Kinetic Theory of Gases — B.3 Gas Laws — IB Physics Notes

4 Lesson 4 of 6 · B.3 Gas Laws

Kinetic Theory of Gases

Pressure, volume and temperature are things we can measure in a lab. This lesson builds the bridge down to the microscopic world — countless invisible particles, colliding with walls billions of times a second — and shows how that picture produces a single equation: $P = \tfrac{1}{3}\rho v^2$.

🎯 Hook – pressure is just a lot of tiny impacts

Stand under a tin roof in a hailstorm. Each hailstone delivers one sharp, separate impact — but with thousands landing every second, the roof doesn't feel individual hits, it feels a steady push. That steady push is exactly what gas pressure is: not a continuous force, but the statistical average of an enormous number of individual particle collisions with a surface, each one lasting a fraction of a microsecond. Zoom in far enough on any "smooth" gas pressure and you find nothing but collisions.

Hailstones striking a corrugated tin roof, each impact marked with a small force arrow

Fig 1 — Individually, each hailstone gives one brief impact. Added up over a second, thousands of impacts feel like one steady force — the same idea that turns particle collisions into gas pressure.

💡 Diagnostic question (think – then read on) — if every collision between gas particles conserved kinetic energy perfectly, would you expect the pressure of a sealed gas at constant temperature to change over time? No. If collisions lost kinetic energy the gas would keep getting colder and its pressure would keep falling; because collisions between particles are taken to be elastic, the gas stays at the same temperature and produces a steady, unchanging pressure.

🧪 What makes a gas "ideal"?

An ideal gas is a gas which obeys the ideal gas law, $PV = nRT$, perfectly. No real gas does this exactly, but most real gases come close enough that the model is extremely useful. Although most gases are molecular, physicists often say particles to keep the model general. The microscopic model makes seven specific assumptions:

The gas contains a very large number of identical particles.
The volume of the particles themselves is negligible compared with the total volume occupied by the gas.
The particles move in random directions, with a wide variety of speeds.
There are no forces between particles, except during a collision.
Because there are no forces, particles have no (electrical) potential energy — any change in internal energy of an ideal gas is assumed to be entirely a change in random kinetic energy.
The motion of the particles obeys Newton's laws of motion.
All collisions between particles are elastic — the total kinetic energy of the particles stays constant at constant temperature.

🔗 Linking question — how do the concepts of force and momentum link mechanics and thermodynamics? This question links with understandings in Topics A.2 and A.3: the whole derivation below is nothing more than Newton's second law, written as rate of change of momentum, applied to particle collisions.

📖 Top tip! For a gas all at one temperature, if collisions between gas particles caused a loss of kinetic energy, the gas would keep getting colder. Collisions with the container walls only transfer kinetic energy on average if there is a temperature difference between gas and wall (thermal conduction).

Container of gas showing many small particles with arrows in random directions, colliding with each other and the walls

Fig 2 — Gas particles moving at random inside a container. Some are near the walls, some in the interior; there is a wide spread of speeds and directions, exactly as assumed above.

📐 One particle, one wall: building the model

Start with just one particle, mass $m$, moving with velocity $v$ perpendicular to the end wall of a rectangular box of dimensions $x$, $y$, $z$. After an elastic collision with the wall it returns along the same path with velocity $-v$.

Change of velocity at the wall $$v-(-v) = 2v$$

Average force on the wall (change of momentum, one impact) $$F = m\times 2v = 2mv$$

The particle only strikes this wall again after travelling to the opposite wall and back, a distance $2x$:

Time between collisions with the same wall $$t = \frac{2x}{v}$$

Average force on end wall $$F = \frac{2mv}{\left(\frac{2x}{v}\right)} = \frac{mv^2}{x}$$

Average pressure on end wall (area $=yz$) $$P = \frac{\left(\frac{mv^2}{x}\right)}{yz} = \frac{mv^2}{xyz} = \frac{mv^2}{V}$$

where $V = xyz$ is the total volume of the box.

Rectangular box of dimensions x, y and z with one particle of mass m moving left toward a shaded end wall

Fig 3 — One particle in a box, bouncing elastically off the shaded end wall. This single-particle result is the seed of the whole derivation.

🌐 From one particle to a real gas

Real containers are rarely rectangular, and a real gas has a huge number of particles $N$, moving in random directions with different speeds, not all charging perpendicularly at one wall. Building the model up in stages:

Step 1 — many particles, still all perpendicular: if there are $N$ particles in the box, all moving perpendicular to the end wall, the pressure on that wall becomes $\dfrac{Nmv^2}{V}$.

