$ | $\frac{5}{2}R \approx 20.8$ | $\frac{5}{3} \approx 1.67$ |
| Diatomic (Rigid) | 5 | $\frac{5}{2}RT$ | $\frac{5}{2}R \approx 20.8$ | $\frac{7}{2}R \approx 29.1$ | $\frac{7}{5} = 1.40$ |
| Diatomic (Vibrational) | 7 | $\frac{7}{2}RT$ | $\frac{7}{2}R \approx 29.1$ | $\frac{9}{2}R \approx 37.4$ | $\frac{9}{7} \approx 1.29$ |
| Polyatomic | $3 + 3 + f$ | $(3+f)RT$ | $(3+f)R$ | $(4+f)R$ | $\frac{4+f}{3+f}$ |
In a solid, each atom is held in a lattice and vibrates about its mean position in 3 dimensions. Each 1D oscillation has 2 quadratic energy terms (kinetic and potential), so a 3D oscillation has 6 quadratic terms.
Mathematical Problem Statement: A cylinder of fixed capacity $44.8\text{ litres}$ contains helium gas at standard temperature and pressure (STP). What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15.0^\circ\text{C}$? (Given $R = 8.31\text{ J mol}^{-1}\text{K}^{-1}$).
Step-by-Step Algorithmic Solution Trace:
- At STP, 1 mole of any ideal gas occupies $22.4\text{ litres}$. Thus, the number of moles of helium is: $$\mu = \frac{44.8}{22.4} = 2\text{ moles}$$
- Since the volume of the cylinder is fixed, the process is isochoric (constant volume). The heat required is given by: $$Q = \mu C_v \Delta T$$
- Helium is a monatomic gas, so its molar specific heat at constant volume is $C_v = \frac{3}{2} R$.
- Substitute the values: $$Q = 2 \times \left(\frac{3}{2} R\right) \times 15.0 = 45 R$$
- Calculate the final value: $$Q = 45 \times 8.31 = 374\text{ J}$$
Final Evaluation Output: The amount of heat needed is $374\text{ J}$.
The mean free path ($l$) is the average distance a gas molecule travels between two successive collisions.
Mathematical Problem Statement: Estimate the mean free path for a water molecule in water vapour at $373\text{ K}$. Use the molecular diameter $d = 2 \times 10^{-10}\text{ m}$ and the number density at STP adjusted for $373\text{ K}$.
Step-by-Step Algorithmic Solution Trace:
- The number density $n$ is inversely proportional to absolute temperature. At STP ($273\text{ K}$), $n_0 \approx 2.7 \times 10^{25}\text{ m}^{-3}$.
- At $373\text{ K}$, the adjusted number density is: $$n = 2.7 \times 10^{25} \times \frac{273}{373} \approx 2 \times 10^{25}\text{ m}^{-3}$$
- Using the exact mean free path formula: $$l = \frac{1}{\sqrt{2} n \pi d^2}$$
- Substitute the values ($d = 2 \times 10^{-10}\text{ m}$): $$l = \frac{1}{1.414 \times (2 \times 10^{25}) \times 3.1416 \times (2 \times 10^{-10})^2}$$ $$l = \frac{1}{1.414 \times 2 \times 10^{25} \times 3.1416 \times 4 \times 10^{-20}} \approx 4 \times 10^{-7}\text{ m}$$
Final Evaluation Output: The estimated mean free path is $4 \times 10^{-7}\text{ m}$.