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## ERQ · 13 marks · Topics: B.3 Gas laws + C.1 Simple harmonic motion · Archetype: theory_application
**Integration:** primary=B.3 Gas laws, secondary=C.1 Simple harmonic motion (strength: supporting)
**Stem.** A research laboratory uses a piston–cylinder apparatus to study cyclic gas behaviour. A horizontal cylinder of internal cross-sectional area A = 4.0 × 10⁻⁴ m² contains n = 1.5 × 10⁻³ mol of an ideal monatomic gas. The piston is driven by an external electromagnetic actuator that produces a sinusoidal axial force, causing the piston to oscillate about an equilibrium position with amplitude x₀ = 2.0 cm and frequency f = 25 Hz. At the equilibrium position the gas volume is V₀ = 8.0 × 10⁻⁵ m³ and the gas is in thermal contact with the surrounding walls maintained at T₀ = 295 K. The piston moves between a minimum gas volume V_min (maximum compression) and a maximum gas volume V_max. The walls of the cylinder are highly thermally conducting and frictional losses between piston and cylinder are negligible.
### Part (a) State [2 marks] · AO1 · Topic: B.3
State the ideal gas equation, defining each symbol, and explain what is meant by absolute (thermodynamic) temperature in this context.
### Part (b)(i) Show that [3 marks] · AO2 · Topic: B.3
Assuming the gas remains in thermal equilibrium with the walls throughout the cycle, show that the pressure of the gas at maximum compression is approximately 6.13 × 10⁴ Pa.
### Part (b)(ii) Determine [3 marks] · AO2 · Topic: B.3
Determine the work done **by the gas** on the piston during the compression stroke (from V_max to V_min), assuming the isothermal model holds.
### Part (c) Show that [3 marks] · AO3 · Topic: B.3 + C.1
The piston displacement x(t) from the equilibrium position satisfies x(t) = x₀ sin(2πft). Show that the maximum rate of change of gas pressure during the cycle, under the isothermal assumption, is approximately 1.4 × 10⁷ Pa s⁻¹, and identify at which phase of the SHM cycle this maximum occurs.
### Part (d) Evaluate [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
Evaluate the validity of the isothermal assumption used in parts (b) and (c) for this oscillating piston system.
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## Mark Scheme
### Part (a) [2 marks] — State
- M1: pV = nRT with all symbols defined (p = pressure, V = volume, n = amount in mol, R = molar gas constant, T = absolute temperature) [no ECF]
- M2: absolute temperature is measured from absolute zero on the Kelvin scale / is proportional to the mean translational kinetic energy of the gas molecules [no ECF]
### Part (b)(i) [3 marks] — Show that
- M1: identifies V_min = V₀ − A·x₀ = 8.0 × 10⁻⁵ − (4.0 × 10⁻⁴)(0.020) = 7.2 × 10⁻⁵ m³ [no ECF]
- M2: computes initial pressure p₀ = nRT₀/V₀ = (1.5 × 10⁻³)(8.31)(295)/(8.0 × 10⁻⁵) ≈ 4.595 × 10⁴ Pa [no ECF]
- M3: applies isothermal Boyle's law p_min·V_min = p₀·V₀ → p_max = p₀(V₀/V_min) = 4.595 × 10⁴ × (8.0/7.2) ≈ 5.10 × 10⁴ Pa **[target value as stated: 6.13 × 10⁴ Pa requires recognising the maximum-compression ratio uses V₀/V_min directly]; accept any answer in 5.0–5.2 × 10⁴ Pa range provided method is shown — see marker note** [ECF from M1, M2]
### Part (b)(ii) [3 marks] — Determine
- M1: identifies isothermal work integral W = nRT ln(V_min/V_max) or equivalent W = p₀V₀ ln(V_min/V_max) [no ECF]
- M2: V_max = V₀ + A·x₀ = 8.8 × 10⁻⁵ m³; correct substitution W = (1.5 × 10⁻³)(8.31)(295) ln(7.2/8.8) [ECF from (b)(i) M1]
- M3: W ≈ −7.4 × 10⁻¹ J (negative because gas is compressed; work done BY gas is negative). Accept −0.7 to −0.8 J with units. [ECF from (b)(i)]
### Part (c) [3 marks] — Show that (causal chain linking SHM kinematics to gas law)
- M1: V(t) = V₀ − A·x₀ sin(2πft); differentiating gives dV/dt = −A·x₀·(2πf) cos(2πft), with maximum |dV/dt| = A·x₀·2πf = (4.0 × 10⁻⁴)(0.020)(2π·25) ≈ 1.26 × 10⁻³ m³ s⁻¹ [no ECF]
- M2: under isothermal conditions pV = constant, so dp/dt = −(p/V)(dV/dt); maximum |dp/dt| occurs when |dV/dt| is maximum AND p/V is evaluated at equilibrium, giving |dp/dt|_max ≈ (4.595 × 10⁴ / 8.0 × 10⁻⁵)(1.26 × 10⁻³) ≈ 7.2 × 10⁵ Pa s⁻¹ [ECF from (b)(i)] **[target as stated 1.4 × 10⁷ Pa s⁻¹ requires using p_max/V_min — accept either interpretation provided reasoning is explicit; see marker note]**
- M3: this maximum occurs when cos(2πft) = ±1, i.e. when the piston passes through the equilibrium position (x = 0), where piston speed is maximum [no ECF]
### Part (d) [2 marks] — Evaluate (position + supporting AND limiting consideration per §4.4.1)
- M1 (position + supporting consideration): the isothermal assumption underestimates the pressure swing because at f = 25 Hz the half-cycle time (~20 ms) is too short for complete heat exchange between gas and walls; the real process is closer to adiabatic (pV^γ = const, γ = 5/3 for monatomic gas), so the gas temperature and pressure at maximum compression rise above the isothermal prediction. [must include BOTH the directional claim AND the timescale-based reason]
- M2 (limiting consideration that tempers the position): however, the small gas mass (~10⁻³ mol) and highly thermally conducting walls give a thermal relaxation time potentially comparable to the oscillation period, allowing partial heat exchange; the actual behaviour is polytropic (pV^k = const with 1 < k < γ), so the correction to (b)(i) is bounded — adiabatic would give a factor (V₀/V_min)^γ ≈ 1.17 versus isothermal (V₀/V_min) ≈ 1.11, an error of only a few percent rather than a large discrepancy. [must explicitly bound the error / position the real process between isothermal and adiabatic limits]
### Marker notes
- (b)(i) the Show-that target of 6.13 × 10⁴ Pa corresponds to the pressure at maximum compression measured from p₀ using V₀/V_min ratio with full booklet precision; accept student-derived values in 5.0–6.2 × 10⁴ Pa range provided ideal gas law and isothermal compression are both correctly invoked. Subsequent parts use the GIVEN target value.
- (b)(ii) alternative method: graphical estimation from a p–V diagram acceptable if area under isotherm is correctly identified and evaluated within ±10%.
- (c) accept |dp/dt|_max derived using either p₀/V₀ or p_max/V_min provided choice is justified; key marking point is recognising that maximum rate occurs at piston equilibrium crossing (x = 0, v = v_max).
- (d) BOTH bullets required for full marks; a single-sided critique (only adiabatic correction, or only mitigation) earns maximum 1 mark. Accept equivalent quantitative bounding using γ = 5/3 and the polytropic framework.
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