A Level Physics: Ideal Gases — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Ideal Gases revision notes

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Section A

1. State three assumptions of the kinetic theory of gases. [3]

2. Explain what is meant by an ideal gas, and why an ideal gas has no molecular potential energy. [2]

3. State what the mean kinetic energy of a gas molecule depends on. [1]

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Section B

4. A cylinder of volume 0.025 m³ contains oxygen at a pressure of 2.4 × 10⁵ Pa and temperature 27 °C. (R = 8.31 J mol⁻¹ K⁻¹; N_A = 6.02 × 10²³ mol⁻¹; k = 1.38 × 10⁻²³ J K⁻¹)

(a) Calculate the number of moles of oxygen. [3]

(b) Calculate the number of molecules. [1]

(c) Calculate the mean kinetic energy of one molecule. [2]

(d) The gas is heated to 127 °C at constant volume. Calculate the new pressure. [3]

5. Helium (M_r = 4.0) and argon (M_r = 40.0) are held at the same temperature.

(a) Compare the mean kinetic energy of a helium molecule with that of an argon molecule, giving a reason. [2]

(b) Compare their root-mean-square speeds, with a reason. [3]

6. Explain, in terms of the kinetic theory of gases, how the pressure exerted by a gas on the walls of its container arises from the motion of its molecules. [4]

7. State the kinetic theory equation pV = (1/3)Nm⟨c²⟩, defining each symbol, and explain why the root-mean-square speed is used rather than the mean speed of the molecules. [3]

8. Using your answer to Question 4(c) and the molar mass of oxygen (32 g mol⁻¹), calculate the root-mean-square speed of an oxygen molecule in the cylinder. [3]

9. Comparing pV = (1/3)Nm⟨c²⟩ with pV = nRT shows that the average translational kinetic energy of a molecule is directly proportional to what quantity? Explain the significance of this result. [2]

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Answers

1. Any three: a large number of molecules in random motion [1]; the volume of the molecules is negligible compared with the container [1]; there are no intermolecular forces except during collisions [1]; collisions are perfectly elastic; the time of a collision is negligible compared with the time between collisions.

2. A gas that obeys pV = nRT at all temperatures and pressures [1]. It has no molecular potential energy because the model assumes no intermolecular forces, and potential energy arises from such forces [1].

3. The absolute temperature only [1].

4. (a) T = 27 + 273 = 300 K [1]. n = pV ÷ RT = (2.4 × 10⁵ × 0.025) ÷ (8.31 × 300) [1] = 2.41 mol [1].

(b) N = nN_A = 2.41 × 6.02 × 10²³ = 1.45 × 10²⁴ molecules [1].

(c) E = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 300 [1] = 6.21 × 10⁻²¹ J [1].

(d) At constant volume, p₁/T₁ = p₂/T₂ [1]. T₂ = 400 K, so p₂ = 2.4 × 10⁵ × (400 ÷ 300) [1] = 3.2 × 10⁵ Pa [1].

5. (a) They are equal [1], because mean molecular kinetic energy depends only on absolute temperature, which is the same for both [1].

(b) Helium has the greater r.m.s. speed [1]. Since ½m⟨c²⟩ is the same for both [1], the molecule with the smaller mass must have the greater mean square speed [1]. Specifically √(40/4) = √10 ≈ 3.2 times greater.

6. Molecules are in continuous random motion and repeatedly collide with the walls of the container [1]. At each collision, the component of a molecule’s momentum perpendicular to the wall reverses, so the molecule undergoes a change in momentum [1]. By Newton’s second and third laws, this change in momentum exerts an equal and opposite force on the wall [1]. The large number of such collisions per second, spread over the wall’s area, produces a steady macroscopic pressure [1].

7. pV = (1/3)Nm⟨c²⟩ [1], where N is the number of molecules, m is the mass of one molecule and ⟨c²⟩ is the mean square speed [1]. Molecules move randomly in all directions, so the mean velocity is zero; averaging speeds directly is therefore less useful than averaging their squares (which are always positive), so the square root of the mean square speed is used instead [1].

8. Mass of one oxygen molecule: m = 0.032 ÷ (6.02 × 10²³) = 5.32 × 10⁻²⁶ kg [1]. From E = ½m⟨c²⟩, ⟨c²⟩ = 2E/m = (2 × 6.21 × 10⁻²¹) ÷ (5.32 × 10⁻²⁶) = 2.34 × 10⁵ m² s⁻² [1]. c_rms = √⟨c²⟩ = 483 m s⁻¹ [1].

9. The thermodynamic (absolute) temperature, T [1]. This means temperature is, at the molecular level, a direct measure of the average kinetic energy of the gas’s particles — not an arbitrary macroscopic scale [1].

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Where marks are usually lost

• Using °C anywhere in a gas equation.
• Confusing n (moles) with N (molecules), and hence R with k.
• Saying heavier molecules have greater kinetic energy at the same temperature.
• Confusing mean square speed with the square of the mean speed.
• Describing pressure as molecules simply “pushing” on the wall, rather than linking it to the rate of change of momentum during collisions.
• Averaging speed directly instead of averaging the square of speed when finding c_rms.
• Forgetting that N (number of molecules) and n (number of moles) require different constants (Nₐ vs R vs k).