IB Physics · Theme B The particulate nature of matter
B.3 Gas laws
Summary to follow. 8 syllabus statements · 19 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 8 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Pressure
Pressure
The force exerted perpendicular to a surface per unit area of that surface: P = F/A, where F is the component of the force perpendicular (normal) to the surface. A force parallel to the surface contributes nothing to the pressure. Pressure is a scalar: a gas at a given pressure pushes perpendicular to every surface it touches, whatever the surface's orientation, and the force on a surface is PA. SI unit: pascal (Pa), where 1 Pa = 1 N m⁻².
The pascal
The SI unit of pressure: 1 Pa is the pressure produced by a force of 1 N acting perpendicular to an area of 1 m² (1 Pa = 1 N m⁻² = 1 kg m⁻¹ s⁻²). Atmospheric pressure at sea level is about 1.0 × 10⁵ Pa. Because areas are often given in cm² or mm², note that 1 cm² = 10⁻⁴ m² and 1 mm² = 10⁻⁶ m²: the linear conversion factor is squared.
Students often think Pressure is always the total force divided by the area, whatever direction the force acts in. In fact No. Only the component of the force perpendicular to the surface produces pressure: P = F⊥/A.
Students often think The component 'into' a surface is always found with sin θ, whatever the angle is measured from. In fact No. The component perpendicular to the surface is along the vertical, which is adjacent to θ, so it is F cos θ.
Amount of substance (n)
Amount of substance (n)
The SI base quantity that counts the number of particles in a sample in units of the Avogadro constant: n = N/N_A, where N is the number of molecules (or atoms, for a monatomic gas). SI unit: mole (mol). One mole of any substance contains N_A particles, whatever the mass or size of those particles; the amount of substance is therefore not a mass and not a volume.
Avogadro constant (N_A)
The number of particles per mole of substance: N_A = 6.02 × 10²³ mol⁻¹. It links the microscopic count of molecules N to the amount of substance n by N = nN_A, and it links the two gas constants by k_B = R/N_A.
Molar mass (M)
The mass of one mole of a substance: M = m/n, so the amount of substance in a sample of mass m is n = m/M. SI unit: kg mol⁻¹, although values are usually quoted in g mol⁻¹ (32 g mol⁻¹ = 0.032 kg mol⁻¹ for O₂). For a gas of diatomic molecules such as O₂ the molar mass is twice that of the atoms.
Students often think The mass can be divided by the molar mass as the numbers stand, whatever units they are in. In fact No. The units must match: convert the mass to grams, or the molar mass to kg mol⁻¹ (32 g mol⁻¹ = 0.032 kg mol⁻¹), before dividing.
Students often think The amount of substance is the mass of the sample, so a heavier sample always contains more moles and more molecules. In fact No. The amount of substance n counts particles (n = N/N_A) and is measured in mol. Equal amounts of different gases contain equal numbers of molecules but have different masses.
Ideal gas
Ideal gas
A modelled gas whose molecules behave exactly as the kinetic model assumes, so that it obeys PV = nRT at every pressure, volume and temperature. No real gas is ideal; the ideal gas is a model used to approximate the behaviour of real gases, and it does so closely at low pressure, low density and high temperature. An ideal gas cannot be liquefied, because it has no intermolecular forces to hold molecules together.
Assumptions of the kinetic model of an ideal gas
The gas consists of a very large number of identical molecules in random motion that obey Newton's laws; the molecules are point particles, so their total volume is negligible compared with the volume of the container; there are no intermolecular forces except during collisions; collisions between molecules and with the walls are elastic, so no kinetic energy is lost; and the duration of a collision is negligible compared with the time between collisions. The molecules have a range of speeds, not a single speed.
Real gas
A gas made of actual molecules, which have a small but non-zero volume and exert weak attractive forces on one another between collisions. A real gas therefore has intermolecular potential energy, can be liquefied by cooling and compression, and deviates from PV = nRT, most noticeably at high pressure, high density and temperatures close to its boiling point.
Students often think All the molecules of a gas at one temperature move with the same speed, and heating raises that speed. In fact No. The molecules have a wide range of speeds, which change at every collision. Temperature fixes the AVERAGE kinetic energy of the molecules, not a single speed.
Students often think Every collision loses a little kinetic energy, so the molecules gradually slow down and the gas cools or its pressure falls. In fact No. The model assumes all collisions are elastic: the total kinetic energy of the molecules is unchanged by collisions.
Absolute (kelvin) temperature
Absolute (kelvin) temperature
Temperature measured on the kelvin scale, whose zero is absolute zero: T/K = θ/°C + 273. The gas laws and the ideal gas law require T in kelvin, because V ∝ T at constant pressure and P ∝ T at constant volume are proportionalities only when temperature is measured from absolute zero. A temperature in degrees Celsius must always be converted before it is used in a gas-law calculation.
Constant-temperature law (Boyle's law)
For a fixed mass of gas at constant temperature, pressure is inversely proportional to volume: PV = constant, so P₁V₁ = P₂V₂. Halving the volume doubles the pressure. It applies only when the temperature and the amount of gas are both unchanged.
