B.3 Gas laws flashcards: IB Physics SL and HL

All 73 B.3 Gas laws cards

1. DefinitionSL & HLData booklet: Yes

   Define pressure and state its SI unit and whether it is a scalar or a vector.

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   Pressure is the force acting normally (perpendicular) per unit area of the surface: p = F/A. SI unit pascal, Pa, where 1 Pa = 1 N m⁻², and pressure is a scalar even though force is a vector. Exam tip: the mark is lost for "force per unit area" without the word normal/perpendicular, and for quoting the unit as N m⁻¹ or N m². In a gas the pressure at a point acts equally in all directions, which is why it has no direction of its own; 1 atm ≈ 1.01 × 10⁵ Pa is worth memorising for gas-law substitutions.

2. DefinitionSL & HL

   Explain, in terms of molecules, what the pressure exerted by a gas on its container is.

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   Gas pressure is the total normal force per unit area exerted by molecules colliding with the container wall. Each collision reverses the molecule's momentum component perpendicular to the wall, so the wall exerts an impulse on the molecule and, by Newton's third law, the molecule exerts an equal and opposite force on the wall. Because collisions are enormously frequent and random, the effect averages to a steady, uniform pressure. Exam tip: answers that say molecules "push" or "hit hard" score nothing — the mark scheme wants rate of change of momentum at the wall, plus the number of collisions per second per unit area.

3. DefinitionSL & HL

   Define the mole and the Avogadro constant, and state the unit of amount of substance.

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   The mole is the SI unit of amount of substance, n; one mole contains N_A = 6.02 × 10²³ elementary entities (atoms, molecules or ions). The Avogadro constant N_A = 6.02 × 10²³ mol⁻¹ is the number of particles per mole. Amount of substance n has unit mol and is a scalar. Exam tip: N_A is a constant with a unit (mol⁻¹), not a pure number — writing "Avogadro's number = 6.02 × 10²³" with no unit loses precision marks. State clearly which entity is being counted: 1 mol of O₂ contains 6.02 × 10²³ molecules but 1.20 × 10²⁴ atoms.

4. DefinitionSL & HL

   Define molar mass and relative molecular mass, and state how they are used to find amount of substance.

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   Molar mass M is the mass of one mole of a substance, unit kg mol⁻¹ in SI (often quoted in g mol⁻¹). Relative molecular mass M_r is the mass of one molecule expressed in unified atomic mass units u (1 u = 1.66 × 10⁻²⁷ kg = 931.5 MeV c⁻²) and is a pure number. Amount of substance is then n = m/M, with m the sample mass in kg. Exam tip: the commonest error in gas-law questions is using M in g mol⁻¹ with R = 8.31 J K⁻¹ mol⁻¹; convert, e.g. helium M = 4.00 g mol⁻¹ = 4.00 × 10⁻³ kg mol⁻¹.

5. DefinitionSL & HL

   Define an ideal gas.

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   An ideal gas is a theoretical gas that obeys the equation of state pV = nRT exactly at all pressures and temperatures, whose molecules have zero volume and exert no intermolecular forces except during collisions, and whose internal energy is entirely random translational kinetic energy (no molecular potential energy). Exam tip: the highest-scoring definitions mention both "obeys pV = nRT under all conditions" and "no intermolecular forces, so internal energy is purely kinetic". A common loss is saying an ideal gas "cannot be liquefied" without the reason — it cannot, precisely because there are no attractive forces to bind molecules together.

6. DefinitionSL & HL

   State the assumptions of the kinetic model of an ideal gas.

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   • Molecules are identical, hard, perfectly elastic spheres whose total volume is negligible compared with the volume of the container (point molecules).
   • Molecules are in continuous random motion with a distribution of speeds.
   • Collisions between molecules and with the walls are perfectly elastic, so kinetic energy is conserved.
   • No intermolecular forces act except during collisions, so molecules move in straight lines between collisions.
   • The duration of a collision is negligible compared with the time between collisions.
   • The number of molecules is very large so statistical averages are valid. Exam tip: "molecules are small" is not enough — say their total volume is negligible compared with the gas volume.
7. DefinitionSL & HLData booklet: No – derive

   State Boyle's law, including the conditions under which it applies.

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   For a fixed mass (fixed amount) of an ideal gas held at constant temperature, the pressure is inversely proportional to the volume: p ∝ 1/V, so pV = constant. Exam tip: full marks require BOTH conditions — constant temperature AND fixed mass/amount of gas — and the word "inversely proportional", not merely "as V increases p decreases". Molecular explanation: halving the volume doubles the number of wall collisions per second per unit area while the mean speed is unchanged, so the pressure doubles. Boyle's law is an empirical law; it follows from pV = nRT with n and T constant.

8. DefinitionSL & HLData booklet: No – derive

   State Charles's law, including the conditions under which it applies.

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   For a fixed mass of an ideal gas held at constant pressure, the volume is directly proportional to the absolute (thermodynamic) temperature: V ∝ T, so V/T = constant. Exam tip: the mark is only awarded if T is stated to be in kelvin — V ∝ θ in °C is false, since V does not double from 10 °C to 20 °C. Molecular explanation: raising T raises the mean molecular kinetic energy and hence the mean speed and momentum change per collision, so the gas must expand to keep the collision rate per unit area, and therefore the pressure, constant.

9. DefinitionSL & HLData booklet: No – derive

   State the pressure law (Gay-Lussac's law), including the conditions under which it applies.

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   For a fixed mass of an ideal gas held at constant volume, the pressure is directly proportional to the absolute temperature: p ∝ T, so p/T = constant. Exam tip: again T must be in kelvin; students who convert only one of the two temperatures in a p₁/T₁ = p₂/T₂ calculation lose the answer mark. Molecular explanation: at higher T molecules move faster, so they strike the walls both more often and with a greater change of momentum per collision, while the wall area is unchanged — hence a greater force per unit area.

10. DefinitionSL & HLData booklet: No – memorise

    Define absolute zero and the kelvin (thermodynamic) temperature scale.

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    Absolute zero, 0 K = −273 °C (−273.15 °C), is the temperature at which the average random kinetic energy of the molecules of an ideal gas is zero and the gas would exert zero pressure; it is the lowest temperature theoretically attainable. The kelvin scale is the absolute temperature scale on which T(K) = θ(°C) + 273, with the same size of degree as the Celsius scale. Exam tip: say "average random kinetic energy is zero", not "all motion stops" (quantum zero-point motion remains). Never write "°K", and never use Celsius in pV = nRT, E_k = 3/2 k_BT or any proportionality involving T.

11. DefinitionSL & HL

    Define temperature in terms of the kinetic model of a gas.