Step 2 — random directions: particles actually move in three dimensions, so on average only $\tfrac13$ of the total is directed along any one of the three perpendicular directions:

$$P = \frac{1}{3}\frac{Nmv^2}{V}$$

Step 3 — a spread of speeds: if the particles have different speeds, $v^2$ here means the average value of their speeds-squared.

Step 4 — any shape of container: it can be shown that the shape of the container does not affect the pressure — only its volume matters.

Pressure of an ideal gas $$P = \frac{1}{3}Nm\frac{v^2}{V}$$

Since $Nm$ is the total mass of the gas, $\dfrac{Nm}{V}$ is its density, $\rho$, giving the compact form used throughout this topic:

Pressure of an ideal gas $$P = \frac{1}{3}\rho v^2$$

Cube with three perpendicular axes x, y and z labelled, each carrying one third of the arrows representing particle velocities

Fig 4 — Why the $\tfrac13$ appears. Random motion in three dimensions means, on average, one third of the particles' velocity is directed along any one axis.

📊 Reading isotherms on a PV diagram

A gas's behaviour is often plotted as pressure against volume at a series of fixed temperatures — each curve is called an isotherm. Because $PV = nRT$, for a fixed amount of gas a higher temperature means a larger value of $PV$, so the whole curve sits further from the origin.

Graph type	Axes (X, Y)	Shape	What to read
Single isotherm	Volume · Pressure	Smooth curve falling from upper-left to lower-right (a hyperbola)	Every point on one curve is at the same temperature; $PV$ is constant along it
Family of isotherms	Volume · Pressure	Several curves of the same shape, fanning outward from the origin	The curve further from the origin — higher $P$ at the same $V$, or larger $V$ at the same $P$ — is the higher temperature
A change of state	Volume · Pressure	A line or arrow joining a starting point to a finishing point, possibly crossing between isotherms	Whether the point moves to a higher or lower isotherm tells you whether $T$ rose, fell, or stayed the same

PV diagram showing a family of isotherm curves fanning outward from the origin, each labelled with a different temperature

Fig 5 — A family of isotherms. Curves further from the origin represent higher temperatures.

PV diagram showing a single isotherm with a point moving along it from a smaller volume and higher pressure to a larger volume and lower pressure

Fig 6 — A change of state along one isotherm: volume increases as pressure falls, temperature unchanged.

✓ Examiner tip: never say a curve "is higher" without saying at what volume — always compare pressures at the same volume (or volumes at the same pressure) when arguing which isotherm has the higher temperature.

✍️ Worked example – nitrogen in a container

Determine (a) a value for the average speed of nitrogen molecules at normal air pressure ($1.0\times10^{5}\ \text{Pa}$) if a mass of 1.4 g of the gas was in a container of volume 1200 cm³; (b) how many moles of nitrogen were in the container; (c) the temperature of the gas; (d) a value for the average molecular speed if the temperature rose to 350 K in the same container.

Identify: use $P=\tfrac13\rho v^2$ for (a), $n = m/M$ for (b), $PV=nRT$ for (c), and $P_1/T_1 = P_2/T_2$ followed by $P=\tfrac13\rho v^2$ again for (d), since density doesn't change.

Data (a): $\rho = \dfrac{1.4\times10^{-3}}{1200\times10^{-6}} = 1.17\ \text{kg m}^{-3}$

Equation (a): $1.0\times10^{5} = \tfrac13\times1.17\times v^2$

Answer (a): $v = 5.1\times10^{2}\ \text{m s}^{-1}$. This is the square root of the average of speeds-squared — not exactly the same thing as the average speed.
(b) $n = \dfrac{1.4}{28} = 0.050\ \text{mol}$
(c) $(1.0\times10^{5})(1200\times10^{-6}) = 0.050\times8.31\times T$, so $T = 289\ \text{K}$
(d) $\dfrac{1.0\times10^{5}}{289} = \dfrac{P_2}{350}$, so $P_2 = 1.21\times10^{5}\ \text{Pa}$; then $1.21\times10^{5} = \tfrac13\times1.17\times v^2$ (density unchanged) gives $v = 560\ \text{m s}^{-1}$.

