Ideal gases and kinetic theory: worked solutions

Answer: n = 0.181 mol; N = 1.09 × 10²³ atoms; atomic mass = 6.64 × 10⁻²⁷ kg; total mass = 0.722 g; density = 0.289 kg/m³.

1. p = 1.80 × 10⁵ Pa; V = 2.50 × 10⁻³ m³; T = 27 + 273 = 300 K.
   n = pV/(RT) = 450/(8.31 × 300) = 0.180505… ≈ 0.181 mol.
2. N = nN_A = 0.180505… × 6.02 × 10²³ = 1.09 × 10²³ atoms.
3. M = 4.00 × 10⁻³ kg mol⁻¹;
   m = M/N_A = (4.00 × 10⁻³)/(6.02 × 10²³) = 6.64 × 10⁻²⁷ kg per atom.
4. m_gas = nM = 7.22022… × 10⁻⁴ kg = 0.722 g;
   ρ = m_gas/V = 0.289 kg m⁻³.The molecular mass m, molar mass M and whole-sample mass mgas are three different quantities.

Check: A gas density below 1 kg/m³ is plausible for helium at these conditions. N has no unit of mol; n does.

Mark checklist · 6 marks

• 2 marks — Correct SI/K conversions and n = pV/(RT), giving approximately 0.181 mol. Award one for a valid equation/substitution, one for the amount with units.
• 1 mark — N = nNA, approximately 1.09 × 10²³ atoms; carry a previous amount error forward.
• 1 mark — M/NA using M in kg/mol, approximately 6.64 × 10⁻²⁷ kg per atom.
• 1 mark — nM gives approximately 7.22 × 10⁻⁴ kg or 0.722 g; carry the earlier amount forward.
• 1 mark — Total sample mass divided by volume gives approximately 0.289 kg/m³; carry earlier values forward.

Common mistake: Putting 27 into the ideal-gas equation, using 2.50 as m³, or treating 4.00 g/mol as the mass of one atom creates large errors.

Exam technique: Name the quantity required before selecting an equation. The checked reports distinguish number of particles N from amount in moles n.

Related examiner evidence: AQA 7408/2, June 2022 — Q02.3; report p. 3 · AQA 7408/2, June 2023 — Q02.1–02.2; report p. 3

Answer: T₂ = 405 K, or 132 °C. The equation uses thermodynamic temperature relative to absolute zero.

1. For fixed n, pV/T = nR is constant.
   T₁ = 300 K;
   T₂ = T₁(p₂V₂)/(p₁V₁) = 300(90.0 × 7.20)/(120 × 4.00) = 405 K.
2. θ₂ = 405 − 273 = 132 °C.
3. The proportional gas relation uses temperature measured from absolute zero. Celsius has an offset: 0 °C still corresponds to 273 K, so multiplying Celsius values does not preserve gas-state ratios.

Check: p₂V₂/p₁V₁ = 1.35, so final kelvin temperature must be 1.35 times the initial value.

Mark checklist · 4 marks

• 1 mark — Identifies fixed amount and uses p₁V₁/T₁ = p₂V₂/T₂.
• 1 mark — Converts the initial temperature to 300 K and correctly obtains 405 K.
• 1 mark — Converts final temperature to 132 °C; carry an earlier temperature answer forward.
• 1 mark — Explains that gas equations use absolute temperature measured from absolute zero, whereas Celsius has an offset.

Common mistake: Using p/T = constant ignores the changed volume. Adding 273 to a temperature difference is also wrong: differences in K and °C have the same numerical value.

Exam technique: State what is held constant before simplifying a gas equation. These new state values are original teaching data.

Answer: pgas = 2.50 × 10⁵ Pa; T₂ = 450 K; work done by the gas = 150 J. Of this, 90 J lifts the load and 60 J displaces the atmosphere.

1. p_gasA = p_atmA + F_load;
   p_gas = 1.00 × 10⁵ + 300/(2.00 × 10⁻³) = 2.50 × 10⁵ Pa.The load supplies a gauge-pressure contribution of 1.50 × 10⁵ Pa, not the whole absolute pressure.
2. At fixed amount and pressure, V/T is constant.
   T₂ = 300(1.80/1.20) = 450 K.
3. ΔV = 0.60 × 10⁻³ m³;
   W_{by gas} = pΔV = (2.50 × 10⁵)(0.60 × 10⁻³) = 150 J.
4. Δx = ΔV/A = 0.300 m;
   W_load = 300(0.300) = 90.0 J;
   W_atmosphere = p_atmΔV = 60.0 J.The gas does work against both external contributions. This is work by the gas, reported as positive for expansion.