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    The thermodynamic temperature of an ideal gas is a measure of the average random translational kinetic energy of its molecules; explicitly, average E_k = 3/2 k_BT, so T ∝ average E_k. Two bodies at the same temperature have the same average molecular kinetic energy and there is no net transfer of thermal energy between them. Exam tip: the mark scheme rejects "temperature is the heat/energy of a body" and "temperature measures the total kinetic energy" — total kinetic energy also depends on the number of molecules. At the same T, a heavy molecule (e.g. CO₂) has the same average E_k as a light one (H₂) but a much smaller mean speed.

12. DefinitionSL & HLData booklet: Yes

    Define the internal energy of an ideal gas and state what it depends on.

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    Internal energy U is the total of the random kinetic energies and the mutual potential energies of all the molecules of a substance. For an ideal monatomic gas the intermolecular potential energy is zero by definition, so U is purely random translational kinetic energy: U = 3/2 Nk_BT = 3/2 nRT. It therefore depends only on the absolute temperature and the amount of gas, not on the pressure or the volume separately. Exam tip: a very common loss is omitting "random" or omitting potential energy from the general definition; another is claiming U changes during an isothermal compression — for an ideal gas ΔT = 0 means ΔU = 0.

13. DefinitionSL & HL

    Define mean square speed and root mean square speed for the molecules of a gas.

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    Mean square speed v̄² is the average of the squares of the molecular speeds, (v₁² + v₂² + … + v_N²)/N, unit m² s⁻². Root mean square speed v_rms = √(v̄²) is the square root of that average, unit m s⁻¹; it is the speed of a molecule having the average kinetic energy. Exam tip: v_rms is NOT the mean speed and is slightly larger than it, because squaring weights the fast molecules more heavily; writing v̄² = (v̄)² is a standard error. It is the mean square speed v̄², not v_rms, that appears in pV = 1/3 Nmv̄² and in E_k = ½mv̄² = 3/2 k_BT, so take the square root only at the very end.

14. DefinitionSL & HL

    Define a real gas and state the conditions under which a real gas deviates most from ideal behaviour.

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    A real gas is an actual gas whose molecules occupy a finite volume and exert intermolecular attractive forces, so it obeys pV = nRT only approximately. Deviations are largest at high pressure (molecular volume is no longer negligible compared with the container volume) and at low temperature, near liquefaction (attractive forces reduce the momentum with which molecules strike the walls, so the measured pressure is lower than predicted). Exam tip: students must name the condition AND the reason; "real gases are not ideal" earns nothing. Real gases approach ideal behaviour at low pressure and high temperature, i.e. low density.

15. DefinitionSL & HL

    Define an isothermal change and an isovolumetric (isochoric) change for a gas.

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    An isothermal change is one that takes place at constant temperature, so for an ideal gas pV = constant and ΔU = 0. An isovolumetric (isochoric) change takes place at constant volume, so p/T = constant and no work is done by or on the gas. An isobaric change is at constant pressure, with V/T = constant. Exam tip: on a p–V diagram an isothermal is a smooth hyperbola, an isovolumetric change is a vertical line and an isobaric change is a horizontal line; students frequently mislabel a straight sloping p–V line as an isothermal. For a slow isothermal compression, thermal energy must leave the gas.

16. DefinitionSL & HLData booklet: No – derive

    Define the molar gas constant R and the Boltzmann constant k_B, and state the relationship between them.

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    The molar gas constant R = 8.31 J K⁻¹ mol⁻¹ is the constant of proportionality in pV = nRT; it is the work-equivalent energy per kelvin per mole. The Boltzmann constant k_B = 1.38 × 10⁻²³ J K⁻¹ is the corresponding constant per molecule, appearing in pV = Nk_BT and E_k = 3/2 k_BT. They are related by R = N_A k_B, since a mole contains N_A molecules. Exam tip: the error that ruins many answers is mixing them — using N (a number of molecules) with R, or n (moles) with k_B. Check: 6.02 × 10²³ × 1.38 × 10⁻²³ = 8.31 J K⁻¹ mol⁻¹.

17. EquationSL & HLData booklet: Yes

    State the defining equation for pressure and explain each symbol, unit and condition of use.

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    p = F/A. p = pressure in pascal (Pa = N m⁻²); F = force acting normally to the surface, in newton (N); A = area over which the force acts, in square metres (m²). Valid whenever the force is uniformly distributed and perpendicular to the surface. Data booklet: yes. Common misuse: using the total weight rather than only the normal component, or leaving A in cm² (1 cm² = 1 × 10⁻⁴ m²). Sanity check: a 60 kg person on one foot of area 1.5 × 10⁻² m² exerts p = 60 × 9.8/1.5 × 10⁻² ≈ 3.9 × 10⁴ Pa, well below atmospheric 1.01 × 10⁵ Pa.

18. EquationSL & HLData booklet: Yes

    State the equation linking amount of substance to number of molecules, with symbols, units and a check.

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    n = N/N_A. n = amount of substance in mol; N = number of molecules (dimensionless); N_A = Avogadro constant = 6.02 × 10²³ mol⁻¹. Data booklet: yes. Useful companion, not in the booklet: n = m/M with m the mass in kg and M the molar mass in kg mol⁻¹ — so N = (m/M)N_A. Common misuse: substituting N as "number of moles", which is off by 10²³. Sanity check: 32 g of O₂ (M = 32 g mol⁻¹) is n = 1.0 mol, so N = 6.02 × 10²³ molecules; 2.0 g of hydrogen gas is 1.0 mol of H₂.

19. EquationSL & HLData booklet: Yes

    State the ideal gas equation in its molar form and explain every symbol, unit and condition.

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    pV = nRT. p = pressure in Pa; V = volume in m³; n = amount of substance in mol; R = molar gas constant = 8.31 J K⁻¹ mol⁻¹; T = absolute temperature in K. Valid for an ideal gas — a fixed amount of gas at low pressure and well above its liquefaction temperature. Data booklet: yes. Common misuse: V in dm³ or litres (1 dm³ = 1 × 10⁻³ m³) or T in °C. Sanity check: 1.00 mol at 273 K and 1.01 × 10⁵ Pa gives V = nRT/p = (1.00 × 8.31 × 273)/1.01 × 10⁵ = 2.25 × 10⁻² m³ = 22.5 dm³, the familiar molar volume.

20. EquationSL & HLData booklet: Yes

    State the ideal gas equation in its molecular form and explain every symbol, unit and condition.

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    pV = Nk_BT. p = pressure in Pa; V = volume in m³; N = number of molecules; k_B = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹; T = absolute temperature in K. Equivalent to pV = nRT because N = nN_A and R = N_A k_B. Data booklet: yes. Common misuse: pairing N with R or n with k_B — always check that the constant matches the counting quantity. Sanity check: at p = 1.0 × 10⁵ Pa and T = 300 K, the number of molecules in 1.0 m³ is N = pV/(k_BT) = 1.0 × 10⁵/(1.38 × 10⁻²³ × 300) ≈ 2.4 × 10²⁵.

21. EquationSL & HLData booklet: Yes

    State the equation for the average kinetic energy of a molecule of an ideal gas and explain it fully.