Constant-pressure law (Charles's law)
For a fixed mass of gas at constant pressure, volume is proportional to absolute temperature: V/T = constant, so V₁/T₁ = V₂/T₂ with T in kelvin. Doubling the kelvin temperature doubles the volume.
Constant-volume law (pressure law)
For a fixed mass of gas at constant volume, pressure is proportional to absolute temperature: P/T = constant, so P₁/T₁ = P₂/T₂ with T in kelvin. A sealed rigid container of gas that is heated from 300 K to 600 K doubles its pressure.
Combined gas law (PV/T = constant)
For a fixed mass of ideal gas, PV/T = constant, so P₁V₁/T₁ = P₂V₂/T₂ with T in kelvin. It follows from the three empirical laws: each of them is the special case in which one of P, V and T is held constant. Because it is a ratio of the same quantities before and after, any consistent units of P and V may be used, but T must be in kelvin.
Pressure–volume (P–V) diagram
A graph of pressure (vertical axis) against volume (horizontal axis) on which each state of a fixed mass of gas is a point and each change of state is a line. A change at constant pressure is a horizontal straight line; a change at constant volume (isovolumetric) is a vertical straight line; a change at constant temperature (isothermal) is a curve on which PV is constant, which gets steeper as V decreases and never reaches either axis. Isothermal curves for higher temperatures lie further from the origin.
Students often think Any temperature scale can be used in gas-law calculations, so temperatures in °C can be substituted directly. In fact No. The gas laws are proportionalities to ABSOLUTE temperature, so T must be in kelvin: T/K = θ/°C + 273.
Students often think Each gas law applies whatever the third variable does, so PV stays constant during any compression, even one that warms the gas. In fact No. Each empirical law holds only when the third variable is constant: PV is constant only at constant temperature. If P, V and T all change, PV/T = constant must be used.
Ideal gas law (PV = nRT) and the molar gas constant R
Ideal gas law (PV = nRT) and the molar gas constant R
The equation of state of an ideal gas: PV = nRT, where P is the pressure in Pa, V the volume in m³, n the amount of substance in mol and T the absolute temperature in K. R is the molar gas constant, 8.31 J mol⁻¹ K⁻¹. Because R is in SI units, P and V must be in Pa and m³ (1 dm³ = 10⁻³ m³, 1 cm³ = 10⁻⁶ m³, 1 kPa = 10³ Pa).
Boltzmann constant (k_B) and PV = Nk_B T
The ideal gas law written in terms of the number of molecules N rather than the amount of substance: PV = Nk_B T. The Boltzmann constant k_B = R/N_A = 1.38 × 10⁻²³ J K⁻¹ is the gas constant per molecule. Since Nk_B = nR, the two forms are identical; one uses a count of molecules, the other a count of moles. At the same P, V and T, equal volumes of ideal gases contain equal numbers of molecules, whatever the gas.
Students often think Volumes can be substituted into PV = nRT in whatever units they are given, including dm³. In fact No. With R = 8.31 J mol⁻¹ K⁻¹, V must be in m³: 1 dm³ = 10⁻³ m³.
Students often think R and k_B are interchangeable gas constants, so PV/(RT) gives the number of molecules just as PV/(k_B T) does. In fact No. R = 8.31 J mol⁻¹ K⁻¹ is the gas constant per mole and gives n from PV = nRT; k_B = 1.38 × 10⁻²³ J K⁻¹ is the constant per molecule and gives N from PV = Nk_B T.
Pressure from molecular collisions
Pressure from molecular collisions
In the kinetic model, a molecule of mass m that strikes a wall with velocity component v_x perpendicular to it rebounds elastically with component −v_x, so its momentum changes by 2mv_x. The wall exerts the force that causes this change, and by Newton's third law the molecule exerts an equal and opposite force on the wall. The pressure is the total rate of change of momentum from all such collisions per unit area of wall. Pressure rises if collisions become more frequent or each transfers more momentum.
Mean square speed (v²)
In P = ⅓ρv², v² is the average of the squares of the translational speeds of the molecules: square each molecule's speed, then take the mean. This is larger than the square of the mean speed (for the speeds of an ideal gas, (mean speed)² is about 0.85 of v²). Its square root is the root mean square (r.m.s.) speed. Unit: m² s⁻².
P = ⅓ρv²
The result of the kinetic analysis of pressure: P = ⅓ρv², where ρ is the density of the gas (kg m⁻³) and v² the mean square speed of its molecules. The factor ⅓ arises because the molecules move randomly in three dimensions, so on average only one third of v² is the square of the velocity component perpendicular to any one wall.
Students often think Gas molecules repel one another, and this repulsion, or the molecules pushing on each other, pushes outwards on the walls. In fact No. In the kinetic model pressure is caused by molecules colliding with the walls and changing momentum; there are no intermolecular forces between collisions.