Sealed cylinder of nitrogen gas at volume 1200 cubic centimetres labelled with pressure and temperature

Fig 7 — The container for the worked example: fixed volume 1200 cm³, mass 1.4 g of nitrogen, pressure and temperature as calculated above.

🌍 Real-world skill: mean and range

"Average particle speed" gets used constantly in this topic, so it's worth being precise about it. Consider timing a ball rolling down a slope: 15, 12, 11, 21, 14, 11, 14, 13 s. The range (data) is the spread from smallest to largest value — here 11 to 21 s. The value 21 s looks anomalous, inconsistent with the rest of the set, and could be called an outlier — it may be a mistake and should be excluded unless checked and confirmed.

An average value is any single number chosen to represent a set of values — the median (13), the most common value (11 or 14), or, most often in physics, the mean: add all the values and divide by how many there are. Here the mean is 12.9 s, better quoted to 2 s.f. as 13 s.

✓ Examiner tip: for gas particles, "average speed" cannot be found as a straightforward mean — the speeds are calculated from average kinetic energies instead, which is why $P=\tfrac13\rho v^2$ uses the average of $v^2$, not the square of the average $v$.

🚀 HL extension – checking the equation with units

A quick unit check catches obvious algebra slips. Pressure is in pascal, equivalent to $\text{N m}^{-2}$, and $\text{N} \equiv \text{kg m s}^{-2}$, so pressure reduces to $\text{kg m}^{-1}\,\text{s}^{-2}$ in SI base units. On the right of $P=\tfrac13\rho v^2$: $\tfrac13$ has no units, $\rho$ has units $\text{kg m}^{-3}$, and $v^2$ has units $\text{m}^2\,\text{s}^{-2}$; multiplying gives $\text{kg m}^{-1}\,\text{s}^{-2}$ — matching the left side.

Challenge: show that $P=\tfrac13 Nm\,v^2/V$ also reduces to the same base units. Answer: $N$ is a pure number; $m$ in kg; $v^2$ in $\text{m}^2\,\text{s}^{-2}$; $V$ in m³. Combining: $\text{kg}\cdot\text{m}^2\,\text{s}^{-2}/\text{m}^3 = \text{kg m}^{-1}\,\text{s}^{-2}$, the same as before — as it must be, since $Nm/V$ is just $\rho$.

⚠️ Trap corner – common mistakes

Misconception	Correct view
Collisions between gas particles gradually lose kinetic energy, like real macroscopic collisions	Ideal-gas collisions are assumed elastic — total kinetic energy stays constant at constant temperature, or the gas would keep cooling
$V$ and $v$ in $P=\tfrac13Nmv^2/V$ are interchangeable — they're both "V"	Upper-case $V$ is volume; lower-case $v$ is speed. Easy to confuse, and worth writing out in full the first time you use the equation
The derivation only works for a cubic or rectangular container	It can be shown that the shape of the container does not affect the result — only the volume does
The "average speed" and the speed you get from $P=\tfrac13\rho v^2$ are the same number	The equation gives $\sqrt{\overline{v^2}}$ (the root-mean-square speed), which is not exactly equal to the mean speed, $\bar v$

⚠️ Most common slip: swapping upper-case $V$ (volume) and lower-case $v$ (speed) midway through a calculation — always identify both quantities explicitly before substituting.

P = ⅓ρv² P = ⅓Nmv²/V ρ = Nm/V elastic collisions ⇒ ΣEk constant ⅓ comes from 3 perpendicular directions shape of container doesn't matter higher isotherm = further from origin √(v²) ≠ v̄ (rms speed ≠ mean speed)

Practice check — multiple choice

1. Which equation correctly links the pressure of an ideal gas to its density?

A $P = \rho v^2$

B $P = \tfrac13\rho v^2$

C $P = 3\rho v^2$

D $P = \tfrac13\rho v$

2. Why does the factor $\tfrac13$ appear in the derivation?

A Because a third of collisions are inelastic

B Because the container has three walls

C Because random motion in three perpendicular directions puts on average a third of the velocity along any one axis