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    E_k = 3/2 k_BT, where E_k is the average random translational kinetic energy per molecule in joule (J); k_B = 1.38 × 10⁻²³ J K⁻¹; T = absolute temperature in K. Equivalent forms: E_k = ½m v̄² and, per mole, E = 3/2 RT. Valid for an ideal gas; the result is independent of the molecular mass and of the pressure. Data booklet: yes. Common misuse: treating E_k as the total kinetic energy of the gas (that is U = N × 3/2 k_BT) or using °C. Sanity check: at T = 300 K, E_k = 1.5 × 1.38 × 10⁻²³ × 300 = 6.2 × 10⁻²¹ J ≈ 0.039 eV.

22. EquationSL & HLData booklet: Yes

    State the kinetic theory expression for the pressure of an ideal gas and explain every symbol and assumption.

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    pV = 1/3 Nm v̄². p = pressure in Pa; V = volume in m³; N = number of molecules; m = mass of ONE molecule in kg; v̄² = mean square speed in m² s⁻². The factor 1/3 arises because molecular motion is random and shared equally between the three perpendicular directions, so v̄ₓ² = v̄²/3. Valid under the ideal-gas assumptions (elastic collisions, negligible molecular volume, no intermolecular forces). Data booklet: yes. Common misuse: putting the total gas mass or the molar mass in place of m, or writing (v̄)² for v̄². Equivalent form: p = 1/3 ρ v̄² with ρ = Nm/V the gas density.

23. EquationSL & HLData booklet: Yes

    State the equation for the internal energy of an ideal monatomic gas and give its equivalent forms.

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    U = 3/2 Nk_BT, equivalently U = 3/2 nRT = 3/2 pV. U = internal energy in joule (J); N = number of molecules; k_B = 1.38 × 10⁻²³ J K⁻¹; n = amount in mol; R = 8.31 J K⁻¹ mol⁻¹; T = absolute temperature in K. Valid only for an ideal monatomic gas, where all internal energy is random translational kinetic energy. Data booklet: yes. Common misuse: applying it to a real or diatomic gas, or forgetting that ΔU = 3/2 nRΔT so an isothermal change gives ΔU = 0. Sanity check: 1.0 mol at 300 K has U = 1.5 × 8.31 × 300 ≈ 3.7 × 10³ J.

24. EquationSL & HLData booklet: No – derive

    State the equation for the root mean square speed of gas molecules and show how it is obtained.

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    From ½m v̄² = 3/2 k_BT: v_rms = √(v̄²) = √(3k_BT/m) = √(3RT/M), where m = mass of one molecule in kg, M = molar mass in kg mol⁻¹, T in K. Unit m s⁻¹. Data booklet: no — derive it from E_k = 3/2 k_BT. Conditions: ideal gas, random motion. Common misuse: substituting M in g mol⁻¹, which makes v_rms about 32 times too large. Sanity check: nitrogen at 300 K, M = 2.8 × 10⁻² kg mol⁻¹, gives v_rms = √(3 × 8.31 × 300/2.8 × 10⁻²) ≈ 5.2 × 10² m s⁻¹, comparable to the speed of sound in air.

25. EquationSL & HLData booklet: No – derive

    State the working form of Boyle's law used in calculations and give its conditions and a check.

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    p₁V₁ = p₂V₂ for a fixed amount of ideal gas at constant temperature. p₁, p₂ = initial and final pressures in Pa (any consistent unit works, since only a ratio matters); V₁, V₂ = initial and final volumes in m³ (again any consistent unit). Data booklet: no — derive from pV = nRT with n, T constant. Common misuse: applying it when the gas is heated, or when gas leaks from the container. Sanity check: a gas at 1.0 × 10⁵ Pa occupying 250 cm³ compressed isothermally to 100 cm³ reaches p₂ = 1.0 × 10⁵ × 250/100 = 2.5 × 10⁵ Pa.

26. EquationSL & HLData booklet: No – derive

    State the working forms of Charles's law and the pressure law, with units and conditions.

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    Charles's law: V₁/T₁ = V₂/T₂ for a fixed amount of gas at constant pressure. Pressure law: p₁/T₁ = p₂/T₂ for a fixed amount of gas at constant volume. V in m³ (any consistent unit), p in Pa (any consistent unit), but T MUST be in kelvin. Data booklet: no — both follow from pV = nRT. Common misuse: leaving temperature in °C, which is the single most frequent error in Paper 2 gas questions. Sanity check: a sealed rigid can at 1.0 × 10⁵ Pa and 27 °C (300 K) heated to 127 °C (400 K) reaches p₂ = 1.0 × 10⁵ × 400/300 = 1.3 × 10⁵ Pa.

27. EquationSL & HLData booklet: No – derive

    State the combined gas equation and explain when it is preferable to pV = nRT.

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    p₁V₁/T₁ = p₂V₂/T₂ for a fixed mass (fixed n) of ideal gas undergoing any change of state. p in Pa, V in m³, T in K; consistent units for p and V cancel, but T must be absolute. Data booklet: no — obtained by writing pV/T = nR = constant. Use it whenever the amount of gas is unknown or not required, which saves a step; use pV = nRT when n, N, m or the density is wanted. Common misuse: applying it when gas escapes or is added. Sanity check: 2.0 dm³ at 1.0 × 10⁵ Pa, 300 K becomes, at 2.0 × 10⁵ Pa and 600 K, V₂ = 2.0 dm³.

28. EquationSL & HLData booklet: No – derive

    State the relationship between R, k_B and N_A, and between density, pressure and mean square speed.

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    R = N_A k_B, so k_B = R/N_A = 8.31/6.02 × 10²³ = 1.38 × 10⁻²³ J K⁻¹. Dividing pV = 1/3 Nm v̄² by V gives p = 1/3 ρ v̄², where ρ = Nm/V is the gas density in kg m⁻³, p in Pa and v̄² in m² s⁻². Data booklet: neither rearrangement is printed — derive both. Conditions: ideal gas. Common misuse: using the density of the container material or of the liquid rather than of the gas. Sanity check: air at ρ = 1.2 kg m⁻³ and v_rms ≈ 5.0 × 10² m s⁻¹ gives p = 1/3 × 1.2 × (5.0 × 10²)² = 1.0 × 10⁵ Pa.

29. EquationSL & HLData booklet: No – memorise

    State how to convert between Celsius and kelvin and why the conversion matters in the gas laws.

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    T/K = θ/°C + 273 (more precisely 273.15); equivalently θ/°C = T/K − 273. A temperature difference is numerically identical on both scales, ΔT/K = Δθ/°C, because the degree sizes are equal. Data booklet: no — memorise. Conditions: applies to all temperatures. Common misuse: converting a temperature CHANGE by adding 273 (wrong), or failing to convert an absolute temperature in pV = nRT, p/T, V/T, E_k = 3/2 k_BT or U = 3/2 nRT (all require kelvin). Sanity check: heating a gas from 27 °C to 327 °C doubles T from 300 K to 600 K, so at constant volume the pressure doubles — not a twelve-fold rise.