Check: The pressure–volume rectangle has area 150 J. The independent force × displacement route gives (300 + 200) × 0.300 = 150 J.

Mark checklist · 7 marks

• 1 mark — Balances gas force against atmosphere plus load.
• 1 mark — Finds absolute pressure approximately 2.50 × 10⁵ Pa.
• 1 mark — Uses V₁/T₁ = V₂/T₂ at constant p and n.
• 1 mark — Obtains 450 K; carry an earlier physically positive value forward.
• 1 mark — Uses constant absolute pressure and volume change in Wby gas = pΔV.
• 1 mark — Obtains 150 J; carry their absolute pressure forward.
• 1 mark — Explains that gas also displaces the atmosphere; load-only work omits that contribution.

Common mistake: Gauge pressure gives only the load contribution here. Using pV₂ rather than p(V₂−V₁) counts work over volume that did not change.

Answer: 20.0% of the original amount escaped; mass lost = 1.01 × 10⁻² kg = 10.1 g.

1. n₂/n₁ = (p₂T₁)/(p₁T₂) = (180 × 320)/(240 × 300) = 0.800.
2. The fraction remaining is 0.800, so the fraction lost is 1−0.800 = 0.200 = 20.0%. The constant V and R cancel in the ratio; n does not remain constant.
3. n₁ = (240 × 10³)(2.00 × 10⁻²)/(8.31 × 320) = 1.80505… mol;
   n₂ = (180 × 10³)(2.00 × 10⁻²)/(8.31 × 300) = 1.44404… mol;
   Δn = 0.361011… mol.
4. M = 0.0280 kg mol⁻¹;
   Δm = ΔnM = 1.01 × 10⁻² kg = 10.1 g.

Check: If there were no leak, cooling alone would give 225 kPa. The measured 180 kPa is lower, so gas loss is required by the stated model.

Mark checklist · 5 marks

• 1 mark — Uses n = pV/(RT) separately or a valid pressure/temperature amount ratio.
• 1 mark — Finds remaining amount ratio 0.800 or equivalent initial/final amounts.
• 1 mark — Reports fraction lost 0.200 or 20.0%, rather than fraction remaining.
• 1 mark — Determines amount lost approximately 0.361 mol; carry an earlier amount error forward.
• 1 mark — Uses M = 0.0280 kg/mol to find approximately 0.0101 kg lost; carry earlier amount lost forward.

Common mistake: A 25% pressure fall is not a 25% gas loss when temperature also falls. Subtract the changes in moles, not the mass of the tank.

Exam technique: The 2023 report stresses using both gas states and the difference in amounts when a mass change is involved; this leak context applies that lesson to new data.

Related examiner evidence: AQA 7408/2, June 2023 — Q02.1–02.2; report p. 3

Answer: Plot p against 1/V: gradient = 3.60 J; n = 1.44 × 10⁻³ mol. Compress slowly, allow temperature to settle and keep the trapped gas sealed.

1. For fixed n and T, p = (nRT)(1/V). Plot absolute pressure p vertically against reciprocal total volume 1/V horizontally. The ideal relationship is a straight line through the origin; the origin is a limiting extrapolation, not an attainable infinite-volume measurement.
2. Using the 20.0 and 50.0 cm³ readings:
   gradient = (180−72.0)/[(1/20.0)−(1/50.0)] = 3600 kPa cm³.
   1 kPa cm³ = (10³ Pa)(10⁻⁶ m³) = 10⁻³ J;
   SI gradient = 3.60 J.
3. n = gradient/(RT) = 3.60/(8.31 × 300) = 1.44 × 10⁻³ mol.Each row independently gives pV = 3.60 J; a best-fit line across several measured readings would reduce sensitivity to random errors in a real experiment.
4. Compress or expand slowly and wait for thermal equilibrium with the room or controlled bath before recording. Monitor temperature rather than assuming that a rapid compression is isothermal. Keep the fittings and syringe airtight, so the amount stays fixed. Use the known total volume, including tubing, and repeat readings to check consistency.

Check: At twice the volume, the data give half the absolute pressure. pV is the same for every supplied reading.

