Kinetic theory of gases

1. Overview

The kinetic theory of gases is a microscopic model that explains the macroscopic properties of gases—such as pressure, volume, and temperature—by considering the motion of individual molecules. It treats a gas as a collection of a vast number of particles (atoms or molecules) in constant, random motion. By applying Newtonian mechanics to these particles, we can derive the ideal gas laws and establish that the absolute temperature of a gas is a direct measure of the average translational kinetic energy of its molecules. This theory provides the physical justification for the behavior of an ideal gas and defines the limits of the gas laws.

---

Key Definitions

• Ideal Gas: A theoretical gas that obeys the equation of state $pV = nRT$ (or $pV = NkT$) at all pressures, temperatures, and volumes. It consists of point masses with no intermolecular forces.
• Root-mean-square (r.m.s.) speed ($c_{\text{rms}}$): The square root of the average of the squares of the speeds of the molecules. It is the effective speed used to calculate the average kinetic energy of the gas particles.
• Mean-square speed ($\langle c^2 \rangle$): The average value of the square of the speeds of all the particles in the gas. Note that $\langle c^2 \rangle$ is not the same as the square of the average speed.
• Boltzmann Constant ($k$): The gas constant per molecule, defined as $k = \frac{R}{N_A}$. It relates the average kinetic energy of the particles to the thermodynamic temperature.
• Absolute Zero: The temperature at which the internal energy of an ideal gas is at its minimum. At $0\text{ K}$, the translational kinetic energy of the molecules is zero.
• Avogadro Constant ($N_A$): The number of constituent particles per mole of substance. Since 2019 its value has been fixed at exactly $6.02214076 \times 10^{23}\text{ mol}^{-1}$, approximately $6.02 \times 10^{23}\text{ mol}^{-1}$.

---

Content

3.1 Assumptions of the Kinetic Theory

To model the behavior of a gas mathematically, we make several simplifying assumptions. These are essential for the derivation of the pressure equation and define the "Ideal Gas" model:

1. Large Number of Particles: The gas contains a very large number of molecules moving in random, rapid motion. This allows for statistical averages to be meaningful.
2. Negligible Volume: The total volume of the molecules themselves is negligible compared to the total volume of the container (the molecules are treated as point masses).
3. Elastic Collisions: All collisions between molecules, and between molecules and the container walls, are perfectly elastic. This means kinetic energy is conserved during collisions.
4. No Intermolecular Forces: There are no forces of attraction or repulsion between molecules, except during the negligible time of a collision. This implies the internal energy of the gas is entirely kinetic.
5. Negligible Collision Duration: The time spent in a collision is negligible compared to the time interval between collisions.
6. Newtonian Mechanics: The molecules obey Newton’s laws of motion.

3.2 Molecular Movement and Pressure

Pressure is a macroscopic property resulting from microscopic collisions.

• Momentum Change: When a gas molecule collides with the wall of a container, it undergoes a change in momentum. For an elastic collision against a stationary wall, the velocity component perpendicular to the wall is reversed.
• Force and Newton’s Second Law: According to Newton’s Second Law, the force exerted on the molecule is the rate of change of momentum ($F = \frac{\Delta p}{\Delta t}$).
• Newton’s Third Law: The molecule exerts an equal and opposite force on the wall.
• Pressure: Since pressure is force per unit area ($P = \frac{F}{A}$), the sum of the forces from millions of collisions per second results in a steady, uniform pressure on the container walls.

3.3 Derivation of the Pressure Equation

The syllabus requires the derivation of $pV = \frac{1}{3}Nm\langle c^2 \rangle$ using a 1D model extended to 3D.

Step 1: Momentum change of one molecule
Consider a cube of side $L$ containing $N$ molecules of mass $m$. A molecule moving with velocity $c$ has components $c_x, c_y, c_z$.
When it hits a wall perpendicular to the $x$-axis:

• Initial momentum = $mc_x$
• Final momentum = $-mc_x$
• Change in momentum $\Delta p = -mc_x - (mc_x) = -2mc_x$
• Momentum imparted to the wall = $2mc_x$

Step 2: Time between collisions
The molecule must travel to the opposite wall and back to hit the same wall again.