30. Graph/diagramSL & HL

    Describe the graph of pressure against volume for a fixed mass of ideal gas at constant temperature.

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    Axes: p/Pa (y) against V/m³ (x). Shape: a smooth rectangular hyperbola (isotherm) that approaches but never touches either axis, since p ∝ 1/V. Every point satisfies pV = constant, and the area under a p–V curve equals the work done by the gas. Isotherms for higher temperatures lie further from the origin because pV = nRT is larger. To extract a quantity: pick any point, evaluate pV, then n = pV/(RT). Changing a parameter: doubling n or T at the same V doubles p, shifting the whole curve outward; the curve never becomes a straight line.

31. Graph/diagramSL & HL

    Describe the linearised graph used to test Boyle's law and how a quantity is obtained from it.

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    Axes: p/Pa (y) against 1/V / m⁻³ (x). Shape: a straight line through the origin, confirming p ∝ 1/V. Gradient = pV = nRT (units Pa m³ = J), so n = gradient/(RT) and N = gradient/(k_BT); the intercept should be zero. Extracting a quantity: measure the gradient over a large triangle and divide by RT to find the amount of gas. Changing a parameter: raising the temperature or adding gas increases the gradient but keeps the line straight through the origin. A non-zero positive intercept usually signals a systematic error such as trapped dead-space volume in the tube or syringe.

32. Graph/diagramSL & HL

    Describe the volume–temperature graph for a fixed mass of ideal gas at constant pressure.

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    Axes: V/m³ (y) against T/K (x). Shape: a straight line through the origin, since V = (nR/p)T. Gradient = nR/p, so n = gradient × p/R. Plotted instead against θ/°C, the line is still straight but has a positive intercept on the V-axis and, extrapolated backwards, cuts the temperature axis at −273 °C — the experimental route to absolute zero. Extracting a quantity: the x-intercept of the extrapolated V–θ line gives absolute zero. Changing a parameter: increasing p reduces the gradient (line becomes shallower); the extrapolated intercept stays at −273 °C for every pressure, which is strong evidence for the absolute scale.

33. Graph/diagramSL & HL

    Describe the pressure–temperature graph for a fixed mass of ideal gas at constant volume.

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    Axes: p/Pa (y) against T/K (x). Shape: a straight line through the origin, since p = (nR/V)T. Gradient = nR/V, so n = gradient × V/R and N = gradient × V/k_B. Against θ/°C the line has a positive p-intercept and extrapolates back to −273 °C, where an ideal gas would exert zero pressure. Extracting a quantity: use the gradient with the known container volume to find the amount of trapped gas. Changing a parameter: reducing the volume steepens the line; real gases curve away from the straight line and liquefy before 0 K, so the low-temperature portion must be extrapolated, not measured.

34. Graph/diagramSL & HL

    Describe the graph of pV against p for a real gas and compare it with an ideal gas.

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    Axes: pV/J (y) against p/Pa (x) at constant temperature. Ideal gas: a horizontal straight line, because pV = nRT is independent of pressure. Real gas: the curve dips below the ideal line at moderate pressures — attractive intermolecular forces reduce the momentum delivered to the walls — then rises above it at high pressures, where the finite molecular volume means the free volume is less than V. All curves converge on the ideal value as p → 0. Extracting a quantity: the limiting intercept as p → 0 equals nRT, giving n. Changing a parameter: at higher temperature the deviations shrink and the curve flattens towards the ideal line.

35. Graph/diagramSL & HL

    Describe the graph of average molecular kinetic energy against absolute temperature for an ideal gas.

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    Axes: E_k/J (y) against T/K (x). Shape: a straight line through the origin, since E_k = 3/2 k_BT. Gradient = 3/2 k_B = 2.07 × 10⁻²³ J K⁻¹, so k_B = 2 × gradient/3; the zero intercept is the graphical statement of absolute zero. The identical line is obtained for every gas, whatever the molecular mass — a key syllabus point. Extracting a quantity: measure the gradient to determine k_B, then N_A = R/k_B. Changing a parameter: plotting total internal energy U = 3/2 Nk_BT instead multiplies the gradient by N, so a larger sample gives a steeper line with the same zero intercept.

36. Graph/diagramSL & HL

    Describe the graph of v_rms against absolute temperature and against molar mass for gas molecules.

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    v_rms against T: axes v_rms/m s⁻¹ (y) against T/K (x); the shape is a curve of decreasing gradient because v_rms = √(3RT/M) ∝ √T — doubling v_rms needs a fourfold rise in T. Linearise by plotting v_rms against √T (straight line through the origin, gradient √(3R/M), so M = 3R/gradient²) or v_rms² against T (gradient 3R/M). v_rms against M at fixed T is an inverse-square-root curve, so light molecules such as H₂ move far faster than heavy ones such as CO₂ at the same temperature, even though both have the same average kinetic energy.

37. Graph/diagramSL & HL

    Describe the log–log graph used to verify Boyle's law and state what the gradient shows.

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    Axes: ln(p/Pa) (y) against ln(V/m³) (x) for a fixed mass of gas at constant temperature. Taking logarithms of p = kV^x gives ln p = x ln V + ln k, so the graph is a straight line of gradient x and y-intercept ln k. Boyle's law is confirmed if the gradient is −1 within uncertainty; the intercept gives k = pV = nRT, hence n. Extracting a quantity: draw maximum and minimum gradient lines through the error bars to quote the gradient with an uncertainty and test whether −1 lies inside the range. Changing a parameter: raising T raises the intercept but leaves the gradient at −1.

38. Concept/explainSL & HL

    Explain, in terms of molecular motion, how a gas exerts pressure on the walls of its container.

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    • Molecules move randomly at high speed with a wide range of speeds
    • Each molecule colliding with a wall rebounds, so its momentum changes
    • By Newton's second law the wall exerts a force on the molecule, and by Newton's third law the molecule exerts an equal and opposite force on the wall
    • Pressure is the total average normal force per unit area, p = F/A
    • The collision rate is enormous (~10²⁷ s⁻¹ per m²), so the force appears steady and uniform
    • p increases if the collision rate or the momentum change per collision increases
    • Exam tip: incomplete answers say molecules push the wall; the marks are for momentum change and Newton's second and third laws.
39. Concept/explainSL & HL

    Outline the assumptions of the ideal gas model and state why each is needed.