Mark checklist · 7 marks

• 1 mark — Chooses p vertically against 1/V horizontally, using absolute pressure.
• 1 mark — Justifies a straight line through the ideal origin from p = nRT(1/V), with n and T constant.
• 1 mark — Determines 3600 kPa cm³ or equivalent scaled gradient from separated points.
• 1 mark — Converts gradient to 3.60 Pa m³ = 3.60 J.
• 1 mark — Uses gradient/(RT) to obtain approximately 1.44 × 10⁻³ mol; carry earlier gradient forward.
• 1 mark — Slow adjustment/wait for thermal equilibrium with temperature monitoring, explicitly to hold T constant.
• 1 mark — An airtight sealed system to keep the gas amount constant.

Common mistake: p against V is a curve, not a straight line. Using only syringe-barrel volume would omit the gas in tubing. Gauge pressure adds an intercept in a p against 1/V plot.

Exam technique: The practical is an original RP8 application. The 2022 graph report illustrates why a correct general shape must also match the numerical gas-state points.

Related examiner evidence: AQA 7408/2, June 2022 — Q02.4; report p. 4

Answer: Gradient = 0.200 cm³/°C (equivalently cm³/K for an interval); extrapolated zero-volume intercept = −273 °C, corresponding to 0 K.

1. a = (66.6−54.6)/(60−0) = 0.200 cm³ °C⁻¹.
   V = 0.200θ + 54.6, with V in cm³ and θ in °C.
2. At the mathematical zero-volume intercept,
   0 = 0.200θ + 54.6;
   θ = −54.6/0.200 = −273 °C.Absolute zero is 0 K, the lower limit of thermodynamic temperature. This classroom data model uses T = θ + 273.
3. The measured points cover 0–60 °C. Extending the line far beyond them gives an ideal-gas extrapolation; a real gas can condense and depart from the model before that point. It is not an instruction to cool gas to zero volume.
4. Keep the gas sealed and dry, with no gas entering or leaving. Let the piston move freely against constant atmospheric pressure; check for sticking and stable ambient pressure. Immerse the whole gas-containing section, stir the bath and wait for temperature and volume to stabilise before reading. Repeat readings over a suitable range and read the volume scale without parallax.

Check: Using kelvin rewrites the fitted line as V = 0.200T. In this ideal data set it would pass through the kelvin origin.

Mark checklist · 7 marks

• 1 mark — Gradient approximately 0.200 cm³/°C, or equivalent cm³/K for the interval.
• 1 mark — Uses V = 0.200θ + 54.6 and sets V = 0, or a correctly constructed extrapolation.
• 1 mark — Obtains approximately −273 °C and relates it to 0 K.
• 1 mark — Identifies extrapolation outside measured range and real-gas condensation/model limitations.
• 1 mark — Keeps the dry trapped amount fixed in a sealed system.
• 1 mark — Uses freely moving low-friction piston and constant atmospheric pressure; addresses sticking/pressure stability.
• 1 mark — Whole gas is equilibrated to the bath temperature, with stirring/waiting before readings.

Common mistake: A negative Celsius intercept is not a negative kelvin temperature. The gradient is unchanged by adding a constant offset to the horizontal axis.

Exam technique: This is an original RP8 data-analysis task; no claim is made that the checked reports assessed this exact apparatus or data.

Answer: Nitrogen molecular mass = 4.65 × 10⁻²⁶ kg; crms = 517 m/s; mean translational KE = 6.21 × 10⁻²¹ J. Helium’s RMS speed is √7 = 2.65 times as large (about 1.37 × 10³ m/s), with the same mean translational KE.

1. m = M/N_A = 0.0280/(6.02 × 10²³) = 4.65 × 10⁻²⁶ kg.
2. c_rms = √(3RT/M) = √[(3 × 8.31 × 300)/0.0280] = 516.817… ≈ 517 m s⁻¹.Equivalently use √(3kT/m); the rounded constants cause a small final-digit difference.
3. ⟨E_k,trans⟩ = (3/2)kT = (3/2)(1.38 × 10⁻²³)(300) = 6.21 × 10⁻²¹ J.
4. c_rms,He/c_rms,N₂ = √(M_N₂/M_He) = √7 = 2.65.The lighter helium atoms have greater RMS speed but the same mean translational KE at the same T. RMS is √⟨c²⟩, not the mean speed and not the speed of every molecule. Rotational and vibrational energies are not being compared.

Check: Using R and molar mass or k and molecular mass gives the same physical result within rounding. A sevenfold mass ratio gives a √7 speed ratio, not seven.

