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Kinetic Theory of Gases



Introduction

The Kinetic Theory of Gases (KTG) is a microscopic model that explains the macroscopic properties of gases, such as pressure, volume, and temperature, in terms of the behavior of their constituent molecules. It bridges the gap between the microscopic world of atoms and molecules and the macroscopic world that we observe and measure.

KTG is based on the idea that gases are composed of a large number of tiny particles (molecules or atoms) that are in constant, random motion. The theory uses statistical methods to relate the average behavior of these particles to the observable properties of the gas.

The development of the kinetic theory, notably by scientists like James Clerk Maxwell and Ludwig Boltzmann, provided a strong theoretical foundation for the empirical gas laws (Boyle's Law, Charles's Law, etc.) and thermodynamics. It also offered insights into concepts like temperature as a measure of molecular kinetic energy and pressure as a result of molecular collisions with container walls.

While the KTG starts with a simplified model of an ideal gas, it provides a powerful framework that can be extended and refined to understand the behavior of real gases and other states of matter at a fundamental level.



Molecular Nature Of Matter

The Kinetic Theory of Gases is built upon the fundamental concept that matter is composed of particles – atoms and molecules – that are in constant motion. This is the core idea of the molecular theory of matter.


Evidence for the Molecular Nature of Matter

While we cannot directly see individual atoms or molecules with the naked eye, there is overwhelming indirect evidence supporting the molecular nature of matter:

These phenomena, and many others, are best explained by the concept that matter is made up of discrete particles in motion.


Size and Number of Molecules

Atoms and molecules are extremely small. Typical atomic sizes are on the order of $10^{-10}$ metres (0.1 nanometre). A single drop of water contains an enormous number of molecules. The number of molecules in one mole of any substance is given by Avogadro's number, $N_A \approx 6.022 \times 10^{23}$. A mole of water (18 grams) is about 18 ml, which is a few drops. Thus, even small amounts of matter contain a vast number of molecules.

The Kinetic Theory of Gases deals with systems containing a large number of molecules, where statistical averages are meaningful and predictable, even though the motion of individual molecules is random.



Behaviour Of Gases

Gases exhibit several characteristic macroscopic behaviours that the Kinetic Theory of Gases aims to explain from a microscopic perspective. These behaviours are distinct from those of solids and liquids.


Key Characteristics of Gases


Distinction from Solids and Liquids

The key differences in the macroscopic behaviour of gases, liquids, and solids arise from the differences in the strength of intermolecular forces and the freedom of motion of the particles:

Property Solid Liquid Gas
Shape Fixed Takes shape of container Takes shape of container
Volume Fixed Fixed (under constant T & P) Takes volume of container
Compressibility Very Low Low High
Intermolecular Forces Strong Moderate Weak
Particle Motion Vibration about fixed positions Random motion, sliding past each other Random, rapid motion, large distances between particles
Diffusion Very Slow Slow Rapid

The simplicity of the interactions in the gaseous state (molecules far apart, weak forces except during collisions) makes it amenable to theoretical modelling using the Kinetic Theory of Gases, especially for ideal gases.



Kinetic Theory Of An Ideal Gas

The Kinetic Theory of Gases aims to derive the macroscopic properties of an ideal gas from the microscopic behaviour of its molecules. It provides a physical explanation for the ideal-gas equation ($PV = nRT$) and the concept of temperature.


Postulates of the Kinetic Theory of Ideal Gases

The theory is based on a set of simplifying assumptions (postulates) about the nature and behaviour of ideal gas molecules:

  1. A gas consists of a very large number of identical molecules, which are treated as point particles (their volume is negligible compared to the container volume).
  2. The molecules are in constant, random motion. They move in straight lines between collisions.
  3. Collisions between molecules and with the walls of the container are perfectly elastic (total kinetic energy is conserved during collisions).
  4. The duration of a collision is negligible compared to the time between collisions.
  5. There are no forces of attraction or repulsion between molecules, except during collisions.
  6. Newton's laws of motion apply to the motion of the molecules.

Pressure Of An Ideal Gas

According to the kinetic theory, the pressure exerted by a gas on the walls of its container is due to the average force per unit area exerted by the molecules as they collide with the walls and bounce back.


Derivation of Pressure Formula

Consider an ideal gas of $N$ molecules, each of mass $m$, enclosed in a cubical container of volume $V = L^3$. Let the molecule's velocity be $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$. When a molecule collides elastically with the wall perpendicular to the x-axis (say, at $x=L$), its x-component of velocity changes from $v_x$ to $-v_x$, while $v_y$ and $v_z$ remain unchanged. The change in momentum of the molecule is $\Delta \vec{p} = (-mv_x) - (mv_x) = -2mv_x$ in the x-direction.