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    • Molecules are point particles: their total volume is negligible compared with the container volume, so all of V is available for motion
    • No intermolecular forces except during collisions, so molecules have no potential energy and internal energy is purely kinetic
    • Collisions with each other and the walls are perfectly elastic, so total kinetic energy and hence pressure are constant
    • Motion is random, with no preferred direction, so pressure is the same on all walls
    • Collision duration is negligible compared with time between collisions
    • Molecules obey Newtonian mechanics and are large in number, so averages are meaningful
    • Exam tip: students list assumptions without justifying them; explain questions need the consequence of each.
40. Concept/explainSL & HL

    Explain, using kinetic theory, why the pressure of a fixed mass of gas at constant volume increases as temperature increases.

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    • Absolute temperature is proportional to the average kinetic energy of molecules, E_k = 3/2 k_B T
    • Raising T raises the mean square speed, so molecules move faster on average
    • Faster molecules hit the walls more frequently, increasing the collision rate
    • Each collision produces a larger momentum change, so a larger average force
    • Volume and hence wall area are unchanged, so p = F/A increases
    • Result: p ∝ T at constant V and constant N (the pressure law)
    • Exam tip: many answers stop at molecules move faster; both the increased rate of collisions and the greater momentum change per collision are required for full marks.
41. Concept/explainSL & HL

    Explain Boyle's law (p ∝ 1/V at constant temperature) in terms of molecular behaviour.

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    • Temperature constant means average kinetic energy and hence mean square speed are constant
    • Reducing the volume shortens the average distance a molecule travels between wall collisions
    • So each molecule strikes the walls more often per second, increasing the collision rate per unit area
    • The momentum change per collision is unchanged because the speeds are unchanged
    • Force per unit area therefore rises in inverse proportion to volume, pV = constant
    • Exam tip: a very common error is to say the molecules speed up when the gas is compressed; at constant temperature their speeds do not change, only the collision frequency does.
42. Concept/explainSL & HL

    Explain, in molecular terms, why the volume of a gas at constant pressure is proportional to its absolute temperature.

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    • Heating raises the average kinetic energy of the molecules, so mean speed and momentum change per collision increase
    • If the volume were fixed, pressure would rise
    • To keep p constant the gas must expand, increasing the wall area and reducing the collision rate per unit area
    • These two effects exactly cancel the increased momentum change, so V ∝ T (Charles's law)
    • Real example: a gas syringe with a free piston, where atmospheric pressure plus piston weight fixes p
    • Exam tip: the answer must explain why expansion restores the original pressure, not simply state that hot gases expand.
43. Concept/explainSL & HL

    Explain the physical meaning of absolute zero and why gas-law calculations must use the Kelvin scale.

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    • Extrapolating V–T or p–T graphs for any fixed mass of gas to zero volume or zero pressure gives the same intercept, θ = −273 °C
    • At absolute zero (0 K) molecular random kinetic energy and hence gas pressure would be zero; molecular motion is minimal
    • T(K) = θ(°C) + 273, so the Kelvin scale has the same size degree but a true zero
    • Gas laws state proportionality (p ∝ T, V ∝ T), which is only valid on a scale with a true zero
    • Absolute zero is unattainable; real gases liquefy before reaching it
    • Exam tip: using °C in pV = nRT is the single most common error in this topic.
44. Concept/explainSL & HL

    Explain why the internal energy of an ideal gas depends only on its absolute temperature.

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    • Internal energy is the total random kinetic energy plus total potential energy of the molecules
    • An ideal gas has no intermolecular forces, so the molecular potential energy is zero
    • Therefore U equals the total random kinetic energy of the molecules
    • Average molecular kinetic energy is E_k = 3/2 k_B T, so for a monatomic ideal gas U = 3/2 N k_B T = 3/2 nRT
    • U is independent of pressure and volume; equal temperatures mean equal U for the same amount of gas
    • In a real gas, work done against intermolecular forces changes potential energy, so U also depends on volume
    • Exam tip: internal energy is not heat; do not call U the heat content of the gas.
45. Concept/explainSL & HL

    Explain the conditions under which a real gas deviates significantly from ideal-gas behaviour.

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    • Deviations are large at high pressure and at low temperature, near liquefaction
    • At high pressure the molecules are close together, so their own volume is no longer negligible compared with the container volume, making the available volume smaller than V
    • At low temperature molecular kinetic energy is small, so attractive intermolecular forces reduce the momentum with which molecules strike the wall, lowering p below the ideal value
    • Consequently pV/nRT departs from 1
    • Gases behave nearly ideally at low pressure, high temperature and low density, and when far above their boiling point
    • Exam tip: state which assumption of the model fails, not simply that the gas is not ideal.
46. Concept/explainSL & HL

    Explain why two different gases at the same temperature have the same average molecular kinetic energy but different root-mean-square speeds.

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    • Absolute temperature is a measure of the average random translational kinetic energy of molecules: E_k = 3/2 k_B T, independent of the type of gas
    • Thermal equilibrium means no net energy transfer, so average kinetic energies must be equal
    • Since E_k = ½ m v̄², a molecule of larger mass m must have a smaller mean square speed
    • Hence v_rms = √(3k_B T/m), so v_rms ∝ 1/√m at fixed T
    • Example: at 300 K helium molecules move about 2.8 times faster than oxygen molecules
    • Exam tip: students wrongly assume same temperature means same speed; temperature fixes energy, not speed.
47. Concept/explainSL & HLData booklet: Yes

    Outline how the kinetic-theory result pV = ⅓Nm v̄² is obtained and what each symbol represents.

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    • Consider N molecules of mass m in a cube of side L moving randomly
    • One molecule hitting a wall perpendicularly reverses its velocity component, giving momentum change 2mv_x
    • It returns after time 2L/v_x, so its average force on that wall is m v_x²/L
    • Summing over all molecules and using randomness, v̄_x² = ⅓v̄², where v̄² is the mean square speed
    • Dividing by area L² and using V = L³ gives p = ⅓ (Nm/V) v̄², i.e. pV = ⅓Nm v̄²
    • Nm is the total mass of gas, so p = ⅓ρv̄²
    • Exam tip: the ⅓ comes from three equivalent directions, not from averaging three speeds.
48. Concept/explainSL & HLData booklet: Yes

    Explain how the ideal gas equation pV = nRT combines the three empirical gas laws.

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    • Boyle: at constant T and n, p ∝ 1/V
    • Charles: at constant p and n, V ∝ T
    • Gay-Lussac (pressure law): at constant V and n, p ∝ T
    • Avogadro: at constant p and T, V ∝ n
    • Combining gives pV ∝ nT, and the constant of proportionality is the universal molar gas constant R = 8.31 J K⁻¹ mol⁻¹
    • Writing n = N/N_A gives pV = N(R/N_A)T = Nk_BT, where k_B = R/N_A = 1.38 × 10⁻²³ J K⁻¹
    • Nature of science: empirical laws found by experiment were later unified and then explained by the microscopic kinetic model
    • Exam tip: R is per mole, k_B is per molecule; mixing them is a standard error.
49. Concept/explainSL & HLData booklet: Yes

    Explain the meaning of the mole, the Avogadro constant and molar mass, and how they connect to n = N/N_A.