Mark checklist · 6 marks

• 1 mark — Converts molar mass and uses M/NA to obtain approximately 4.65 × 10⁻²⁶ kg.
• 1 mark — Uses crms = √(3RT/M) with M in kg/mol, or the valid molecular-mass form.
• 1 mark — Obtains approximately 517 m/s; carry a previous molecular-mass error forward if using it consistently.
• 1 mark — Uses 3kT/2 to find approximately 6.21 × 10⁻²¹ J per molecule.
• 1 mark — Correct RMS speed ratio √7, approximately 2.65.
• 1 mark — Same mean translational kinetic energy because both gases have the same thermodynamic temperature.

Common mistake: Using M = 28 in SI or confusing mean speed with RMS speed gives an incorrect result. Equal temperature does not mean equal molecular speed.

Exam technique: The 2022 amount question combined molecular energy with the macroscopic gas equation. Keep per-particle, per-mole and whole-sample quantities distinct.

Related examiner evidence: AQA 7408/2, June 2022 — Q02.3; report p. 3

Answer: Δpx = −2mcₓ; Δt = 2L/cₓ; mean force on the wall = mcₓ²/L. Summing and dividing by area gives pV = Nm⟨cₓ²⟩ = ⅓Nmcrms².

1. p_x,before = +mc_x; p_x,after = −mc_x;
   Δp_x = p_after−p_before = −2mc_x.Elastic reflection keeps the speed while reversing the normal component.
2. The round-trip distance to the same wall is 2L.
   Δt = 2L/c_x.This is not the time to reach either wall and is not a wave-period equation.
3. ⟨F_{x,on molecule}⟩ = Δp_x/Δt = −mc_x²/L.
   By Newton III, ⟨F_{x,on wall}⟩ = +mc_x²/L.Newton II links the collision momentum change to the force on the molecule; Newton III gives the equal opposite force on the wall.
4. F_wall = (m/L)Σc_x,i² = (Nm/L)⟨c_x²⟩;
   p = F/L² = Nm⟨c_x²⟩/L³;
   pV = Nm⟨c_x²⟩.
5. For no preferred direction, ⟨c_x²⟩ = ⟨c_y²⟩ = ⟨c_z²⟩.
   ⟨c²⟩ = ⟨c_x²⟩ + ⟨c_y²⟩ + ⟨c_z²⟩ = 3⟨c_x²⟩;
   c_rms² ≡ ⟨c²⟩, so pV = ⅓Nmc_rms².
6. Two suitable additional assumptions are: the particles’ total physical volume is negligible compared with the container volume; and there are no intermolecular forces except during collisions. Alternatively, collision duration is negligible compared with the interval between collisions, or Newtonian mechanics applies. Do not merely repeat the conditions supplied in the question.

Check: The wall pressure is positive despite the molecule’s negative momentum change. (Nmcrms²) has units kg m²/s², the same as pV.

Mark checklist · 8 marks

• 1 mark — Explains before +mcx, after −mcx, hence final−initial = −2mcx.
• 1 mark — Uses round-trip 2L and the normal speed component to obtain 2L/cx.
• 1 mark — Uses Newton II with signed momentum change per interval for the mean particle force.
• 1 mark — Uses Newton III to obtain positive wall force mcx²/L.
• 1 mark — Sums the squared components and divides by L² to obtain pV = Nm⟨cx²⟩.
• 1 mark — Explains equal mean squared components for isotropic motion and crms² = ⟨c²⟩, producing factor ⅓.
• 2 marks — Two distinct further assumptions with the comparison or qualification stated where necessary. One mark each; do not repeat elastic collisions, random motion or given identical molecules.

Common mistake: The factor ⅓ concerns the mean squared velocity component, not three separate gas pressures. Mean velocity can be zero while RMS speed is nonzero.

Exam technique: The 2023 report asks for an explained momentum sign and the distance/time stage. The 2022 report stresses applying Newton’s laws to the wall and molecule, rather than quoting a law in isolation.

Related examiner evidence: AQA 7408/2, June 2023 — Q01.2–01.4; report p. 3 · AQA 7408/2, June 2022 — Q02.1–02.2; report p. 3

Answer: Internal energy is the total random translational KE of all atoms. Mean energy change = 3.11 × 10⁻²¹ J per atom; heat transferred = 374 J.

1. For this ideal monatomic gas, intermolecular potential energy is neglected and each atom has only translational kinetic energy. Internal energy is the total random KE of all atoms, not a mean per atom.
2. ΔT = 450−300 = 150 K;
   Δ⟨E_k⟩ = (3/2)kΔT = (3/2)(1.38 × 10⁻²³)(150) = 3.105 × 10⁻²¹ ≈ 3.11 × 10⁻²¹ J.
3. ΔU = (3/2)nRΔT = (3/2)(0.200)(8.31)(150) = 373.95 ≈ 374 J.The rigid boundary has ΔV = 0, so Wby gas = 0 and Q = ΔU under the stated energy-transfer assumptions.