By Newton's Third Law, the momentum imparted to the wall is $2mv_x$. The time taken for the molecule to travel across the box and back to collide with the same wall is $\Delta t = 2L/v_x$ (assuming no collisions with other molecules, a simplification that doesn't affect the average force). The rate at which momentum is transferred to this wall by this single molecule is $\frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}$.

The average force exerted by this molecule on the wall is $\frac{mv_x^2}{L}$. To find the total force on the wall due to all $N$ molecules, we sum this over all molecules and take the average:

$ F_{total, x} = \sum_{i=1}^{N} \frac{m(v_{xi})^2}{L} = \frac{m}{L} \sum_{i=1}^{N} v_{xi}^2 $

The average value of the square of the x-component of velocity is $\overline{v_x^2} = \frac{1}{N} \sum_{i=1}^{N} v_{xi}^2$. So, $\sum v_{xi}^2 = N \overline{v_x^2}$.

$ F_{total, x} = \frac{m}{L} N \overline{v_x^2} $

The pressure on the wall perpendicular to the x-axis is the force divided by the area of the wall, $A = L^2$.

$ P = \frac{F_{total, x}}{A} = \frac{m N \overline{v_x^2}/L}{L^2} = \frac{m N \overline{v_x^2}}{L^3} = \frac{m N \overline{v_x^2}}{V} $

Since the motion is random, the average kinetic energy associated with motion in each direction is the same: $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$. The mean square speed $\overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2} = 3\overline{v_x^2}$. Thus, $\overline{v_x^2} = \frac{1}{3}\overline{v^2}$.

Substituting this into the pressure formula:

$ P = \frac{m N (\frac{1}{3}\overline{v^2})}{V} = \frac{1}{3} \frac{N m \overline{v^2}}{V} $

This is the formula for the pressure exerted by an ideal gas based on the kinetic theory. $m\overline{v^2}$ is related to the average kinetic energy of the molecules. The term $\overline{v^2}$ is the mean square speed, and $\sqrt{\overline{v^2}}$ is the root-mean-square (rms) speed of the molecules.

The average translational kinetic energy of a molecule is $\overline{KE}_{trans} = \frac{1}{2}m\overline{v^2}$. So, $m\overline{v^2} = 2 \overline{KE}_{trans}$.

Substituting this into the pressure formula:

$ P = \frac{1}{3} \frac{N (2 \overline{KE}_{trans})}{V} = \frac{2}{3} \frac{N \overline{KE}_{trans}}{V} $

$ PV = \frac{2}{3} N \overline{KE}_{trans} $

This important result relates the macroscopic product PV to the total translational kinetic energy of the gas molecules ($N \overline{KE}_{trans}$).


Kinetic Interpretation Of Temperature

Comparing the result from the kinetic theory ($PV = \frac{2}{3} N \overline{KE}_{trans}$) with the ideal-gas equation ($PV = N k_B T$), we can establish a relationship between the average translational kinetic energy of the gas molecules and the absolute temperature of the gas.

$ \frac{2}{3} N \overline{KE}_{trans} = N k_B T $

Cancel $N$ from both sides:

$ \frac{2}{3} \overline{KE}_{trans} = k_B T $

Rearranging for $\overline{KE}_{trans}$:

$ \overline{KE}_{trans} = \frac{3}{2} k_B T $

This is a fundamental result of the kinetic theory. It shows that the average translational kinetic energy of a molecule in an ideal gas is directly proportional to the absolute temperature of the gas.

This provides a microscopic interpretation of temperature: Temperature is a measure of the average translational kinetic energy of the molecules in a gas.

At absolute zero (T = 0 K), the average translational kinetic energy is zero. This does not mean the molecules stop moving entirely, but their random translational motion is at a minimum (related to zero-point energy in quantum mechanics). At higher temperatures, molecules move faster on average.

The total internal energy ($U$) of an ideal monatomic gas (like Helium, Neon, Argon) is solely due to the translational kinetic energy of its atoms. For $N$ atoms:

$ U = N \overline{KE}_{trans} = N \left(\frac{3}{2} k_B T\right) = \frac{3}{2} N k_B T $

Using $N k_B = n R$: $ U = \frac{3}{2} n R T $.

For polyatomic gases, the internal energy also includes rotational and vibrational kinetic energy, as explained by the Law of Equipartition of Energy.