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    • One mole is the amount of substance containing N_A = 6.02 × 10²³ elementary entities (molecules, atoms)
    • The number of moles is therefore n = N/N_A, where N is the number of molecules
    • Molar mass M (kg mol⁻¹) is the mass of one mole; the mass of one molecule is m = M/N_A
    • Total mass of gas = nM = Nm
    • The molar mass in g mol⁻¹ is numerically equal to the relative molecular mass, e.g. O₂ has M = 32 g mol⁻¹ = 0.032 kg mol⁻¹
    • Amount of substance is one of the SI base quantities
    • Exam tip: n is not the number of molecules, and molar mass must be converted to kg mol⁻¹ when using R = 8.31 J K⁻¹ mol⁻¹.
50. Concept/explainSL & HLData booklet: No – derive

    Explain what the root-mean-square speed is and why it, rather than the mean speed, appears in kinetic theory.

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    • Molecules have a wide spread of speeds; the mean velocity is zero because motion is random in all directions
    • Kinetic energy depends on v², so the physically relevant average is the mean square speed v̄², obtained by squaring each speed, averaging, and the rms speed is v_rms = √(v̄²)
    • From ½m v̄² = 3/2 k_B T, v_rms = √(3k_BT/m) = √(3RT/M)
    • rms speed is slightly larger than the mean speed because squaring weights fast molecules more heavily
    • Raising T by a factor of 4 only doubles v_rms
    • Exam tip: v̄² is the mean of the squares, not the square of the mean; the two are not equal. v_rms = √(3k_BT/m) is not printed in the data booklet — derive it from E_k = 3/2 k_BT.
51. Concept/explainSL & HL

    Discuss how the gas laws and kinetic theory illustrate the relationship between empirical laws and theoretical models in science.

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    • Boyle, Charles and Gay-Lussac obtained their laws purely from measurement, with no knowledge of molecules
    • The kinetic model was then proposed and, using Newtonian mechanics and statistics, derived pV = ⅓Nm v̄², reproducing the empirical laws and explaining why they hold
    • This gave the microscopic interpretation of temperature as average molecular kinetic energy
    • The model makes testable predictions (v_rms values, Brownian motion, diffusion rates) that were confirmed, raising confidence in the existence of molecules
    • Observed deviations at high p and low T show the limits of the model and motivated improved equations of state
    • Exam tip: a good answer names both the empirical evidence and the explanatory power and limitations of the model.
52. Worked problemSL & HLData booklet: Yes

    A sealed syringe contains 250 cm³ of air at 1.0 × 10⁵ Pa. The plunger is pushed in slowly until the volume is 1.0 × 10² cm³, with the temperature unchanged. Calculate the new pressure.

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    Principle: constant T and fixed mass, so Boyle's law p₁V₁ = p₂V₂. Substitution: (1.0 × 10⁵ Pa)(250 cm³) = p₂(1.0 × 10² cm³). Rearranging: p₂ = 1.0 × 10⁵ × 250/100. Intermediate: ratio V₁/V₂ = 2.5. Answer: p₂ = 2.5 × 10⁵ Pa (2 s.f.). Check/Trap: because both volumes appear as a ratio, cm³ need not be converted to m³ here — but they must be converted for pV = nRT. Pushing the plunger slowly keeps the process isothermal; a fast compression heats the gas and gives a higher pressure than Boyle's law predicts.

53. Worked problemSL & HLData booklet: Yes

    A balloon holds 2.0 × 10⁻³ m³ of gas at 27 °C. It is warmed to 87 °C at constant atmospheric pressure. Determine the new volume.

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    Principle: constant p and fixed mass, so Charles's law V₁/T₁ = V₂/T₂. Convert: T₁ = 27 + 273 = 300 K, T₂ = 87 + 273 = 360 K. Substitution: V₂ = 2.0 × 10⁻³ × 360/300. Intermediate: 360/300 = 1.20. Answer: V₂ = 2.4 × 10⁻³ m³ (2 s.f.). Check/Trap: using Celsius would give 2.0 × 10⁻³ × 87/27 = 6.4 × 10⁻³ m³, more than double the correct value — always convert to kelvin. Note the temperature rose by 20 % in kelvin, so the volume rose by 20 %, not by the 220 % a Celsius ratio suggests.

54. Worked problemSL & HLData booklet: Yes

    A car tyre of fixed volume contains air at 2.2 × 10⁵ Pa when the temperature is 17 °C. After driving, the air temperature is 47 °C. Calculate the new pressure.

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    Principle: constant V and fixed mass, so the pressure law p₁/T₁ = p₂/T₂. Convert: T₁ = 290 K, T₂ = 320 K. Substitution: p₂ = 2.2 × 10⁵ × 320/290. Intermediate: 320/290 = 1.103. Answer: p₂ = 2.4 × 10⁵ Pa (2 s.f.). Check/Trap: a 30 °C rise gives only about a 10 % pressure rise because 30/290 ≈ 0.10 — students who use 47/17 get 6.1 × 10⁵ Pa, which is absurd. Also note the tyre volume actually increases slightly, so the real pressure is a little lower than calculated.

55. Worked problemSL & HLData booklet: Yes

    A container of volume 2.5 × 10⁻² m³ holds an ideal gas at 1.0 × 10⁵ Pa and 300 K. Determine the amount of gas in moles and the number of molecules.

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    Principle: pV = nRT, then N = nN_A. Substitution: n = pV/(RT) = (1.0 × 10⁵ Pa × 2.5 × 10⁻² m³)/(8.31 J K⁻¹ mol⁻¹ × 300 K). Intermediate: numerator = 2.5 × 10³ J, denominator = 2.49 × 10³ J mol⁻¹. So n = 1.00 mol. Then N = 1.00 × 6.02 × 10²³ = 6.0 × 10²³ molecules (2 s.f.). Check/Trap: this is the familiar result that one mole occupies about 25 dm³ at room temperature and atmospheric pressure — a useful sanity check. Volume must be in m³ and pressure in Pa for R = 8.31 J K⁻¹ mol⁻¹.

56. Worked problemSL & HLData booklet: Yes

    A gas cylinder of volume 1.5 × 10⁻³ m³ contains gas at 2.0 × 10⁵ Pa and 320 K. Calculate the number of molecules present.

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    Principle: pV = Nk_BT with k_B = 1.38 × 10⁻²³ J K⁻¹. Substitution: N = pV/(k_BT) = (2.0 × 10⁵ × 1.5 × 10⁻³)/(1.38 × 10⁻²³ × 320). Intermediate: numerator = 3.0 × 10² J; denominator = 4.42 × 10⁻²¹ J. Answer: N = 6.8 × 10²² molecules (2 s.f.). Cross-check with moles: n = N/N_A = 6.8 × 10²²/6.02 × 10²³ = 0.11 mol, and pV/(RT) = 300/(8.31 × 320) = 0.113 mol — consistent. Check/Trap: using R instead of k_B here gives 0.11 and students then quote that as a number of molecules.