Check: Initial U ≈ 748 J and final U ≈ 1122 J; their difference is 374 J. Multiplying the per-atom change by nNA agrees within rounding.

Mark checklist · 5 marks

• 1 mark — Total random translational kinetic energy of all atoms; excludes a mean/single-atom answer and intermolecular potential energy for this model.
• 1 mark — Uses ΔT = 150 K and Δ⟨Ek⟩ = 3kΔT/2.
• 1 mark — Finds approximately 3.11 × 10⁻²¹ J per atom.
• 1 mark — Uses ΔU = 3nRΔT/2, or particle number times per-atom energy change.
• 1 mark — Finds approximately 374 J heat into the gas and identifies zero expansion work; allow small constant-rounding differences.

Common mistake: Adding 273 to the 150 K interval, or calling 3.11 × 10⁻²¹ J the sample’s internal-energy change, confuses absolute temperature or total with mean. The formula U = 3nRT/2 is not a general total-energy formula for gases with rotational/vibrational modes.

Exam technique: The 2023 report distinguishes total gas internal energy from mean particle energy. The 2025 report warns against adding 273 to a temperature interval.

Related examiner evidence: AQA 7408/2, June 2023 — Q01.1; report p. 3 · AQA 7408/2, June 2025 — Q02.4; report p. 5

Answer: The observed object is the smoke grain, buffeted unequally by unseen air molecules. Molecular motion has a range of speeds and no preferred direction. Heating raises mean translational KE; at fixed p, N and m, volume increases. Gas laws summarise measured relationships, while kinetic theory explains them through a model.

1. The microscope observes the smoke grain, not individual air molecules. Many unseen molecules strike it; the impacts are not balanced at every instant, causing irregular changes in its momentum. This provides evidence for microscopic molecular motion.
2. In the equilibrium model, molecules have a range of speeds and their directions are equally likely, with no preferred direction. “Random” does not mean that Newton’s laws cease to apply. It is not the same definition as radioactive randomness.
3. ⟨E_k,trans⟩ = 3kT/2 and c_rms² = 3kT/m, so heating increases mean translational KE and c_rms².
   V = Nm c_rms²/(3p).p, N and molecular mass m remain constant, so V must increase. The symbols mean absolute pressure, number of particles and mass per particle. This is a kinetic-theory argument rather than merely quoting the empirical V/T relation.
4. In collision language, faster particles would transfer more momentum per collision and strike the walls more frequently at the old volume. Increasing the volume reduces the collision rate per wall area enough to restore the original pressure. It is not enough to say “the gas pushes outward”: it did that before heating as well.
5. Empirical gas laws were inferred from measurements of p, V, T and gas amount. Kinetic theory is a theoretical microscopic model with stated assumptions that explains those relationships. Brownian evidence supports the particle interpretation; agreement does not make all ideal-model assumptions exactly true for every real gas.

Check: A random-direction gas can have zero mean velocity while its mean squared speed and pressure remain positive.

Mark checklist · 8 marks

• 1 mark — Identifies the observed moving object as the smoke grain.
• 1 mark — Explains unequal/uneven impacts from unseen moving air molecules cause its irregular motion.
• 1 mark — Molecules have a range of speeds.
• 1 mark — No preferred direction/all directions equally likely, not merely several directions.
• 1 mark — Heating raises mean translational KE or crms².
• 1 mark — Uses the kinetic-pressure equation to infer increasing V, or a correct contextual momentum-transfer/collision-rate argument.
• 1 mark — Names and states p, N and m as constant; for an equivalent word-based route explicitly identifies constant pressure, amount and particle mass.
• 1 mark — Empirical gas laws summarise measured behaviour; kinetic theory is an explanatory model based on microscopic assumptions.

Common mistake: Saying “we can see air molecules” misidentifies Brownian evidence. Saying only that warmer gas expands does not explain the requested microscopic mechanism.

Exam technique: The 2025 report requires no preferred direction, not just different directions, and insists that named constants accompany an equation-based explanation.

Related examiner evidence: AQA 7408/2, June 2025 — Q02.1–02.2; report pp. 4–5 · AQA 7408/2, June 2025 — Q02.3; report p. 5