IB DP Physics: Ideal gases: pressure, the gas laws and kinetic theory (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover topic B.3 (Gas laws) of the IB Physics guide, first assessment 2025, for Standard Level and Higher Level; none of B.3 is marked additional higher level. Constants are given in the questions where needed. Use R = 8.31 J K⁻¹ mol⁻¹, kB = 1.38 × 10⁻²³ J K⁻¹ and NA = 6.02 × 10²³ mol⁻¹, and convert temperatures with 273.

Learn the ideas first in the study guide and use the revision notes for formulas. Course page: IB DP Physics. Checklist: IB DP Physics checklist. Weak spots? Try the free 10-minute diagnostics.

Questions

1. A force of 450 N acts perpendicular to a piston of area 1.5 × 10⁻³ m². Calculate the pressure on the piston. [2]

2. (a) A sample contains 9.03 × 10²³ molecules. Calculate the amount of substance. [1] (b) Calculate the number of molecules in 0.300 mol of gas. [2]

3. (a) State two features of the ideal gas model used in kinetic theory. [2] (b) State the conditions of pressure, density and temperature under which a real gas behaves most like an ideal gas, and explain why. [3]

4. A fixed mass of air is trapped in a syringe.

(a) The volume is reduced from 60 cm³ to 24 cm³ at constant temperature. The initial pressure is 1.0 × 10⁵ Pa. Calculate the final pressure. [2] (b) A fixed mass of gas in a rigid can is at 2.2 × 10⁵ Pa and 293 K. It is heated to 353 K. Calculate the new pressure. [2]

5. A fixed mass of gas has volume 3.0 × 10⁻⁴ m³ at 1.6 × 10⁵ Pa and 300 K. It is compressed to 1.0 × 10⁻⁴ m³ and its temperature rises to 330 K. Calculate the final pressure. [3]

6. A gas cylinder of volume 12 litres holds an ideal gas at 1.5 × 10⁶ Pa and 295 K.

(a) Calculate the amount of gas. [3] (b) Calculate the number of molecules. [1] (c) The cylinder cools to 280 K. Calculate the new pressure. [2]

7. Calculate the number of molecules in 2.0 m³ of an ideal gas at 1.0 × 10⁵ Pa and 300 K, using an equation that contains kB. [3]

8. Oxygen has molar mass 0.0320 kg mol⁻¹. Treat it as an ideal gas at 1.00 × 10⁵ Pa and 300 K.

(a) Show that its density is about 1.28 kg m⁻³. [3] (b) Use P = (1/3)ρv² to calculate the root mean square speed of the molecules. [2]

9. The temperature of a gas in a sealed rigid container rises. Explain, in terms of molecules, why the pressure of the gas rises. [4]

10. A monatomic ideal gas of 1.50 mol is at 350 K.

(a) Calculate its internal energy. [2] (b) Calculate the mean translational kinetic energy of one atom. [2] (c) Calculate the decrease in internal energy when the gas is cooled to 280 K. [2]

11. An ideal gas of 0.080 mol goes round the cycle A to B to C to A.

A is 2.0 × 10⁵ Pa and 1.5 × 10⁻³ m³. A to B is a rise in pressure to 4.0 × 10⁵ Pa at constant volume. B to C is an expansion at constant temperature to 3.0 × 10⁻³ m³. C to A is a compression at constant pressure.

(a) Calculate the temperature at A, B and C. [3] (b) Describe the shape of each of the three lines on a graph of pressure (vertical) against volume. [3] (c) Show that the pressure at C is 2.0 × 10⁵ Pa. [2]

12. A sealed container of volume 6.0 × 10⁻³ m³ holds 0.40 mol of helium, a monatomic ideal gas, at 330 K. The molar mass of helium is 4.00 × 10⁻³ kg mol⁻¹.

(a) Calculate the pressure of the gas. [2] (b) Calculate the internal energy. [2] (c) Calculate the density of the gas, and then the root mean square speed of its atoms. [3] (d) The temperature is raised to 660 K at constant volume. State the new pressure and internal energy, and the factor by which the root mean square speed changes. [2]

Answers

1. P = F/A [1] = 450 / 1.5 × 10⁻³ = 3.0 × 10⁵ Pa [1] Examiner insight: Write the equation and the unit; a bare number with no unit may not earn full credit, and either pascals or N m⁻² states the unit correctly.

2. (a) n = N/NA = 9.03 × 10²³ / 6.02 × 10²³ = 1.50 mol [1] (b) N = nNA [1] = 0.300 × 6.02 × 10²³ = 1.81 × 10²³ [1] Examiner insight: The rearrangement N = nNA is the method step. State the final answer to three significant figures.

3. (a) Any two from: molecules are treated as points or have negligible volume; molecules exert no forces on each other apart from during collisions; collisions are elastic; molecules are in constant random motion. [1] [1] (b) Low pressure and low density, and high temperature well above the condensation point [1]. At low density the molecules are far apart, so intermolecular forces and molecular volume matter little [1]. At high temperature the molecules move fast, so attractions have little effect on their motion [1] Examiner insight: A reason is needed alongside each condition. Answers that give the conditions but no explanation tend to stop at the first point.

4. (a) Constant temperature, so P₁V₁ = P₂V₂ [1]. P₂ = 1.0 × 10⁵ × 60 / 24 = 2.5 × 10⁵ Pa [1] (b) Constant volume, so P₁/T₁ = P₂/T₂ [1]. P₂ = 2.2 × 10⁵ × 353 / 293 = 2.65 × 10⁵ Pa [1] Examiner insight: Volume units cancel in part (a). In part (b) the ratio must use kelvin; using 20 °C and 80 °C gives a wrong value and the final answer will not earn credit.

