Every formula, definition, trick, and trap you need for IB Physics HL Topic B.3 — kinetic theory, Boyle/Charles/Gay-Lussac, ideal gas law, internal energy, real vs ideal gases.
Gas Laws is the most algebra-heavy topic in Theme B of IB Physics. Every Paper 1 features at least one multiple-choice question on $PV = nRT$ or its particle-form cousin $PV = N k_B T$, and Paper 2 routinely asks for an extended energy-balance calculation that mixes kinetic theory with the ideal gas law. The kinetic-model assumptions also sit at the heart of any Paper 3 essay on thermodynamics or astrophysics.
This cheatsheet condenses every formula, definition, trick, and trap from Topic B.3 into one revisable page — kinetic model, the three empirical laws, the unified ideal gas law, $\bar{E}_k = \tfrac{3}{2} k_B T$, and the real-vs-ideal comparison. The worked example walks through a Boyle/Charles fusion problem in IB mark-scheme rhythm, and the FAQ section drills the conceptual misconceptions our students hit most often in mock exams.
A large number of identical particles in constant random motion.
Particle volume is negligible compared to the container.
Collisions are perfectly elastic (KE conserved).
No intermolecular forces between particles.
Collision times are negligible compared to time between collisions.
All directions of motion are equally probable.
TrickThe factor $\tfrac{1}{3}$ in $P = \tfrac{1}{3}\rho \langle v^2 \rangle$ comes from 3D random motion — only one third of molecules contribute to each direction at a time.
TrapPressure counts only forces perpendicular to the wall area. Tangential (parallel) forces don't generate pressure.
The kinetic model of a gas. Molecules in constant random motion collide elastically with the walls; the perpendicular momentum they transfer, per unit area per second, is the pressure.
From the Photon question bank
What examiners actually test on this concept:
P = ⅓ρ⟨v²⟩Kinetic-model assumptionsPerpendicular forces onlyElastic collisions
Pressure is the rate of perpendicular momentum transfer in elastic wall collisions — tangential forces contribute nothing. frequent
Know the assumptions verbatim: many identical particles in random motion, negligible molecular volume, perfectly elastic collisions, no intermolecular forces.
The $\tfrac13$ in $P=\tfrac13\rho\langle v^2\rangle$ comes from motion being shared equally over three dimensions.
§2 — Amount of Substance Topic B.3
Key formulae
Moles from $N$:$n = \dfrac{N}{N_A}$
Moles from mass:$n = \dfrac{m}{M_r}$
Number from mass:$N = \dfrac{m N_A}{M_r}$
Boltzmann from R:$k_B = \dfrac{R}{N_A} = 1.38 \times 10^{-23}\,\text{J K}^{-1}$
NoteTo find which gas sample (from equal masses) has the most molecules, pick the one with the smallest molar mass $M_r$ — smallest $M_r$ ⇒ most moles ⇒ most molecules.
From the Photon question bank
What examiners actually test on this concept:
n = N/N_An = m/M_rk_B = R/N_AMost molecules
For equal masses, the smallest molar mass has the most moles → most molecules.frequent
Memorise $N_A=6.02\times10^{23}$, $R=8.31$, $k_B=1.38\times10^{-23}$ and $k_B=R/N_A$.
Convert mass ↔ moles ↔ particles cleanly: $N = mN_A/M_r$.
§3 — Empirical Gas Laws Topic B.3
The three classical laws
Law
Constant
Relation
Form
Boyle
$T$
$P \propto 1/V$
$P_1 V_1 = P_2 V_2$
Charles
$P$
$V \propto T$
$V_1/T_1 = V_2/T_2$
Gay-Lussac
$V$
$P \propto T$
$P_1/T_1 = P_2/T_2$
TrapTemperature must be in kelvin: $T(\text{K}) = \theta({}^\circ\text{C}) + 273$. Never substitute Celsius directly into a gas-law ratio.