D Because only a third of the particles reach the wall

3. Which of these is not one of the ideal gas assumptions?

A All collisions between particles are elastic

B Particles attract each other weakly even when not colliding

C The volume of the particles is negligible compared with the volume of the gas

D The particles move in random directions with a range of speeds

4. A single particle collides elastically, head-on, with a container wall. Immediately after the collision, its speed is

A The same as before, direction reversed

B Reduced, direction reversed

C The same as before, same direction

D Zero, since it has come to rest against the wall

5. On a PV diagram, isotherm $T_5$ lies further from the origin than isotherm $T_2$. This tells you that

A $T_5$ represents a higher temperature than $T_2$

B $T_5$ represents a lower temperature than $T_2$

C $T_5$ and $T_2$ must be the same temperature

D Nothing can be concluded without knowing $n$

Practice check — fill in the blanks

Complete the statements:

An ideal gas is assumed to contain a very large number of particles, whose own volume is compared with the volume of the gas. There are no forces between particles except during a , and all such collisions are assumed to be . In the derivation, the factor of $\tfrac13$ arises because the particles move in directions. The quantity $Nm/V$ is the gas's , usually written $\rho$. On a PV diagram, a curve of constant temperature is called an .

Exam‑style questions

Exam question · Estimate

Q14. Helium gas is widely used to fill balloons. If the density of the helium in a balloon is $2.0\times10^{-4}\ \text{g cm}^{-3}$, estimate its (a) volume, (b) mass, (c) amount in moles, (d) number of atoms. [5 marks]

✓ (a) Estimate: a typical party balloon is roughly spherical, diameter ≈ 30 cm ($r=0.15\ \text{m}$): $V=\tfrac43\pi r^3 \approx 1.4\times10^{-2}\ \text{m}^3$.
✓ (b) $\rho = 2.0\times10^{-4}\ \text{g cm}^{-3} = 0.20\ \text{kg m}^{-3}$; $m=\rho V \approx 0.20\times1.4\times10^{-2} \approx 2.8\times10^{-3}\ \text{kg}\ (2.8\ \text{g})$.
✓ (c) Molar mass of He $=4\ \text{g mol}^{-1}$: $n = 2.8/4 \approx 0.71\ \text{mol}$.
✓ (d) $N = nN_{\text A} \approx 0.71\times6.02\times10^{23} \approx 4.3\times10^{23}\ \text{atoms}$.
✓ Any reasonable balloon diameter is acceptable — this is an estimation question, so the method matters more than the exact assumed size.

Exam question · Determine

Q15. A container of gas has a volume of 120 cm³. At a temperature of 300 K the gas has a pressure of $1.5\times10^{6}\ \text{Pa}$. Determine the amount of gas in the container. [3 marks]

✓ $PV=nRT$, so $n = \dfrac{PV}{RT}$.
✓ $n = \dfrac{(1.5\times10^{6})(120\times10^{-6})}{8.31\times300} = \dfrac{180}{2493}$.
✓ $n \approx 0.072\ \text{mol}$.

Exam question · Explain

Q16. How can you be sure that an isotherm $T_5$ is representing a higher temperature than an isotherm $T_2$ on a PV diagram? [2 marks]

✓ Compare the two curves at the same volume: whichever curve has the higher pressure there is the higher temperature (equivalently, compare volumes at the same pressure).
✓ This follows because $PV=nRT$ — for the same fixed amount of gas, a larger value of the product $PV$ can only mean a larger $T$, so the higher-temperature isotherm sits further from the origin.

Exam question · Describe

Q17. A fixed amount of gas is represented by a point S on a PV diagram. What changes must be made to move it to a different point X on the diagram? [3 marks]

✓ Read off the pressure and volume at S and at X directly from the diagram.
✓ If X lies on the same isotherm as S, only $P$ and $V$ change (in opposite senses) and $T$ is constant; if X lies on a different isotherm, $T$ has also changed — higher isotherm means a temperature increase, lower isotherm a decrease.
✓ State explicitly which of $P$, $V$ and $T$ increased, decreased, or stayed the same, based on the actual positions of S and X on the given diagram.

Exam question · Sketch

Q18. Represent the following changes to a fixed mass of gas on a PV diagram: (a) a gas expanding at constant pressure, as its temperature increases; (b) a compression at constant temperature; (c) a gas cooled so that its volume and pressure both decreased. [3 marks]

✓ (a) A horizontal line moving to the right (increasing $V$ at constant $P$), crossing from a lower isotherm to a higher one.
✓ (b) A curve moving left along a single isotherm — volume decreases, pressure increases, temperature unchanged.
✓ (c) A line moving down and to the left, from a point on a higher isotherm to a point on a lower isotherm — both $P$ and $V$ fall as $T$ falls.