57. Worked problemSL & HLData booklet: Yes

    Calculate the average random kinetic energy of a molecule of any ideal gas at 27 °C, and state the total random kinetic energy of 0.50 mol of a monatomic gas at this temperature.

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    Principle: E_k = 3/2 k_BT and U = 3/2 nRT. Convert: T = 27 + 273 = 300 K. Substitution: E_k = 1.5 × 1.38 × 10⁻²³ J K⁻¹ × 300 K. Answer: E_k = 6.2 × 10⁻²¹ J (2 s.f.), which is about 0.039 eV. Total: U = 1.5 × 0.50 mol × 8.31 J K⁻¹ mol⁻¹ × 300 K = 1.9 × 10³ J. Check/Trap: E_k is per molecule and is independent of the gas used, so helium and carbon dioxide molecules at 300 K have the same average kinetic energy. Multiplying E_k by nN_A must reproduce U: 6.21 × 10⁻²¹ × 0.50 × 6.02 × 10²³ = 1.9 × 10³ J.

58. Worked problemSL & HLData booklet: No – derive

    Helium has a molar mass of 4.00 g mol⁻¹. Determine the root-mean-square speed of a helium atom at 300 K.

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    Principle: ½m v̄² = 3/2 k_BT, so v_rms = √(3k_BT/m). Mass of one atom: m = 4.00 × 1.66 × 10⁻²⁷ kg = 6.64 × 10⁻²⁷ kg. Substitution: v_rms = √(3 × 1.38 × 10⁻²³ × 300/6.64 × 10⁻²⁷). Intermediate: numerator = 1.24 × 10⁻²⁰ J; ratio = 1.87 × 10⁶ m² s⁻². Answer: v_rms = 1.37 × 10³ m s⁻¹ (3 s.f.). Alternative route: v_rms = √(3RT/M) = √(3 × 8.31 × 300/0.00400) = 1.37 × 10³ m s⁻¹. Check/Trap: molar mass must be in kg mol⁻¹; the answer should be of the same order as the speed of sound in that gas — about 1000 m s⁻¹ for helium, which it is. The booklet gives E_k = 3/2 k_BT; the v_rms expression must be derived from it.

59. Worked problemSL & HLData booklet: Yes

    2.0 mol of a monatomic ideal gas is heated from 300 K to 450 K. Calculate the change in internal energy.

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    Principle: for a monatomic ideal gas U = 3/2 nRT, so ΔU = 3/2 nRΔT. Substitution: ΔU = 1.5 × 2.0 mol × 8.31 J K⁻¹ mol⁻¹ × (450 − 300) K. Intermediate: ΔT = 150 K; 1.5 × 2.0 × 8.31 = 24.9 J K⁻¹. Answer: ΔU = 3.7 × 10³ J (2 s.f.), an increase. Check/Trap: ΔT is the same in kelvin and in degrees Celsius, so a temperature difference need not be converted — but an absolute temperature always must. ΔU depends only on ΔT, so the same answer applies whether the gas is heated at constant volume, constant pressure or any other way.

60. Worked problemSL & HLData booklet: Yes

    An ideal gas occupies 4.0 × 10⁻³ m³ at 1.0 × 10⁵ Pa and 300 K. It is compressed to 1.0 × 10⁻³ m³ and its temperature rises to 450 K. Determine the final pressure.

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    Principle: fixed mass, so p₁V₁/T₁ = p₂V₂/T₂ (combined gas law from pV = nRT). Rearrange: p₂ = p₁ × (V₁/V₂) × (T₂/T₁). Substitution: p₂ = 1.0 × 10⁵ × (4.0 × 10⁻³/1.0 × 10⁻³) × (450/300). Intermediate: volume factor = 4.0; temperature factor = 1.5. Answer: p₂ = 6.0 × 10⁵ Pa (2 s.f.). Check/Trap: both effects push the pressure up, so the factors multiply — a common error is to invert the temperature ratio or the volume ratio. Sanity check with n: n = pV/RT = 400/(8.31 × 300) = 0.160 mol before, and 600/(8.31 × 450) = 0.161 mol after, confirming no gas escaped.

61. Worked problemSL & HLData booklet: No – derive

    Air has a mean molar mass of 0.029 kg mol⁻¹. Calculate the density of air at 1.0 × 10⁵ Pa and 300 K.

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    Principle: from pV = nRT with n = m/M, p V = (m/M)RT, so the density ρ = m/V = pM/(RT). Substitution: ρ = (1.0 × 10⁵ Pa × 0.029 kg mol⁻¹)/(8.31 J K⁻¹ mol⁻¹ × 300 K). Intermediate: numerator = 2.9 × 10³; denominator = 2.49 × 10³. Answer: ρ = 1.2 kg m⁻³ (2 s.f.). Check/Trap: the accepted density of air at room temperature is about 1.2 kg m⁻³, so the result is sensible. The derivation ρ ∝ p/T explains why hot air rises and why aircraft need longer runways on hot days or at altitude.

62. Worked problemSL & HLData booklet: Yes

    A gas of mass 0.64 g occupies 5.0 × 10⁻⁴ m³ at 1.0 × 10⁵ Pa and 300 K. Determine its molar mass and suggest the gas.

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    Principle: n = pV/(RT), then M = mass/n. Substitution: n = (1.0 × 10⁵ × 5.0 × 10⁻⁴)/(8.31 × 300) = 50/2493. Intermediate: n = 2.01 × 10⁻² mol; mass = 0.64 g = 6.4 × 10⁻⁴ kg. Then M = 6.4 × 10⁻⁴/2.01 × 10⁻² = 3.2 × 10⁻² kg mol⁻¹. Answer: M = 32 g mol⁻¹, so the gas is most likely oxygen, O₂ (relative molecular mass 32). Check/Trap: convert grams to kilograms before dividing, or the molar mass comes out 1000 times too large. Number of molecules present: N = nN_A = 2.01 × 10⁻² × 6.02 × 10²³ = 1.2 × 10²² molecules.

63. Worked problemSL & HLData booklet: Yes

    A rigid flask is sealed at 27 °C and 1.0 × 10⁵ Pa. It is then heated to 127 °C while a valve allows gas to escape so the pressure stays at 1.0 × 10⁵ Pa. Determine the percentage of the original gas that escapes.

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    Principle: pV = nRT with p and V constant, so n ∝ 1/T. Convert: T₁ = 300 K, T₂ = 400 K. Ratio: n₂/n₁ = T₁/T₂ = 300/400 = 0.750. So 75 % of the gas remains. Fraction escaped = 1 − 0.750 = 0.250. Answer: 25 % of the original gas escapes (2 s.f.). Check/Trap: the flask is rigid, so V is constant, but the gas is not a fixed mass — the combined gas law p₁V₁/T₁ = p₂V₂/T₂ does not apply because n changes. Students who use it get the wrong physics even though the arithmetic looks similar.