Students often think Molecules give up their kinetic energy to the wall when they hit it, and this energy delivered per second is the pressure. In fact No. In elastic collisions with a stationary wall the molecules keep their kinetic energy. Pressure is a force per unit area, produced by the rate of change of momentum of the molecules.
Internal energy of an ideal monatomic gas
Internal energy of an ideal monatomic gas
The total energy of the molecules of the gas. In an ideal gas there are no intermolecular forces, so there is no intermolecular potential energy, and in a monatomic gas the only energy is translational kinetic energy. Hence U = (3/2)Nk_B T = (3/2)RnT. U depends only on the number of molecules and the absolute temperature: at constant T it does not change, whatever happens to P and V. SI unit: joule (J).
Average translational kinetic energy of a molecule
For an ideal gas the average translational kinetic energy of one molecule is (3/2)k_B T, which is proportional to the absolute temperature and is the same for every gas at the same temperature. Multiplying by N gives the internal energy of a monatomic ideal gas, U = (3/2)Nk_B T. Molecules of a heavier gas at the same temperature therefore move more slowly on average.
Students often think Because PV = nRT has units of joules, nRT (or PV) is the internal energy of the gas. In fact No. For an ideal monatomic gas U = (3/2)nRT = (3/2)PV. PV has units of energy, but it is two thirds of the internal energy, not all of it.
Students often think When a gas is compressed, its molecules are closer together, so they store more potential energy and the internal energy increases even at constant temperature. In fact No. An ideal gas has no intermolecular forces, so it has no intermolecular potential energy at any separation. Its internal energy is entirely the kinetic energy of its molecules.
Conditions for a real gas to approximate an ideal gas
Conditions for a real gas to approximate an ideal gas
A real gas behaves most nearly as an ideal gas at low pressure, low density and high temperature (well above its boiling point). At low density the molecules are far apart, so their own volume is a negligible fraction of the container's volume and the time they spend close enough to attract one another is negligible; at high temperature their kinetic energy is large compared with the intermolecular potential energy. Deviations are largest at high pressure and density and near the point of liquefaction.
Students often think Cooling makes a gas behave more ideally, because slower molecules collide less often and so interact less. In fact No. Real gases deviate MORE from ideal behaviour at low temperature, especially close to their boiling point.
Students often think The kinetic model is about molecular collisions, so conditions with more collisions (high pressure, high density, high temperature) make a gas fit the model better. In fact No. At high pressure and density the molecules are close together, so their own volume and their mutual attractions are no longer negligible, and the gas deviates more from ideal behaviour.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 A sealed container holds a gas at a pressure of 1.0 × 10⁵ Pa. Which statement about this pressure is correct?
Answer and reasoning
The gas presses mainly on the base of the container, because gas pressure acts downwards. — A student who thinks gas pressure is caused by the weight of the gas picks this. The weight of the gas in a container is negligible; the pressure comes from molecules colliding with every wall, and it is the same on the top, sides and base.
Pressure is a vector, so the pressures acting on opposite walls of the container cancel out. — A student who treats pressure as a vector picks this. Pressure is a scalar. The FORCES on opposite walls are opposite in direction, but the pressure on each wall is the same positive 1.0 × 10⁵ Pa.
The gas exerts a force of 1.0 × 10⁵ N on each wall, whatever the area of that wall. — A student who treats pressure and force as the same quantity picks this. The force depends on the area: F = PA, so a wall of 0.20 m² receives 2.0 × 10⁴ N and a wall of 2.0 m² receives 2.0 × 10⁵ N.
The gas pushes perpendicular to every wall, exerting 1.0 × 10⁵ N on each square metre of wall. — Pressure is force per unit area, so F = PA: every 1 m² of wall receives a force of 1.0 × 10⁵ N, directed perpendicular to the wall. Because the molecules move randomly in all directions, this is true for the top, the sides and the base alike.
Working P = F/A, so F = PA = (1.0 × 10⁵ Pa)(1 m²) = 1.0 × 10⁵ N on each square metre, perpendicular to the wall.
2 A cylinder contains 0.64 kg of oxygen gas, which consists of O₂ molecules. The molar mass of oxygen atoms is 16 g mol⁻¹, and N_A = 6.02 × 10²³ mol⁻¹. How many oxygen molecules are in the cylinder?
Answer and reasoning
1.2 × 10²⁵ — The molar mass of O₂ is 2 × 16 = 32 g mol⁻¹ = 0.032 kg mol⁻¹, so n = 0.64 ÷ 0.032 = 20 mol. Then N = nN_A = 20 × 6.02 × 10²³ = 1.2 × 10²⁵ molecules.
1.2 × 10²² — A student who divides 0.64 kg by 32 g mol⁻¹ without converting units gets n = 0.020 mol, 1000 times too small. Convert first: 32 g mol⁻¹ = 0.032 kg mol⁻¹, giving n = 20 mol.
3.9 × 10²³ — A student who treats the mass, 0.64, as the amount of substance multiplies it by N_A. The amount of substance is a count of particles in moles, found from n = m/M = 20 mol, not the mass itself.