TrickCombined gas law for any change of state with fixed mass: $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$. Cancel whichever variable stays constant.
Molecular explanations
↓V at constant T ⇒ ↑P: molecules hit walls more often (shorter distance between collisions).
↑T at constant V ⇒ ↑P: greater speed ⇒ more collisions per second and greater momentum transfer.
↑T at constant P ⇒ ↑V: greater speed ⇒ volume must expand to restore the original collision rate per area.
The three empirical gas laws. Boyle: $P \propto 1/V$; Charles: $V \propto T$ (extrapolating to $-273\,{}^\circ$C); Gay-Lussac: $P \propto T$ — each holds one quantity constant.
From the Photon question bank
What examiners actually test on this concept:
Boyle PV = constCharles V ∝ TGay-Lussac P ∝ TKelvin only
Temperature MUST be in kelvin in any gas-law ratio — the number-one B.3 error. very frequent
The combined law $P_1V_1/T_1 = P_2V_2/T_2$ covers all three; cancel whichever quantity is held constant.
Be ready to explain each law molecularly (collision frequency + momentum transfer per collision).
Equivalence:$nR = N k_B \;\Leftrightarrow\; k_B = R/N_A$
TrickUse $PV = nRT$ when given mass or moles. Use $PV = N k_B T$ when given the number of particles. They give identical results — pick the form that matches the data.
Trap$PV/T = nR$ is constant only for fixed mass. If gas is added or escapes, $n$ changes and the ratio is no longer constant.
Trick$\langle E_K \rangle = \tfrac{3}{2} k_B T$ depends on temperature only — independent of gas type. Equal $T$ ⇒ equal average KE per molecule, regardless of which gas.
NoteFor a monatomic ideal gas, $U$ is purely kinetic — no PE, no rotational or vibrational energy. So $U \propto T$ exactly, with no offset.
TrapHeavier molecules at the same $T$ move slower, not faster. From $\tfrac{1}{2} m v_\text{rms}^2 = \tfrac{3}{2} k_B T$, larger $m$ gives smaller $v_\text{rms}$.
Speed distribution at fixed temperature. Both gases have the same average KE, but the lighter gas has a broader curve peaking at higher speed since $v_{\text{rms}} \propto 1/\sqrt m$.
Average KE per molecule depends only on $T$ — equal $T$ means equal $\langle E_K\rangle$ for ANY gas. frequent
For a monatomic ideal gas $U$ is purely kinetic: $U = \tfrac32 nRT$, so $U \propto T$ with no offset.
At the same $T$, heavier molecules move SLOWER ($v_{\text{rms}} \propto 1/\sqrt m$) — a classic distractor.
§6 — Ideal vs Real Gases Topic B.3
Comparison table
Ideal
Real
Molecular volume
negligible
finite
Intermolecular forces
none
attractive / repulsive
$PV = nRT$
exact
approximate
Compressibility $Z = PV/nRT$
$= 1$
$\neq 1$
Internal energy
KE only
KE + PE
Condensation
never
possible
Ideal approximation valid when…
Low pressure — molecules far apart; finite volume and forces become negligible.
High / moderate temperature — KE $\gg$ intermolecular PE; far from condensation.
Far from the gas's boiling point.
TrapAt high $P$: $Z > 1$ (repulsive forces and finite molecular volume dominate). At low $T$ with moderate $P$: $Z < 1$ (attractive forces pull molecules together, lowering pressure below the ideal value).
TrickOn a real-gas P–V diagram, a flat (constant-$P$) section of an isotherm represents a phase change (condensation). An ideal-gas isotherm is a smooth hyperbola — no flat region.
Ideal vs real isotherm. The ideal gas follows a smooth hyperbola; a real gas has a flat plateau where it condenses to a liquid.
From the Photon question bank
What examiners actually test on this concept:
Z = PV/nRTWhen ideal is validHigh P vs low TCondensation plateau
Ideal is a good approximation at LOW pressure and HIGH temperature (far from condensation). frequent
$Z>1$ at high $P$ (finite molecular volume + repulsion); $Z<1$ at low $T$ (attractions lower the pressure).