Exam question · Outline

Q19. Make a list of all the basic physics principles used to derive the equation $P=\tfrac13\rho v^2$. [4 marks]

✓ Elastic collisions: a particle rebounds off a wall with the same speed, direction reversed, so $\Delta v=2v$.
✓ Newton's second law as rate of change of momentum, $F=\Delta p/\Delta t$, to find the average force from repeated collisions.
✓ Simple geometry (distance $=2x$, speed $v$) to find the time between successive collisions with the same wall.
✓ The definition of pressure, $P=F/\text{area}$, and of density, $\rho=Nm/V$; plus the statistical step of splitting random 3-D motion into three equal perpendicular components (giving the factor $\tfrac13$) and using the average of $v^2$ for a spread of particle speeds.

Exam question · Determine

Q20. What pressure would be created by $5.0\times10^{23}$ ideal gas molecules, each of mass $5.3\times10^{-26}\ \text{kg}$, in a volume of $0.010\ \text{m}^3$, if the average speed of the molecules was 500 m s⁻¹? [3 marks]

✓ $P=\dfrac13\dfrac{Nmv^2}{V}$, with $Nm = (5.0\times10^{23})(5.3\times10^{-26})=2.65\times10^{-2}\ \text{kg}$.
✓ $P = \dfrac13\times\dfrac{(2.65\times10^{-2})\times500^2}{0.010}$.
✓ $P \approx 2.2\times10^{5}\ \text{Pa}$.

Exam question · Determine

Q21. An ideal gas has a density of 2.4 kg m⁻³. If it creates a pressure of $1.5\times10^{5}\ \text{Pa}$ on its container, determine a value for the average speed of its molecules. [3 marks]

✓ $P=\tfrac13\rho v^2$, so $v^2 = \dfrac{3P}{\rho} = \dfrac{3\times1.5\times10^{5}}{2.4}$.
✓ $v^2 \approx 1.9\times10^{5}\ \text{m}^2\,\text{s}^{-2}$.
✓ $v \approx 4.3\times10^{2}\ \text{m s}^{-1}$.

Exam question · Determine

Q22. At room temperature the average speed of oxygen molecules is 500 m s⁻¹. What pressure will these molecules create if the density of oxygen was the same as in air (1.3 kg m⁻³)? Assume that oxygen behaves as an ideal gas. [2 marks]

✓ $P = \tfrac13\rho v^2 = \tfrac13\times1.3\times500^2$.
✓ $P \approx 1.1\times10^{5}\ \text{Pa}$.

Exam question · Calculate

Q23. Three particles have speeds 440 m s⁻¹, 480 m s⁻¹ and 520 m s⁻¹. (a) What is their average speed? (b) What is the square of their average speed? (c) What is the average of their speed-squared? (d) What is the square root of the average of their speed-squared? [4 marks]

✓ (a) $\bar v = \dfrac{440+480+520}{3} = 480\ \text{m s}^{-1}$.
✓ (b) $\bar v^2 = 480^2 = 2.30\times10^{5}\ \text{m}^2\,\text{s}^{-2}$.
✓ (c) $\overline{v^2} = \dfrac{440^2+480^2+520^2}{3} = 2.31\times10^{5}\ \text{m}^2\,\text{s}^{-2}$.
✓ (d) $\sqrt{\overline{v^2}} \approx 481\ \text{m s}^{-1}$ — slightly greater than the mean speed in (a), which is why $P=\tfrac13\rho v^2$ uses the root-mean-square speed, not the plain mean.

Score: 0 pts

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Ideal Gas Equation

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Derivation of the Kinetic Theory Equation

Misconception	Correct view
Collisions between gas particles gradually lose kinetic energy, like real macroscopic collisions	Ideal-gas collisions are assumed elastic — total kinetic energy stays constant at constant temperature, or the gas would keep cooling
\(V\) and \(v\) in \(P=\tfrac13Nmv^2/V\) are interchangeable — they're both "V"	Upper-case \(V\) is volume; lower-case \(v\) is speed. Easy to confuse, and worth writing out in full the first time you use the equation
The derivation only works for a cubic or rectangular container	It can be shown that the shape of the container does not affect the result — only the volume does
The "average speed" and the speed you get from \(P=\tfrac13\rho v^2\) are the same number	The equation gives \(\sqrt{\overline{v^2}}\) (the root-mean-square speed), which is not exactly equal to the mean speed, \(\bar v\)