Example 1. Calculate the average translational kinetic energy of a molecule in an ideal gas at 27°C. (Boltzmann constant $k_B = 1.38 \times 10^{-23}$ J/K).

Answer:

Temperature, $T_C = 27^\circ\text{C}$. Convert to Kelvin: $T_K = T_C + 273.15 = 27 + 273.15 = 300.15$ K. (Use 300 K for calculation).

Boltzmann constant, $k_B = 1.38 \times 10^{-23}$ J/K.

The average translational kinetic energy per molecule is given by $\overline{KE}_{trans} = \frac{3}{2} k_B T$.

$ \overline{KE}_{trans} = \frac{3}{2} \times (1.38 \times 10^{-23} \text{ J/K}) \times (300 \text{ K}) $

$ \overline{KE}_{trans} = 1.5 \times 1.38 \times 300 \times 10^{-23} $ J

$ \overline{KE}_{trans} = 1.5 \times 414 \times 10^{-23} $ J

$ \overline{KE}_{trans} = 621 \times 10^{-23} $ J

$ \overline{KE}_{trans} = 6.21 \times 10^{-21} $ J.

Using $T = 300.15$ K gives a slightly different value. The average translational kinetic energy of a molecule at 27°C is approximately $6.21 \times 10^{-21}$ Joules. This is a very small amount of energy per molecule, but a macroscopic amount when summed over Avogadro's number of molecules.



Law Of Equipartition Of Energy

The Law of Equipartition of Energy is a principle from statistical mechanics that helps determine how the total internal energy of a system is distributed among the various degrees of freedom of its constituent particles. It is particularly useful in extending the results of the kinetic theory from simple monatomic gases to more complex polyatomic gases and even solids.


Statement of the Law

The Law of Equipartition of Energy states that, in thermal equilibrium, the total energy of a system is equally distributed among all its degrees of freedom, and the energy associated with each degree of freedom per molecule is $\frac{1}{2} k_B T$, where $T$ is the absolute temperature and $k_B$ is the Boltzmann constant.

For a system of $N$ molecules in thermal equilibrium at temperature $T$, the total energy associated with a particular mode of energy (translational, rotational, vibrational) is distributed equally among all degrees of freedom for that mode.


Degrees of Freedom (f)

A degree of freedom is an independent mode by which a molecule can possess energy. For a molecule, these modes correspond to its possible ways of moving or storing energy.

The total number of degrees of freedom $f$ for a molecule is the sum of its translational, rotational, and vibrational degrees of freedom.

Total average energy of a molecule = $f \times \frac{1}{2} k_B T$.

Total internal energy of $N$ molecules = $U = N \times f \times \frac{1}{2} k_B T = \frac{f}{2} N k_B T$.

Using $N k_B = n R$: $ U = \frac{f}{2} n R T $.

The Law of Equipartition of Energy is a powerful result for calculating the internal energy of gases and predicting their specific heat capacities, provided the relevant degrees of freedom are active at that temperature.



Specific Heat Capacity

The specific heat capacity of a substance tells us how much heat is needed to raise the temperature of a unit mass (or unit mole) by one degree. Using the concept of internal energy from the kinetic theory and the Law of Equipartition of Energy, we can theoretically predict the specific heat capacities of gases and solids.


Specific Heat Capacities of Gases (from KTG)

For gases, we typically use molar heat capacities, $C_v$ (at constant volume) and $C_p$ (at constant pressure).

The internal energy of $n$ moles of an ideal gas is $U = \frac{f}{2} n R T$, where $f$ is the total number of active degrees of freedom per molecule.

From the First Law of Thermodynamics, for a process at constant volume, $Q = \Delta U + W$. Since volume is constant, work done $W = 0$. So, $Q = \Delta U$. The molar heat capacity at constant volume is $C_v = \frac{1}{n} \left(\frac{\partial Q}{\partial T}\right)_V = \frac{1}{n} \left(\frac{\partial U}{\partial T}\right)_V$.

$ C_v = \frac{1}{n} \frac{\partial}{\partial T} \left(\frac{f}{2} n R T\right) = \frac{1}{n} \frac{f}{2} n R \frac{\partial T}{\partial T} = \frac{f}{2} R $

$ \mathbf{C_v = \frac{f}{2} R} $

For a process at constant pressure, $Q = \Delta U + W$. The work done by the gas during expansion is $W = P \Delta V$. From the ideal gas law, $PV = nRT$, so $P \Delta V = nR \Delta T$ for a process at constant pressure. The First Law becomes $Q = \Delta U + nR \Delta T$. The molar heat capacity at constant pressure is $C_p = \frac{1}{n} \left(\frac{\partial Q}{\partial T}\right)_P = \frac{1}{n} \left(\left(\frac{\partial U}{\partial T}\right)_P + nR\right)$. For an ideal gas, internal energy depends only on temperature, so $\left(\frac{\partial U}{\partial T}\right)_P = \frac{dU}{dT} = \frac{d}{dT} \left(\frac{f}{2} n R T\right) = \frac{f}{2} n R$.

