Kinetic theory of temperature and pressure

Gas Atoms in Motion: Collisions and Forces

Every gas consists of enormous numbers of atoms moving randomly in all directions. Understanding how these atoms interact — with each other and with container walls — explains where pressure and temperature come from at the microscopic level.

Atoms in a gas undergo constant, random motion. They collide with one another and with the walls of whatever container holds them. Each collision involves a contact force between the two colliding objects, whether that is two atoms or an atom and a wall. Between collisions, ideal gas atoms travel in straight lines at constant speed because no net force acts on them.

These atomic collisions obey conservation of momentum. In the absence of a net external force, the total momentum of the colliding pair is constant before and after the collision. For a pair of gas atoms, you analyze the collision exactly as you would any two-object collision problem — using momentum conservation in one or two dimensions.

When an atom strikes a rigid container wall (a fixed object), the wall’s enormous mass means the atom essentially bounces back while the wall barely moves. The wall exerts an impulse on the atom, reversing the atom’s perpendicular velocity component. By Newton’s third law, the atom exerts an equal-magnitude, opposite-direction force on the wall.

MisconceptionStudents often think gas atoms only push on the container walls. Atoms also collide with and exert forces on each other throughout the volume of the gas.

Exam TipWhen asked about atomic interactions, mention both atom–atom and atom–wall collisions.

Pressure from Atomic Collisions

Key Equations

Pressure (definition):

$$P=\frac{{F}_{⟂}}{A}$$

Variables:

• $P$ = pressure (Pa)
• ${F}_{⟂}$ = total perpendicular force on the surface (N)
• $A$ = area of the surface (m²)

SI unit: pascal (Pa), where 1 Pa = 1 N/m²

Rearrangements:

$${F}_{⟂}=PA$$

$$A=\frac{{F}_{⟂}}{P}$$

ProportionalityPressure is directly proportional to the total perpendicular force and inversely proportional to area.

Reference sheet status: On the reference sheet.

Pressure is not some mysterious property — it arises from a clear causal chain at the atomic level. Atoms move randomly → they collide with a surface → each collision exerts a small perpendicular force on that surface → billions of collisions occur per unit area per unit time → the total perpendicular force per unit area equals the pressure.

More precisely, the pressure a gas exerts on a surface equals the sum of the magnitudes of the perpendicular components of the forces from all atomic collisions on that surface, divided by the surface’s area. Only the force component perpendicular to the surface contributes to pressure. A glancing collision contributes less pressure than a head-on collision because the perpendicular component is smaller.

A critical point: pressure exists throughout the gas itself, not just at the boundary between gas and container. Any imaginary surface drawn inside the gas would experience equal atomic bombardment from both sides. This is why a pressure gauge placed anywhere inside the gas reads the same value (for a gas in equilibrium at uniform density).

If the gas atoms move faster on average, each collision delivers a larger impulse. If the gas is compressed into a smaller volume, atoms strike each wall more frequently. Both effects increase pressure.

Examiner InsightFRQs often ask you to connect macroscopic pressure to microscopic atomic behavior. Always state the full chain: random motion → collisions → force → force per unit area.

Exam TipNever skip the “perpendicular component” detail — it distinguishes a complete answer from a partial one.

Worked Example: Estimating Pressure from Atomic Forces

Scenario

The atoms of a gas exert a combined perpendicular force of 450 N on a container wall that has an area of 0.030 m².

Step-by-step solution

Equation used — pressure definition

$$P=\frac{{F}_{⟂}}{A}$$

Given

$${F}_{⟂}=450\text{ N}$$

$$A=0.030{\text{ m}}^{2}$$

Working — substitution

$$P=\frac{450}{0.030}$$

Answer

$$P=15,000\text{ Pa}=1.50\times {10}^{4}\text{ Pa}$$

Interpretation

Each atom contributes a tiny force, but the combined effect of enormous numbers of collisions per second produces a pressure of 15.0 kPa on the wall.

Temperature and Average Kinetic Energy

Key Equations

Average kinetic energy of a gas atom:

$${K}_{\text{avg}}=\frac{3}{2}{k}_{B}T$$

Variables:

• ${K}_{\text{avg}}$ = average translational kinetic energy per atom (J)
• ${k}_{B}$ = Boltzmann constant, $1.38\times {10}^{-23}$ J/K
• $T$ = absolute temperature (K)

SI unit: joule (J)

Rearrangements:

$$T=\frac{2{K}_{\text{avg}}}{3{k}_{B}}$$

ProportionalityAverage kinetic energy is directly proportional to absolute temperature. Doubling the absolute temperature doubles the average kinetic energy.

Reference sheet status: On the reference sheet.