2.4 × 10²⁵ — A student who uses 16 g mol⁻¹ has used the molar mass of oxygen ATOMS. The gas consists of O₂ molecules, so its molar mass is 32 g mol⁻¹ and there are half as many molecules as this answer suggests.
Working M(O₂) = 2 × 16 g mol⁻¹ = 32 g mol⁻¹ = 0.032 kg mol⁻¹. n = m/M = 0.64 kg ÷ 0.032 kg mol⁻¹ = 20 mol. N = nN_A = 20 mol × 6.02 × 10²³ mol⁻¹ = 1.2 × 10²⁵.
3 Which of the following is an assumption of the kinetic model of an ideal gas?
Answer and reasoning
All the molecules move with the same speed, which increases with temperature. — A student who pictures every molecule with the same speed picks this. The model has molecules in random motion with a range of speeds; temperature fixes the average kinetic energy, which is why P = ⅓ρv² uses a MEAN square speed.
Molecules lose a small amount of kinetic energy in every collision with one another. — A student who thinks molecular collisions are like everyday inelastic ones picks this. The model assumes all collisions are elastic, so the total kinetic energy of the molecules is unchanged by collisions.
Molecules exert no forces on one another except during collisions. — This is one of the model's assumptions. Together with point-particle molecules, elastic collisions of negligible duration and random motion, it defines the ideal gas; it also means an ideal gas has no intermolecular potential energy.
The molecules are packed closely together and fill most of the container's volume. — A student who pictures gas molecules as crowded picks this. The model treats the molecules as point particles whose total volume is negligible compared with the volume of the container; a gas is mostly empty space.
4 For a fixed mass of gas, the ideal gas law PV/T = constant can be derived from three empirical gas laws. Which statement of those three laws is correct?
Answer and reasoning
PV is constant at constant θ; V/θ is constant at constant P; P/θ is constant at constant V (θ in °C) — A student who thinks any temperature scale will do picks this. V ∝ T and P ∝ T are proportionalities only when T is measured from absolute zero; with θ in °C, V/θ is not constant (it would be infinite at 0 °C).
PV, V/T and P/T each stay constant (T in K), whatever happens to the remaining variable — A student who drops the conditions from the laws picks this. Each law holds only when the third variable is fixed: for example PV is constant only at constant temperature. If all three change, only PV/T is constant.
PV is constant at constant T; V/T is constant at constant P; P/T is constant at constant V (T in K) — These are the three empirical laws, with temperature measured from absolute zero. Each is the special case of PV/T = constant in which one of P, V and T is held fixed, which is why they combine into the ideal gas law.
PV is constant at constant T; VT is constant at constant P; PT is constant at constant V (T in K) — A student who assumes every gas law has the inverse form of Boyle's law picks this. Heating a gas at constant pressure makes it expand, so V increases with T (V/T constant), and heating at constant volume raises P (P/T constant).
5 A rigid storage tank of internal volume 200 dm³ contains an ideal gas at a pressure of 3.00 × 10⁵ Pa and a temperature of 27 °C. R = 8.31 J mol⁻¹ K⁻¹. What amount of gas is in the tank?
Answer and reasoning
2.7 × 10² mol — A student who uses 27 for the temperature gets an answer about eleven times too large. The ideal gas law needs absolute temperature: 27 °C = 300 K.
2.4 × 10⁴ mol — A student who substitutes V = 200 without converting dm³ gets an answer 1000 times too large. With R in J mol⁻¹ K⁻¹, V must be in m³: 200 dm³ = 0.200 m³.
2.4 × 10¹ mol — Convert to SI: V = 200 dm³ = 0.200 m³ and T = 300 K. Then n = PV/(RT) = (3.00 × 10⁵ × 0.200) ÷ (8.31 × 300) = 24 mol = 2.4 × 10¹ mol.
2.4 × 10³ mol — A student who converts 200 dm³ to 20 m³ has divided by 10, the factor for lengths (10 dm = 1 m). For a volume the factor is cubed: 1 dm³ = 10⁻³ m³, so V = 0.200 m³.
Working V = 200 dm³ = 200 × 10⁻³ m³ = 0.200 m³; T = 27 + 273 = 300 K. n = PV/(RT) = (3.00 × 10⁵ Pa × 0.200 m³) ÷ (8.31 J mol⁻¹ K⁻¹ × 300 K) = 6.00 × 10⁴ J ÷ 2493 J mol⁻¹ = 24.1 mol ≈ 2.4 × 10¹ mol.
6 Which statement correctly explains how a gas exerts a pressure on the walls of its container?
Answer and reasoning
The weight of the molecules presses down on the container, mostly on its base. — A student who thinks gas pressure is caused by the weight of the gas picks this. The weight of a gas in a container is negligible, and the pressure is the same on the top and sides as on the base, because molecules strike every wall.
Each molecule's momentum changes as it rebounds from a wall, so it exerts a force on the wall. — The wall exerts a force on each molecule to reverse its perpendicular velocity (a momentum change of 2mv for an elastic rebound), and by Newton's third law the molecule exerts an equal and opposite force on the wall. The total force from all collisions per unit area is the pressure.