$ C_p = \frac{1}{n} \left(\frac{f}{2} n R + nR\right) = \frac{f}{2} R + R = \left(\frac{f}{2} + 1\right) R $

$ \mathbf{C_p = \left(\frac{f}{2} + 1\right) R} $

From these, we recover the relationship $C_p - C_v = R$ for an ideal gas.

The ratio of specific heats is $\gamma = \frac{C_p}{C_v} = \frac{(f/2 + 1)R}{(f/2)R} = 1 + \frac{2}{f}$. The value of $\gamma$ depends on the number of active degrees of freedom $f$.


Monatomic Gases

Monatomic gases (He, Ne, Ar, etc.) have 3 translational degrees of freedom ($f=3$). Rotational and vibrational degrees of freedom are negligible.

$ C_v = \frac{3}{2} R $

$ C_p = \left(\frac{3}{2} + 1\right) R = \frac{5}{2} R $

$ \gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3} \approx 1.67 $

These predictions match experimental values for monatomic gases well.


Diatomic Gases

Diatomic gases (O$_2$, N$_2$, H$_2$, etc.) have 3 translational and 2 rotational degrees of freedom ($f_{trans}=3, f_{rot}=2$). At ordinary temperatures, vibrational degrees of freedom are often inactive ($f_{vib}=0$). Total $f = 3 + 2 + 0 = 5$.

$ C_v = \frac{5}{2} R $

$ C_p = \left(\frac{5}{2} + 1\right) R = \frac{7}{2} R $

$ \gamma = \frac{C_p}{C_v} = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4 $

At higher temperatures, vibrational modes become active. A diatomic molecule has one vibrational mode (stretching). Each mode has 2 degrees of freedom (kinetic and potential). So, $f_{vib}=2$. Total $f = 3 + 2 + 2 = 7$.

At high temperatures: $ C_v = \frac{7}{2} R, C_p = \frac{9}{2} R, \gamma = \frac{9}{7} \approx 1.28 $.

The specific heat capacities of diatomic gases vary with temperature due to the activation of vibrational modes.


Polyatomic Gases

Polyatomic gases (like CO$_2$, NH$_3$, CH$_4$) have more degrees of freedom, depending on their structure (linear or non-linear).

For polyatomic gases, the number of active vibrational modes depends strongly on temperature. At ordinary temperatures, vibrations are often partially or fully inactive, leading to specific heats lower than predicted by including all vibrational degrees of freedom.

In general, for a gas with $f$ active degrees of freedom per molecule: $ C_v = \frac{f}{2} R, C_p = \left(\frac{f}{2} + 1\right) R, \gamma = 1 + \frac{2}{f} $.


Specific Heat Capacity (per unit mass)

The specific heat capacity per unit mass ($c$) is related to the molar heat capacity ($C_m$) by $C_m = c \times M_{molar}$, where $M_{molar}$ is the molar mass of the substance (mass of one mole in kg/mol). For example, $c_v = C_v/M_{molar}$.


Specific Heat Capacity Of Solids

In a solid, atoms are fixed in position and vibrate about their equilibrium positions. A solid can be modelled as a collection of atoms connected by springs. In 3D, each atom can vibrate in three independent directions (x, y, z). Each vibrational mode has both kinetic and potential energy. According to the equipartition principle, each vibrational degree of freedom contributes $2 \times \frac{1}{2}k_B T = k_B T$ ( $\frac{1}{2}k_B T$ for kinetic energy and $\frac{1}{2}k_B T$ for potential energy).

For a solid consisting of $N$ atoms, there are $3N$ possible vibration directions, so $3N$ vibrational degrees of freedom. Assuming all modes are active (valid at high temperatures, e.g., room temperature for many solids), the total internal energy is:

$ U = 3N \times k_B T $

For $n$ moles, $N = n N_A$. $U = 3 n N_A k_B T = 3 n R T$.

