Cambridge International AS & A Level · 15 · structured
A full kinetic-pressure derivation
Derive gas pressure from momentum transfer · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 12
- Topics
- 1
- Answer
- Complete
A cube of side L contains N ideal-gas molecules of mass m. Starting with an elastic collision of one molecule with a wall, derive pV = (1/3)Nm⟨c²⟩. Explicitly justify the collision impulse, revisit time, summation over unequal molecular velocities and the factor 1/3. State two model assumptions.
- (a)
Derive the one-molecule average wall force.
3 marks - (b)
Sum to N molecules and extend to three dimensions.
4 marks - (c)
State two ideal-gas model assumptions.
2 marks - (d)
Use the derived relation to calculate the pressure of an ideal gas containing 1.20 × 10²³ molecules, each of mass 6.60 × 10⁻²⁶ kg, in volume 0.0200 m³. Its root-mean-square molecular speed is 4.00 × 10² m s⁻¹.
3 marks
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(a)
Derive the one-molecule average wall force.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
Elastic reversal from +mc_x to −mc_x gives wall impulse 2m|c_x| (1).
- 2
The molecule must make a round trip to revisit the same wall: time 2L/|c_x| (1).
- 3
Mean force = impulse / revisit time = mc_x²/L (1).
(b)
Sum to N molecules and extend to three dimensions.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
Sum over unequal molecular components: F=(m/L)Σc_x,i²=Nm⟨c_x²⟩/L (1).
- 2
p=F/L² and V=L³ give pV=Nm⟨c_x²⟩ (1).
- 3
Isotropy gives ⟨c_x²⟩=⟨c_y²⟩=⟨c_z²⟩; therefore ⟨c²⟩=3⟨c_x²⟩ (1).
- 4
Substitute to obtain pV=(1/3)Nm⟨c²⟩ (1).
(c)
State two ideal-gas model assumptions.
Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.
- 1
Award one mark each for any two independent assumptions: negligible molecular volume; no intermolecular forces between collisions; elastic wall collisions; negligible collision duration; random isotropic motion.
- 2
Do not credit equal speeds for all molecules: the derivation explicitly averages different squared speeds.
(d)
Use the derived relation to calculate the pressure of an ideal gas containing 1.20 × 10²³ molecules, each of mass 6.60 × 10⁻²⁶ kg, in volume 0.0200 m³. Its root-mean-square molecular speed is 4.00 × 10² m s⁻¹.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
Use ⟨c²⟩ = c_rms² = (4.00 × 10²)² = 1.60 × 10⁵ m² s⁻² (1).
- 2
Substitute p = (1.20 × 10²³)(6.60 × 10⁻²⁶)(1.60 × 10⁵)/(3 × 0.0200) (1).
- 3
Obtain 21120 Pa ≈ 2.11 × 10⁴ Pa (1).
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- (a) Elastic reversal from +mc_x to −mc_x gives wall impulse 2m|c_x| (1). Returning to the same wall takes a round trip 2L/|c_x| (1). Mean force is impulse divided by revisit time, mc_x²/L (1).
- (b) Sum unequal squared components: F = (m/L)Σc_x,i² = Nm⟨c_x²⟩/L (1). Divide by area L² and use V=L³ to get pV=Nm⟨c_x²⟩ (1). Isotropy gives equal mean-square components and ⟨c²⟩=3⟨c_x²⟩ (1). Thus pV=(1/3)Nm⟨c²⟩ (1).
- (c) Any two: molecular volume negligible compared with gas volume; no intermolecular forces between collisions; elastic wall collisions; collision duration negligible compared with time between collisions; random isotropic motion (1 each). Equal speeds for all molecules is not an assumption.
- (d) Use ⟨c²⟩ = c_rms² = (4.00 × 10²)² = 1.60 × 10⁵ m² s⁻² (1).
- (d) Substitute p = (1.20 × 10²³)(6.60 × 10⁻²⁶)(1.60 × 10⁵)/(3 × 0.0200) (1).
- (d) Obtain 21120 Pa ≈ 2.11 × 10⁴ Pa (1).
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