5. P₁V₁/T₁ = P₂V₂/T₂ [1]. P₂ = P₁V₁T₂/(T₁V₂) = (1.6 × 10⁵ × 3.0 × 10⁻⁴ × 330) / (300 × 1.0 × 10⁻⁴) [1] = 5.3 × 10⁵ Pa [1] (5.28 × 10⁵) Examiner insight: Show the rearranged equation before substituting, so that a slip in one number is separate from a wrong method.

6. (a) V = 12 × 10⁻³ m³ [1]. n = PV/(RT) = (1.5 × 10⁶ × 12 × 10⁻³) / (8.31 × 295) [1] = 7.34 mol [1] (b) N = nNA = 7.34 × 6.02 × 10²³ = 4.42 × 10²⁴ [1] (c) Constant V and n, so P₂ = 1.5 × 10⁶ × 280 / 295 [1] = 1.42 × 10⁶ Pa [1] Examiner insight: The litre-to-m³ conversion is a common place to drop a mark. Write it as its own line so the conversion is visible.

7. PV = NkBT [1]. N = PV/(kBT) = (1.0 × 10⁵ × 2.0) / (1.38 × 10⁻²³ × 300) [1] = 4.8 × 10²⁵ [1] (4.83 × 10²⁵) Examiner insight: Using nR instead of kB is a different route; it works only if you then convert n to molecules.

8. (a) Mass of one mole = 0.0320 kg, so ρ = nM/V [1]. From PV = nRT, n/V = P/(RT) [1]. ρ = PM/(RT) = (1.00 × 10⁵ × 0.0320) / (8.31 × 300) = 3200 / 2493 = 1.28 kg m⁻³ [1] (b) v² = 3P/ρ = 3 × 1.00 × 10⁵ / 1.28 [1]; v = 4.8 × 10² m s⁻¹ [1] Examiner insight: On a “show that” question, give the unrounded value (1.284) or more digits than the target, then state the target value.

9. Mean kinetic energy of the molecules is proportional to the kelvin temperature, so the molecules move faster [1]. Wall collisions become more frequent [1]. Each collision has a greater change in momentum [1]. So the rate of change of momentum, which is the force on the wall, rises, and force per area is greater, so pressure rises [1] Examiner insight: Naming both the rate of collisions and the momentum change per collision is what separates a full answer from “they collide harder”.

10. (a) U = (3/2)nRT [1] = 1.5 × 1.50 × 8.31 × 350 = 6.54 × 10³ J [1] (b) (3/2)kBT [1] = 1.5 × 1.38 × 10⁻²³ × 350 = 7.25 × 10⁻²¹ J [1] (c) ΔU = (3/2) × 1.50 × 8.31 × 70 [1] = 1.31 × 10³ J [1] Examiner insight: Part (b) asks about one atom, so it uses kB, not R. Using nR here gives a value for a mole and loses the marks.

11. (a) T = PV/(nR). At A, PV = 300 J, so T = 300 / (0.080 × 8.31) = 451 K [1]. At B, PV = 4.0 × 10⁵ × 1.5 × 10⁻³ = 600 J, so T = 903 K [1]. B to C is isothermal, so at C T = 903 K [1] (b) A to B: a vertical straight line [1]. B to C: a curve (hyperbola) where PV is constant, falling from B [1]. C to A: a horizontal straight line [1] (c) B to C is at constant temperature, so PV is constant: 4.0 × 10⁵ × 1.5 × 10⁻³ = P × 3.0 × 10⁻³ [1]; P = 600 / 3.0 × 10⁻³ = 2.0 × 10⁵ Pa [1] Examiner insight: Use the information the question gives about each process to fix the temperature at the next point, and give the shape of each line with its direction, not just its name.

12. (a) P = nRT/V = (0.40 × 8.31 × 330) / 6.0 × 10⁻³ [1] = 1.83 × 10⁵ Pa [1] (b) U = (3/2)nRT = 1.5 × 0.40 × 8.31 × 330 [1] = 1.65 × 10³ J [1] (c) Mass = 0.40 × 4.00 × 10⁻³ = 1.60 × 10⁻³ kg, so ρ = 1.60 × 10⁻³ / 6.0 × 10⁻³ = 0.267 kg m⁻³ [1]. v² = 3P/ρ = 3 × 1.83 × 10⁵ / 0.267 [1]; v = 1.43 × 10³ m s⁻¹ [1] (d) Pressure doubles to 3.66 × 10⁵ Pa, U doubles to 3.29 × 10³ J [1], and the speed rises by a factor of √2, 1.41 [1] Examiner insight: Part (d) can be answered by proportion without recalculating: P and U are both proportional to T at fixed volume, and speed is proportional to the square root of T.

Where marks are usually lost

• Substituting degrees Celsius into a gas equation.
• Leaving volumes in litres or cm³ when the pressure is in pascals and the equation contains R.
• Using R with a number of molecules, or kB with moles.
• Quoting U = (3/2)nRT without saying the gas is monatomic.
• Explaining pressure only as molecules “hitting harder”, with no mention of collision rate or rate of change of momentum.
• Listing conditions for ideal behaviour with no reason.
• Treating the mean of the speeds as the v in P = (1/3)ρv² instead of the root mean square speed.
• Describing a constant-temperature curve on a P-V diagram as a straight line.

Next steps

• Review the formulas in the revision notes.
• Re-read the derivations in the study guide.
• Course page: IB DP Physics and the checklist.
• Take the free 10-minute diagnostics to find what to revise next.
• Book a free trial class.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This page covers section B.3 (Gas laws).