The molecules repel one another, and this repulsion pushes outwards on the walls. — A student who thinks of gas pressure as molecules pushing each other apart picks this. In the kinetic model there are no forces between molecules except during collisions; the pressure comes from momentum changes at the walls.
Molecules give their kinetic energy to the walls, and this energy is the pressure. — A student who confuses force with energy picks this. In an elastic rebound from a stationary wall a molecule keeps its kinetic energy; what changes is its momentum, and force is the rate of change of momentum.
7 A container holds 8.0 g of helium, a monatomic gas of molar mass 4.0 g mol⁻¹, at 27 °C. Treat helium as an ideal gas. R = 8.31 J mol⁻¹ K⁻¹. What is the internal energy of the helium?
Answer and reasoning
7.5 × 10³ J — n = 8.0 g ÷ 4.0 g mol⁻¹ = 2.0 mol and T = 300 K, so U = (3/2)RnT = 1.5 × 8.31 × 2.0 × 300 = 7.5 × 10³ J.
5.0 × 10³ J — A student who calculates nRT = 4986 J has left out the 3/2. nRT equals PV, which is two thirds of the internal energy of a monatomic ideal gas: U = (3/2)nRT.
6.7 × 10² J — A student who uses 27 for the temperature gets an answer about eleven times too small. U is proportional to the absolute temperature, 300 K.
3.0 × 10⁴ J — A student who uses the mass, 8.0, as the amount of substance gets four times the correct value. The amount of substance is n = m/M = 8.0 g ÷ 4.0 g mol⁻¹ = 2.0 mol.
Working n = m/M = 8.0 g ÷ 4.0 g mol⁻¹ = 2.0 mol; T = 27 + 273 = 300 K. U = (3/2)RnT = 1.5 × 8.31 J mol⁻¹ K⁻¹ × 2.0 mol × 300 K = 7479 J ≈ 7.5 × 10³ J.
8 Under which conditions does a real gas behave most nearly as an ideal gas?
Answer and reasoning
High temperature, low pressure and low density — At low pressure and density the molecules are far apart, so their own volume and their mutual attractions are negligible; at high temperature their kinetic energy is large compared with the intermolecular potential energy. These are the conditions assumed by the model.
Low temperature, low pressure and low density — A student who thinks slower molecules interact less picks this. At low temperature the molecules' kinetic energy is small compared with the intermolecular potential energy, so attractions matter more and the gas deviates more, eventually liquefying.
High temperature, high pressure and high density — A student who thinks more collisions means a better fit to the kinetic model picks this. At high pressure and density the molecules are crowded, so their volume and attractions are no longer negligible, and the gas deviates from ideal behaviour.
Any temperature and pressure, provided the gas is helium — A student who thinks some gases are ideal gases picks this. Whether a real gas approximates an ideal gas depends on the conditions; helium is close to ideal at ordinary conditions but deviates at high pressure and near 4 K, where it liquefies.
9 A student pushes on a horizontal table top through a flat pad of area 8.0 cm². The force exerted by the pad on the table is 40 N, directed at 30° to the vertical. What pressure does the pad exert on the table?
Answer and reasoning
5.0 × 10⁴ Pa — A student who divides the whole 40 N by the area gets 5.0 × 10⁴ Pa. The horizontal component of the force pushes along the table and produces no pressure; only the perpendicular component, 40 cos 30° = 34.6 N, counts.
4.3 × 10⁴ Pa — Only the component perpendicular to the table produces pressure. The vertical component is 40 cos 30° = 34.6 N, and A = 8.0 cm² = 8.0 × 10⁻⁴ m², so P = 34.6 ÷ 8.0 × 10⁻⁴ = 4.3 × 10⁴ Pa.
2.5 × 10⁴ Pa — A student who uses 40 sin 30° = 20 N has taken the component parallel to the table. The angle is measured from the vertical, so the component perpendicular to the table is the adjacent one, 40 cos 30° = 34.6 N.
4.3 × 10² Pa — A student who converts 8.0 cm² to 8.0 × 10⁻² m² has divided by 100, the factor for lengths. For an area the factor is squared: 1 cm² = 10⁻⁴ m², so A = 8.0 × 10⁻⁴ m² and P is 100 times larger.
Working Perpendicular component of the force: F⊥ = 40 N × cos 30° = 34.6 N. Area: A = 8.0 cm² = 8.0 × 10⁻⁴ m². P = F⊥/A = 34.6 N ÷ 8.0 × 10⁻⁴ m² = 4.33 × 10⁴ Pa ≈ 4.3 × 10⁴ Pa.
10 Two sealed flasks each contain 0.50 mol of gas. Flask X holds helium, a monatomic gas of molar mass 4.0 g mol⁻¹. Flask Y holds nitrogen, which consists of N₂ molecules; the molar mass of nitrogen atoms is 14 g mol⁻¹. Which statement correctly compares the two samples?