The molar heat capacity at constant volume ($C_v$) for a solid is the rate of change of internal energy with temperature (volume of solid is nearly constant with temperature compared to gases):

$ C_v = \frac{1}{n} \frac{dU}{dT} = \frac{1}{n} \frac{d}{dT}(3nRT) = 3R $

This is the Dulong-Petit Law, which states that the molar heat capacity at constant volume for most solids is approximately $3R \approx 3 \times 8.314 = 24.9$ J/mol·K at room temperature and above. This law holds well for many metals and other solid elements at relatively high temperatures.

At low temperatures, the heat capacity of solids deviates from the Dulong-Petit law and approaches zero as temperature approaches 0 K. This behaviour is explained by quantum mechanics (Einstein and Debye theories), which show that vibrational modes are quantized and require a minimum energy to be excited, so they are 'frozen out' at low temperatures.


Specific Heat Capacity Of Water

Water is a liquid, and its specific heat capacity is significantly higher than most other common substances, especially solids and gases (per unit mass). The specific heat capacity of liquid water is approximately 4186 J/kg·°C (or J/kg·K) at 1 atmosphere and 15°C. This means it takes 4186 Joules of energy to raise the temperature of 1 kg of water by 1°C.

The high specific heat of water is attributed to the strong hydrogen bonding between water molecules. A considerable amount of energy is required to break or disrupt these hydrogen bonds as temperature increases, in addition to increasing the kinetic energy of the molecules. This energy absorption capability makes water an excellent heat storage medium and coolant.

The specific heat capacity of water varies slightly with temperature and pressure, but 4186 J/kg·°C is a commonly used value. The specific heat capacities of ice (solid water) and steam (gaseous water) are lower than that of liquid water.



Mean Free Path

In the Kinetic Theory of Gases, molecules are assumed to be in constant random motion, colliding with each other and with the walls of the container. The distance a molecule travels between successive collisions is called its free path. Because the molecules are moving randomly and colliding frequently, the free paths vary. The average distance travelled by a molecule between successive collisions is called the mean free path ($\lambda$).


Factors Affecting Mean Free Path

The mean free path depends on:


Formula for Mean Free Path

Assuming molecules are hard spheres of diameter $d$, and considering their relative velocities, the theoretical formula for the mean free path is:

$ \lambda = \frac{1}{\sqrt{2} n \pi d^2} $

where $n$ is the number density of molecules (number of molecules per unit volume, $n = N/V$).

Using the ideal gas law $PV = Nk_B T$, the number density $n = N/V = P/(k_B T)$. Substituting this into the formula for $\lambda$:

$ \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} $

This formula shows that the mean free path is:

At standard temperature and pressure (STP), the mean free path of air molecules is typically on the order of $10^{-7}$ metres (about 100 nanometres). This is much larger than the size of the molecules themselves (around $10^{-10}$ m), which justifies the ideal gas assumption that the volume of molecules is negligible compared to the volume of the container.


Significance of Mean Free Path

Example 1. Estimate the mean free path of air molecules at standard temperature and pressure (STP: 0°C and 1 atm). Assume the effective diameter of an air molecule is $3 \times 10^{-10}$ m. (STP pressure $P = 1.013 \times 10^5$ Pa, $k_B = 1.38 \times 10^{-23}$ J/K, $T = 0^\circ\text{C} = 273.15$ K).

Answer:

Pressure, $P = 1.013 \times 10^5$ Pa.

Temperature, $T = 273.15$ K.

Molecular diameter, $d = 3 \times 10^{-10}$ m.

Boltzmann constant, $k_B = 1.38 \times 10^{-23}$ J/K.

Using the formula for mean free path $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$:

$ \lambda = \frac{(1.38 \times 10^{-23} \text{ J/K}) \times (273.15 \text{ K})}{\sqrt{2} \times 3.14159 \times (3 \times 10^{-10} \text{ m})^2 \times (1.013 \times 10^5 \text{ Pa})} $

$ \lambda = \frac{376.95 \times 10^{-23}}{1.414 \times 3.14159 \times (9 \times 10^{-20}) \times (1.013 \times 10^5)} $ m

$ \lambda = \frac{376.95 \times 10^{-23}}{4.44 \times 9 \times 1.013 \times 10^{-15}} $ m

$ \lambda = \frac{376.95 \times 10^{-23}}{40.55 \times 10^{-15}} $ m

$ \lambda \approx 9.296 \times 10^{(-23 - (-15))} $ m

$ \lambda \approx 9.296 \times 10^{-8} $ m.

The estimated mean free path of air molecules at STP is approximately $9.3 \times 10^{-8}$ metres, or 93 nanometres. This is indeed much larger than the molecular size, supporting the assumptions of KTG at these conditions.