Root-mean-square speed relation:

$${K}_{\text{avg}}=\frac{1}{2}m{v}_{\text{rms}}^{2}$$

Variables:

• $m$ = mass of one atom (kg)
• ${v}_{\text{rms}}$ = root-mean-square speed [the square root of the average of the squared speeds] (m/s)

Combining the two expressions:

$$\frac{3}{2}{k}_{B}T=\frac{1}{2}m{v}_{\text{rms}}^{2}$$

Rearrangement for ${v}_{\text{rms}}$:

$${v}_{\text{rms}}=\sqrt{\frac{3{k}_{B}T}{m}}$$

ProportionalityThe rms speed is proportional to the square root of absolute temperature and inversely proportional to the square root of atomic mass. Doubling the absolute temperature increases ${v}_{\text{rms}}$ by a factor of $\sqrt{2}$.

Reference sheet status: On the reference sheet.

Temperature is not simply a number on a thermometer — it has a precise microscopic meaning. The temperature of a system is characterized by the average translational kinetic energy of the atoms within that system. A hotter gas has atoms that move faster on average. A colder gas has slower-moving atoms on average.

The relationship ${K}_{\text{avg}}=\frac{3}{2}{k}_{B}T$ makes this quantitative. Because temperature must be in kelvin (an absolute scale), zero kelvin corresponds to zero average kinetic energy — atoms would have no translational motion.

The root-mean-square speed (${v}_{\text{rms}}$) is the single speed that corresponds to the average kinetic energy. It is not the simple average of all speeds; it is the square root of the mean of the squared speeds. This distinction matters because squaring emphasizes higher speeds. For a given temperature, lighter atoms have a higher ${v}_{\text{rms}}$ than heavier atoms because the same average kinetic energy requires a larger speed when $m$ is smaller.

MisconceptionStudents often think all atoms in a gas move at the same speed. In reality, speeds vary enormously. Temperature determines the average kinetic energy, not the speed of any individual atom.

Exam TipAlways say “average kinetic energy” — never imply a single uniform speed.

Worked Example: Finding the rms Speed of Helium Atoms

Scenario

Helium gas ($m=6.64\times {10}^{-27}$ kg per atom) is at a temperature of 350 K. Determine the rms speed of the helium atoms.

Step-by-step solution

Equation used

$${v}_{\text{rms}}=\sqrt{\frac{3{k}_{B}T}{m}}$$

Given

$${k}_{B}=1.38\times {10}^{-23}\text{ J/K}$$

$$T=350\text{ K}$$

$$m=6.64\times {10}^{-27}\text{ kg}$$

Working — numerator first

$$3{k}_{B}T=3\times 1.38\times {10}^{-23}\times 350=1.449\times {10}^{-20}\text{ J}$$

Dividing by mass:

$$\frac{3{k}_{B}T}{m}=\frac{1.449\times {10}^{-20}}{6.64\times {10}^{-27}}=2.183\times {10}^{6}{\text{ m}}^{2}{\text{/s}}^{2}$$

Taking the square root:

$${v}_{\text{rms}}=\sqrt{2.183\times {10}^{6}}$$

Answer

$${v}_{\text{rms}}=1.48\times {10}^{3}\text{ m/s}$$

Interpretation

Helium atoms at 350 K have an rms speed of about 1480 m/s. Helium is very light, so its atoms move very fast at moderate temperatures.

The Maxwell–Boltzmann Distribution

Not all atoms in a gas share the same speed. The Maxwell–Boltzmann distribution is a graphical representation that shows how atomic speeds (or kinetic energies) are distributed at a given temperature.

The horizontal axis shows speed (or kinetic energy). The vertical axis shows the number of atoms (or fraction of atoms) at each speed. The curve is not symmetric — it rises steeply from zero, reaches a peak, and then tails off gradually toward high speeds. The peak of the curve represents the most probable speed. The area under the entire curve equals the total number of atoms (or 1, if normalized).

When temperature increases, the distribution changes in two important ways. First, the peak shifts to the right — the most probable speed increases. Second, the peak height decreases and the curve broadens, because a wider range of speeds becomes populated. The total area under the curve stays the same because the number of atoms has not changed.

When temperature decreases, the opposite occurs. The peak shifts left (lower most probable speed), the peak becomes taller, and the curve narrows.

BoundaryThe functional (mathematical) form of the Maxwell–Boltzmann distribution is not required. Only qualitative features and their relationship to temperature are assessed.

This is outside the scope of the AP exam.

Examiner InsightAP questions commonly show two Maxwell–Boltzmann curves and ask you to identify which is at a higher temperature. The higher-temperature curve is always broader and shorter, with its peak shifted right.

Exam TipIf asked to sketch the distribution at a new temperature, keep the total area the same.