Answer and reasoning
The number of N atoms equals the number of He atoms, and the nitrogen has 3.5 times the mass. — A student who takes the molar mass of nitrogen gas to be that of its atoms, 14 g mol⁻¹, gets a mass of 7.0 g and a ratio of 3.5. Nitrogen gas is made of N₂ molecules of molar mass 28 g mol⁻¹, so the sample has a mass of 14 g, and it contains twice as many atoms as the helium flask, not the same number.
The number of N₂ molecules equals the number of He atoms, and the nitrogen has seven times the mass. — The amount of substance is a count of particles, N = nN_A, so equal amounts contain equal numbers of particles: 0.50 mol of N₂ molecules and 0.50 mol of He atoms. The masses differ because the molar masses differ: 0.50 × 28 = 14 g of nitrogen against 0.50 × 4.0 = 2.0 g of helium, a ratio of 7.
The two samples have equal masses, because each flask holds the same amount of substance. — A student who identifies the amount of substance with mass picks this. Amount of substance is a count of particles in moles; the mass of a sample is nM, so equal amounts of substances with different molar masses have different masses: 2.0 g of helium and 14 g of nitrogen here.
The nitrogen has more molecules than the helium has atoms, since a mole of heavier particles holds more. — A student who thinks the number of particles in a mole depends on the substance picks this. One mole of anything contains N_A = 6.02 × 10²³ particles, so 0.50 mol of nitrogen contains exactly as many molecules as 0.50 mol of helium contains atoms. The heavier molecules make the sample heavier, not more numerous.
Working The amount of substance counts particles: N = nN_A, so 0.50 mol of He atoms and 0.50 mol of N₂ molecules contain the same number of particles, 0.50 × 6.02 × 10²³ = 3.0 × 10²³ each.
Mass of helium = nM = 0.50 mol × 4.0 g mol⁻¹ = 2.0 g.
Molar mass of N₂ = 2 × 14 = 28 g mol⁻¹, so mass of nitrogen = 0.50 mol × 28 g mol⁻¹ = 14 g.
Ratio of masses = 14/2.0 = 7.0, so the nitrogen has seven times the mass of the helium.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
9 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 At room temperature and atmospheric pressure, measurements on helium agree closely with PV = nRT. A student concludes: 'Helium is an ideal gas.' Which evaluation of this conclusion is correct?
Answer and reasoning
Justified: helium is monatomic and unreactive, and this makes it one of the ideal gases. — A student who thinks some gases simply are ideal picks this. No real gas is ideal; helium can be liquefied, which is impossible for an ideal gas. 'Ideal' describes a model that helium approximates well under some conditions.
Not justified: helium atoms collide with one another, but ideal gas molecules do not collide. — A student who reads 'no intermolecular forces' as 'no collisions' picks this. Ideal gas molecules do collide, elastically and briefly; the model excludes forces between molecules only while they are apart.
Justified: agreement with PV = nRT shows that every assumption of the model is exactly true for helium. — A student who thinks a model that predicts well must be literally true picks this. The model's assumptions (point particles, no forces between collisions) are never exactly true; agreement shows only that the simplifications have negligible effect under these conditions.
Not justified: helium is a real gas that the model approximates closely under these conditions. — The ideal gas is a model. Helium atoms have a small volume and exert weak attractions, so helium is a real gas; under these conditions the effects are tiny and the model fits well. At high pressure or near 4 K, where helium liquefies, it deviates.
2 Air in a sealed syringe has a volume of 60 cm³ at a pressure of 1.00 × 10⁵ Pa and a temperature of 17 °C. The plunger is pushed in quickly until the volume is 20 cm³, and the air warms to 47 °C. Treat the air as an ideal gas. What is its new pressure?
Answer and reasoning
3.00 × 10⁵ Pa — A student who applies Boyle's law, P₁V₁ = P₂V₂, gets 3.00 × 10⁵ Pa. Boyle's law needs constant temperature; here the air warms from 290 K to 320 K, so the pressure rises by a further factor of 320/290.
3.31 × 10⁵ Pa — Using P₁V₁/T₁ = P₂V₂/T₂ with T in kelvin: P₂ = 1.00 × 10⁵ × (60/20) × (320/290) = 3.31 × 10⁵ Pa. Both the compression and the warming raise the pressure.
8.29 × 10⁵ Pa — A student who uses the Celsius temperatures multiplies by 47/17 = 2.76. The gas laws need absolute temperatures: 17 °C = 290 K and 47 °C = 320 K, a ratio of only 1.10.
3.68 × 10⁴ Pa — A student who multiplies the pressure by 'new over old' for each variable uses 20/60 for the volume and gets a pressure LOWER than at the start. Compression raises the pressure: P ∝ 1/V, so the volume ratio must be 60/20.
Working T₁ = 17 + 273 = 290 K; T₂ = 47 + 273 = 320 K. P₁V₁/T₁ = P₂V₂/T₂ gives P₂ = P₁ × (V₁/V₂) × (T₂/T₁) = 1.00 × 10⁵ Pa × (60 cm³/20 cm³) × (320 K/290 K) = 3.31 × 10⁵ Pa. (The volume units cancel in the ratio, so cm³ may be used.)