• Distance = $2L$
• Time interval $\Delta t = \frac{2L}{c_x}$

Step 3: Force from one molecule
Using $F = \frac{\Delta p}{\Delta t}$:
$F = \frac{2mc_x}{\frac{2L}{c_x}} = \frac{mc_x^2}{L}$

Step 4: Total pressure for $N$ molecules
Pressure $P = \frac{\text{Total Force}}{\text{Area}} = \frac{\sum F}{L^2}$.
$P = \frac{\sum \frac{mc_x^2}{L}}{L^2} = \frac{m}{L^3} \sum c_x^2$
Since $L^3 = V$ (Volume) and the mean-square speed in the $x$-direction is $\langle c_x^2 \rangle = \frac{\sum c_x^2}{N}$:
$P = \frac{Nm\langle c_x^2 \rangle}{V}$

Step 5: Extension to three dimensions
By Pythagoras' theorem: $c^2 = c_x^2 + c_y^2 + c_z^2$.
Because the motion is random, there is no preferred direction: $\langle c_x^2 \rangle = \langle c_y^2 \rangle = \langle c_z^2 \rangle$.
Therefore, $\langle c^2 \rangle = 3\langle c_x^2 \rangle$, which means $\langle c_x^2 \rangle = \frac{1}{3}\langle c^2 \rangle$.

Final Equation:
$pV = \frac{1}{3}Nm\langle c^2 \rangle$

3.4 Root-Mean-Square Speed ($c_{\text{rms}}$)

The $c_{\text{rms}}$ is a useful "average" speed because it relates directly to the pressure and kinetic energy of the gas.
$c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$

From the pressure equation, we can also derive:
$p = \frac{1}{3}\rho \langle c^2 \rangle$ (where density $\rho = \frac{Nm}{V}$)
This shows that for a constant density, the pressure is directly proportional to the mean-square speed of the molecules.

3.5 Average Translational Kinetic Energy

We can link the microscopic motion to the macroscopic temperature by comparing the two ideal gas equations:

1. $pV = NkT$ (Macroscopic/Thermodynamic)
2. $pV = \frac{1}{3}Nm\langle c^2 \rangle$ (Microscopic/Kinetic)

Equating them:
$NkT = \frac{1}{3}Nm\langle c^2 \rangle$
$kT = \frac{1}{3}m\langle c^2 \rangle$

The formula for kinetic energy is $E_k = \frac{1}{2}mv^2$. To find the average translational kinetic energy $\langle E_k \rangle$:
$\langle E_k \rangle = \frac{1}{2}m\langle c^2 \rangle$

Rearranging the equated expression:
$\frac{3}{2}kT = \frac{1}{2}m\langle c^2 \rangle$
$\langle E_k \rangle = \frac{3}{2}kT$

Crucial Insight: The average translational kinetic energy of a molecule is directly proportional to the thermodynamic temperature $T$ (in Kelvin). This relationship is independent of the mass of the molecule. At a given temperature, a heavy molecule (like $CO_2$) will have the same average kinetic energy as a light molecule (like $H_2$), but the lighter molecule will have a higher $c_{\text{rms}}$.

---

Worked Example 1 — Calculating RMS Speed

A container holds Neon gas (molar mass $M = 20.2\text{ g mol}^{-1}$) at a temperature of $100^{\circ}\text{C}$. Calculate the root-mean-square speed of the Neon atoms.

Step 1: Convert units to SI
$T = 100 + 273.15 = 373.15\text{ K}$
$M = 20.2 \times 10^{-3}\text{ kg mol}^{-1}$

Step 2: Find the mass of one atom ($m$)
$m = \frac{M}{N_A} = \frac{20.2 \times 10^{-3}}{6.02 \times 10^{23}} = 3.355 \times 10^{-26}\text{ kg}$

Step 3: Use the kinetic energy relationship
$\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT$
$\langle c^2 \rangle = \frac{3kT}{m}$
$\langle c^2 \rangle = \frac{3 \times (1.38 \times 10^{-23}) \times 373.15}{3.355 \times 10^{-26}}$
$\langle c^2 \rangle = 4.61 \times 10^5\text{ m}^2\text{ s}^{-2}$

Step 4: Calculate $c_{\text{rms}}$
$c_{\text{rms}} = \sqrt{4.61 \times 10^5} = 679\text{ m s}^{-1}$

---

Worked Example 2 — Internal Energy of a Gas

Calculate the total internal energy of $2.5\text{ moles}$ of a monatomic ideal gas at a temperature of $300\text{ K}$.