A flat section on a real isotherm is a phase change (condensation); an ideal isotherm is a smooth hyperbola with no flat region.
§7 — Exam Attack Plan All sections
When you see this in the question — reach for that:
A sealed cylinder contains $0.020\,\text{mol}$ of an ideal monatomic gas at $T_1 = 27\,{}^\circ\text{C}$, pressure $P_1 = 1.0 \times 10^{5}\,\text{Pa}$ and volume $V_1$. The gas is heated to $T_2 = 327\,{}^\circ\text{C}$ while a piston allows the volume to expand isobarically. (a) Find $V_1$. (b) Find the new volume $V_2$. (c) Find the change in internal energy of the gas.
Solution
Convert to kelvin: $T_1 = 300\,\text{K}$, $T_2 = 600\,\text{K}$. (R1)
Monatomic ideal gas: $\Delta U = \tfrac{3}{2} nR \Delta T = \tfrac{3}{2}(0.020)(8.31)(300)$. (M1)
Evaluate: $\Delta U \approx 74.8\,\text{J} \approx 75\,\text{J}$. (A1)
Examiner's note: The biggest losers in this question type forget to convert to kelvin, then back-substitute Celsius into the ratio $V_2/T_2 = V_1/T_1$ — answers come out negative, with no physical meaning. Always convert temperatures first, then chase the algebra.
Common Student Questions
Why must temperature be in kelvin for gas laws?
All the gas laws (Boyle, Charles, Gay-Lussac, ideal gas) are derived from extrapolating $P$ or $V$ vs $T$ to absolute zero, which sits at 0 K. Substituting Celsius gives nonsensical results — for example, a temperature of $0\,{}^\circ\text{C}$ would imply zero pressure or volume. Always convert with $T(\text{K}) = \theta({}^\circ\text{C}) + 273$ before substituting into any ratio or product.
When should I use $PV = nRT$ vs $PV = N k_B T$?
Use $PV = nRT$ when the problem gives mass or moles ($n$), with $R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$. Use $PV = N k_B T$ when the problem gives the number of molecules ($N$), with $k_B = 1.38 \times 10^{-23}\,\text{J K}^{-1}$. They are mathematically identical because $nR = N k_B$. Pick the form that matches the data — converting wastes time and risks rounding errors.
Do heavier gas molecules at the same temperature move faster?
No — heavier molecules move slower at the same temperature. From $\tfrac{1}{2} m v_\text{rms}^2 = \tfrac{3}{2} k_B T$, a larger $m$ gives a smaller $v_\text{rms}$. At equal $T$, all gases share the same average kinetic energy per molecule, but the heavier ones are slower. This is why oxygen molecules in air move slower than helium atoms at the same temperature.
When does the ideal gas approximation break down?
At high pressure, finite molecular volume and intermolecular forces become significant — the compressibility factor $Z = PV/nRT$ drifts above 1. At low temperature (or near a phase change), attractive intermolecular forces become comparable to KE, and $Z$ drops below 1. The flat region of a real-gas isotherm at low $T$ is condensation; an ideal gas would never show this.
Which gas sample has the most molecules from equal masses?
The gas with the smallest molar mass $M_r$. From $n = m/M_r$, equal mass means more moles for smaller $M_r$, and therefore more molecules ($N = n N_A$). For example, 1 g of H$_2$ contains roughly $6\times$ more molecules than 1 g of N$_2$. This is a recurring multiple-choice format in Paper 1.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy Gas Laws library (only available to enrolled students or via the resource library subscription) adds:
Notes PDF — kinetic-model derivations, P–V diagram intuition, and full ideal-gas walkthroughs.
Tutorial booklet — IB-style questions sequenced from foundation to full HL difficulty.
Tutorial Solutions — full mark-scheme-style solutions with R1/M1/A1 annotations.
Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