3 On a pressure–volume diagram (pressure on the vertical axis, volume on the horizontal axis), a fixed mass of ideal gas moves from state X to state Y along a horizontal straight line. At X the volume is 3.0 × 10⁻³ m³ and the temperature is 27 °C; at Y the volume is 6.0 × 10⁻³ m³. What is the temperature at Y, and under what condition does the change X → Y take place?
Answer and reasoning
327 °C, at constant pressure — A horizontal line on a P–V diagram means the pressure is constant, so V/T is constant with T in kelvin. The volume doubles, so T doubles from 300 K to 600 K, which is 327 °C.
54 °C, at constant pressure — A student who doubles the Celsius temperature gets 54 °C. V ∝ T only for absolute temperature: 27 °C = 300 K doubles to 600 K, which is 327 °C.
27 °C, at constant temperature — A student who reads a horizontal line as 'no change' takes the change to be at constant temperature. Along a horizontal line on a P–V diagram the pressure is fixed; constant-temperature changes lie on curves on which PV is constant.
−123 °C, at constant pressure — A student who treats volume and temperature as inversely related, like pressure and volume, halves the kelvin temperature to 150 K = −123 °C. At constant pressure a gas expands when heated: V/T is constant, so doubling V doubles T.
Working Horizontal line on a P–V diagram: P constant. V/T = constant with T in K: T_X = 27 + 273 = 300 K; T_Y = T_X × (V_Y/V_X) = 300 K × (6.0 × 10⁻³/3.0 × 10⁻³) = 600 K = 327 °C.
4 A flask of volume 250 cm³ contains an ideal gas at a pressure of 1.0 × 10⁵ Pa and a temperature of 300 K. k_B = 1.38 × 10⁻²³ J K⁻¹, R = 8.31 J mol⁻¹ K⁻¹ and N_A = 6.02 × 10²³ mol⁻¹. How many molecules are in the flask?
Answer and reasoning
1.0 × 10⁻² — A student who uses R instead of k_B has calculated PV/(RT) = 1.0 × 10⁻² mol, the amount of substance, not the number of molecules. R is the constant per mole; k_B is the constant per molecule.
6.0 × 10²⁵ — A student who converts 250 cm³ to 2.5 m³ has divided by 100, the factor for lengths. For a volume the factor is cubed: 1 cm³ = 10⁻⁶ m³, so V = 2.5 × 10⁻⁴ m³.
3.6 × 10⁴⁵ — A student who multiplies PV/(k_B T) by N_A treats the N in PV = Nk_B T as a number of moles. N is already the number of molecules; multiplying again by N_A gives an impossible 10⁴⁵ molecules in a small flask.
6.0 × 10²¹ — PV = Nk_B T gives N directly: V = 250 cm³ = 2.5 × 10⁻⁴ m³, so N = (1.0 × 10⁵ × 2.5 × 10⁻⁴) ÷ (1.38 × 10⁻²³ × 300) = 6.0 × 10²¹.
Working V = 250 cm³ = 2.5 × 10⁻⁴ m³. N = PV/(k_B T) = (1.0 × 10⁵ Pa × 2.5 × 10⁻⁴ m³) ÷ (1.38 × 10⁻²³ J K⁻¹ × 300 K) = 25 J ÷ 4.14 × 10⁻²¹ J = 6.0 × 10²¹. (Check: n = PV/(RT) = 1.0 × 10⁻² mol, and nN_A = 6.0 × 10²¹.)
5 Two identical rigid containers are at the same temperature and the same pressure. One holds hydrogen (H₂) and the other holds carbon dioxide (CO₂). Both gases behave as ideal gases. How do the numbers of molecules in the two containers compare?
Answer and reasoning
They are equal, as equal pressure, volume and temperature mean equal numbers. — PV = Nk_B T contains no property of the molecules, so equal P, V and T mean equal N. A CO₂ molecule is heavier, but at the same temperature it moves more slowly, and the two effects on pressure cancel.
There are more H₂ molecules, as smaller molecules leave room for more of them. — A student who pictures the molecules filling the container picks this. In a gas the molecules occupy a negligible fraction of the volume, so their size does not limit how many fit; N depends only on P, V and T.
There are more CO₂ molecules, as the heavier gas is the larger amount of substance. — A student who treats the amount of substance as a mass picks this. The amount of substance counts molecules; from PV = nRT, equal P, V and T give equal n and so equal numbers of molecules, whatever their mass.
There are fewer CO₂ molecules, as each heavier molecule strikes the walls harder. — A student who forgets that heavier molecules move more slowly at the same temperature picks this. Each CO₂ molecule does transfer more momentum per collision, but it collides less often; at the same T the average pressure contribution per molecule is the same.
6 A sample of helium at 300 K has a density of 0.160 kg m⁻³. The mean speed of its atoms is 1.26 × 10³ m s⁻¹, and the mean of the squares of their speeds is 1.87 × 10⁶ m² s⁻². Treat helium as an ideal gas. What is its pressure?