Step 1: Identify the principle
For an ideal gas, there are no intermolecular forces, so potential energy is zero. For a monatomic gas, the internal energy $U$ is the sum of the translational kinetic energies of all $N$ atoms.
$U = N \times \langle E_k \rangle = N \times \frac{3}{2}kT$

Step 2: Use the molar relationship
Since $N = nN_A$ and $k = \frac{R}{N_A}$, then $Nk = nR$.
$U = \frac{3}{2}nRT$

Step 3: Substitution
$U = \frac{3}{2} \times 2.5 \times 8.31 \times 300$
$U = 1.5 \times 2.5 \times 8.31 \times 300$
$U = 9348.75\text{ J}$

Answer:
$U = 9350\text{ J}$ (to 3 s.f.)

---

Key Equations

Equation	Symbols	SI Units	Status
$pV = \frac{1}{3}Nm\langle c^2 \rangle$	$N$: number of molecules, $m$: mass of one molecule	$p$: Pa, $V$: m³, $m$: kg	Data Sheet
$\langle E_k \rangle = \frac{3}{2}kT$	$\langle E_k \rangle$: avg translational KE, $k$: Boltzmann constant	$\langle E_k \rangle$: J, $k$: J K⁻¹	Memorise
$k = \frac{R}{N_A}$	$R$: molar gas constant, $N_A$: Avogadro constant	$R$: J K⁻¹ mol⁻¹	Memorise
$c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$	$c_{\text{rms}}$: root-mean-square speed	m s⁻¹	Memorise
$p = \frac{1}{3}\rho \langle c^2 \rangle$	$\rho$: density of the gas	kg m⁻³	Memorise
$U = \frac{3}{2}NkT = \frac{3}{2}nRT$	$U$: Total internal energy of a monatomic ideal gas	J	Memorise

---

Common Mistakes to Avoid

• ❌ Wrong: Using Celsius ($^{\circ}\text{C}$) for temperature in any kinetic theory equation.
  • ✅ Right: Always use Kelvin ($T = \theta + 273.15$). A change in temperature $\Delta T$ is the same in both scales, but the absolute value $T$ is not.
• ❌ Wrong: Confusing $m$ (mass of one molecule) with $M$ (molar mass).
  • ✅ Right: In $pV = \frac{1}{3}Nm\langle c^2 \rangle$, $m$ is the mass of one single molecule. If given molar mass in grams, divide by $1000$ to get kg, then divide by $N_A$.
• ❌ Wrong: Forgetting to square the speed or take the square root.
  • ✅ Right: $\langle c^2 \rangle$ is the mean-square speed. If a question asks for $c_{\text{rms}}$, you must take the square root of $\langle c^2 \rangle$.
• ❌ Wrong: Thinking that different gases at the same temperature have the same speed.
  • ✅ Right: They have the same average kinetic energy. Since $E_k = \frac{1}{2}mv^2$, the gas with the smaller mass $m$ must have a higher speed $v$.
• ❌ Wrong: Assuming "Internal Energy" includes potential energy for an ideal gas.
  • ✅ Right: By definition, an ideal gas has zero intermolecular forces, so its internal energy is entirely kinetic energy (for a monatomic gas, all of it translational).

---

Exam Tips

1. The "Why" of Pressure: If an exam question asks "How do molecules exert a pressure?", your answer must include: random motion, collision with walls, change in momentum, force = rate of change of momentum, and pressure = force / area.
2. Proportionality is Key: Many multiple-choice questions rely on $c_{\text{rms}} \propto \sqrt{T}$. If the Kelvin temperature doubles, the $c_{\text{rms}}$ increases by a factor of $\sqrt{2} \approx 1.41$.
3. Internal Energy vs. Temperature: Remember that for an ideal gas, $U \propto T$. If the temperature is constant (isothermal), the internal energy is constant.
4. Derivation Steps: When deriving $pV = \frac{1}{3}Nm\langle c^2 \rangle$, don't skip the step explaining why $\langle c_x^2 \rangle = \frac{1}{3}\langle c^2 \rangle$. State clearly that this is due to random motion in three dimensions.
5. Mass of a molecule: If you need the mass of a molecule and you are given the gas (e.g., Oxygen, $O_2$), remember that oxygen is diatomic. The mass $m$ is the mass of the $O_2$ molecule, not a single $O$ atom. Use $M = 32\text{ g mol}^{-1}$, not $16\text{ g mol}^{-1}$.