Answer and reasoning
8.47 × 10⁴ Pa — A student who squares the mean speed uses (1.26 × 10³)² = 1.59 × 10⁶ m² s⁻². The equation needs the mean of the squares, which is larger than the square of the mean because fast atoms count for more when speeds are squared first.
2.99 × 10⁵ Pa — A student who leaves out the ⅓ gets a pressure three times too large. The atoms move randomly in three dimensions, so on average only a third of v² comes from the velocity component perpendicular to any one wall.
1.50 × 10⁵ Pa — A student who writes P = ½ρv² has borrowed the ½ from E_k = ½mv². The ⅓ in the pressure equation has a different origin: averaging random motion over three dimensions.
9.97 × 10⁴ Pa — In P = ⅓ρv², v² is the mean of the squared speeds, 1.87 × 10⁶ m² s⁻². P = ⅓ × 0.160 × 1.87 × 10⁶ = 9.97 × 10⁴ Pa.
Working P = ⅓ρv², with v² the mean of the squared speeds: P = ⅓ × 0.160 kg m⁻³ × 1.87 × 10⁶ m² s⁻² = 9.97 × 10⁴ Pa. The mean speed, 1.26 × 10³ m s⁻¹, is not needed.
7 A sealed, rigid container of ideal gas is heated. What happens to the pressure of the gas, and why?
Answer and reasoning
It rises: the molecules expand when they are heated, so they strike the walls more often. — A student who thinks the molecules themselves grow when heated picks this. Molecules do not change size; heating raises their speed, and that is what increases the frequency and momentum change of collisions.
It stays the same: the number of molecules per unit volume has not changed at all. — A student who thinks pressure depends only on how crowded the molecules are picks this. Pressure also depends on molecular speed (P = ⅓ρv²); at constant volume it rises in proportion to the kelvin temperature.
It rises: faster molecules hit the walls more often, each with a greater change in momentum. — The volume and number of molecules are fixed, but heating raises the mean square speed. Faster molecules cross the container more often, so they strike each wall more frequently, and each collision reverses a larger momentum. Both effects raise the force per unit area.
It rises: the molecules collide more often with one another, and so push harder on the walls. — A student who thinks pressure comes from molecules pushing on one another picks this. Collisions between molecules do not push on the walls; the pressure comes from the momentum change of molecules rebounding from the walls, which increases because they move faster.
8 A fixed mass of ideal monatomic gas is compressed slowly to half its volume while its temperature is kept constant. What happens to the internal energy U of the gas?
Answer and reasoning
It increases, as the closer molecules store more intermolecular potential energy. — A student who pictures compression like squeezing a spring picks this. An ideal gas has no intermolecular forces and therefore no intermolecular potential energy, whatever the separation of its molecules.
It does not change, as the temperature and so the mean kinetic energy are fixed. — For an ideal monatomic gas the internal energy is only the translational kinetic energy of the molecules, U = (3/2)Nk_B T. With N and T fixed, U is unchanged; the work done on the gas is transferred out of it thermally.
It increases, because work is done on the gas while it is being compressed. — A student who thinks work done on a gas must raise its internal energy picks this. In a slow compression at constant temperature, an equal amount of energy leaves the gas thermally, so U does not change.
It doubles, because the pressure doubles and molecules hit the walls twice as often. — A student who links internal energy to pressure picks this. The pressure does double, because collisions with the walls become twice as frequent, but the molecules' speeds and kinetic energies are unchanged, so U is unchanged.
Working For an ideal monatomic gas U = (3/2)Nk_B T. N is fixed (fixed mass) and T is constant, so ΔU = (3/2)Nk_B ΔT = 0.
9 A real gas at a temperature a little above its boiling point is compressed to a high density. Its measured pressure is found to be lower than the value that PV = nRT predicts for the same n, V and T. Which is the best explanation?
Answer and reasoning
The molecules take up part of the volume, which leaves less space for the gas and so lowers its pressure. — A student who reasons that less space means less gas picks this. The volume of the molecules reduces the free space they move in, so they strike the walls MORE often; on its own this raises the pressure above the ideal value.
Collisions between real molecules are inelastic, so the molecules slow down and the pressure falls. — A student who thinks molecular collisions lose energy picks this. If they did, a sealed gas would cool and its pressure fall with nothing leaving it, which is not observed; the deviation is caused by attractive forces, not by energy loss.
Attractive forces between the molecules pull those near a wall back, so they strike it with less momentum. — Close to the boiling point and at high density, intermolecular attractions are significant. A molecule approaching a wall is pulled back towards the bulk of the gas, so it delivers less momentum to the wall, and the pressure is below the ideal value.
Near the boiling point the molecules have almost stopped moving, so few of them reach the walls. — A student who links 'cold' with 'stationary' picks this. The molecules still move at hundreds of metres per second near the boiling point. The deviation arises because attractions become significant compared with their kinetic energy.
That was your twenty minutes. Real practice on B.3 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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