64. Worked problemSL & HLData booklet: Yes

    A vertical cylinder of cross-sectional area 2.5 × 10⁻³ m² is closed by a freely moving frictionless piston of mass 5.0 kg. Atmospheric pressure is 1.0 × 10⁵ Pa. Calculate the pressure of the gas trapped below the piston.

    Show answer

    Principle: the piston is in equilibrium, so the gas pressure supports atmospheric pressure plus the piston weight; p = p_atm + F/A with F = mg. Substitution: F = 5.0 kg × 9.8 m s⁻² = 49 N; F/A = 49 N/2.5 × 10⁻³ m² = 1.96 × 10⁴ Pa. Total: p = 1.0 × 10⁵ + 0.196 × 10⁵. Answer: p = 1.2 × 10⁵ Pa (2 s.f.). Check/Trap: the extra pressure from the piston is only about 20 % of atmospheric, so the answer must be a little above 1.0 × 10⁵ Pa. If the cylinder were inverted with the gas above the piston, the weight would be subtracted, giving 0.80 × 10⁵ Pa.

65. Worked problemSL & HLData booklet: Yes

    A fixed mass of ideal gas is taken from 27 °C to 327 °C while its volume is halved. Determine the factor by which its pressure changes.

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    Principle: p₁V₁/T₁ = p₂V₂/T₂, so p₂/p₁ = (V₁/V₂)(T₂/T₁). Convert: T₁ = 300 K, T₂ = 600 K. Substitution: V₁/V₂ = 2 and T₂/T₁ = 600/300 = 2. Answer: p₂/p₁ = 2 × 2 = 4, so the pressure increases by a factor of 4. Check/Trap: this is a classic Paper 1 multiple-choice item where the distractor 12 comes from using 327/27 for the temperature ratio. In kelvin the temperature only doubles, even though the Celsius value increases twelve-fold — always convert before forming a ratio.

66. Exam technique/trapSL & HL

    Identify the single most common error in gas-law questions and how to avoid losing marks for it.

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    The trap: substituting a Celsius temperature into pV = nRT, p/T = constant, V/T = constant or E_k = 3/2 k_BT. Why students fall for it: data in questions is usually quoted in °C, and Celsius works fine in Q = mcΔT, so the habit carries over. Correct approach: convert every absolute temperature with T(K) = θ(°C) + 273 before any substitution, and show the conversion in your working so the marker sees it. A temperature difference ΔT does not need conversion because 1 K = 1 °C in size — so ΔU = 3/2 nRΔT can take ΔT directly. Rule of thumb: if T appears alone or in a ratio, it must be in kelvin; if it appears only as Δ, either scale works.

67. Exam technique/trapSL & HLData booklet: Yes

    Explain the distinction between n, N and N_A in gas calculations and the errors caused by confusing them.

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    The trap: putting the number of molecules N into pV = nRT, or the number of moles n into pV = Nk_BT — an error of a factor of 6.02 × 10²³. Why: both equations look alike and both constants are printed in the data booklet. Correct approach: R = 8.31 J K⁻¹ mol⁻¹ is per mole and pairs with n; k_B = 1.38 × 10⁻²³ J K⁻¹ is per molecule and pairs with N; the link is n = N/N_A and k_B = R/N_A. Check the size of your answer: n is usually between 0.01 and 100, while N is around 10²²–10²⁵. Also remember mass = nM = Nm, where m = M/N_A is the mass of one molecule.

68. Exam technique/trapSL & HLData booklet: Yes

    State the units and conversions that must be checked before substituting into pV = nRT, and the significant-figure convention expected.

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    The trap: leaving volumes in cm³ or dm³, pressures in kPa, atm or bar, or molar masses in g mol⁻¹. Correct approach: with R = 8.31 J K⁻¹ mol⁻¹ use SI throughout — p in Pa, V in m³, T in K, M in kg mol⁻¹. Conversions: 1 cm³ = 1 × 10⁻⁶ m³, 1 dm³ = 1 litre = 1 × 10⁻³ m³, 1 kPa = 10³ Pa, 1 atm ≈ 1.0 × 10⁵ Pa. Exception: in a ratio such as p₁V₁ = p₂V₂ consistent non-SI volume units cancel. Significant figures: quote the final answer to the same number as the least precise datum, usually 2 or 3 s.f., and keep full precision in intermediate steps to avoid rounding errors.

69. Exam technique/trapSL & HL

    Outline how to answer command terms correctly in gas-law questions: state, explain, deduce, sketch and suggest.

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    State: give the law or value with no reasoning, e.g. p ∝ 1/V at constant temperature. Explain: give a causal molecular chain — change in speed or collision rate, then momentum change, then force, then pressure; each link is a mark. Deduce: reach a conclusion with supporting reasoning or calculation, e.g. deduce from a p–V graph whether the gas is ideal by testing pV = constant. Sketch: a labelled shape with axes named and units, correct curvature and intercepts, but no accurate plotting. Suggest: propose a plausible physical reason, e.g. suggest why the measured pressure is lower than predicted. Trap: writing an explain-level answer for state wastes time, while a one-line answer to explain caps you at one mark of three.

70. Exam technique/trapSL & HL

    Explain the traps in interpreting p–V isotherms and p–T or V–T graphs for an ideal gas.

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    Trap 1: p–V isotherms are hyperbolae, not straight lines; a higher isotherm corresponds to a higher temperature, and isotherms never cross. Trap 2: a straight line through the origin is only obtained by plotting p against 1/V, which is the standard linearisation for Boyle's law. Trap 3: p–T and V–T graphs for a fixed mass are straight lines that pass through −273 °C, not through the Celsius origin; sketching them through the origin on a Celsius axis is a common lost mark. Trap 4: on a kelvin axis the same lines do pass through the origin. Trap 5: the area under a p–V curve is work done, not pressure. Always label axes with quantity and unit before sketching.

71. Exam technique/trapSL & HL

    Describe an experiment to verify Boyle's law using a gas syringe and a pressure sensor, including variables, limitations and improvements.

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    Apparatus: sealed gas syringe connected by short thick-walled tubing to a pressure sensor and datalogger, clamped in a stand. Method: record the trapped air volume from the syringe scale and the pressure; move the plunger in stages, waiting about 30 s each time, and record about 8 pairs of readings over a wide range. Independent variable V, dependent variable p, controlled: fixed mass of gas (check for leaks) and constant temperature. Analysis: plot p against 1/V; a straight line through the origin verifies p ∝ 1/V, and the gradient equals nRT. Limitations: compression warms the gas, the connecting tube adds dead volume, friction in the plunger, leaks. Improvements: compress slowly and pause for thermal equilibrium, minimise tubing volume, add the dead volume to each reading, repeat and average.

72. Exam technique/trapSL & HL

    Describe how to determine absolute zero experimentally using the pressure law, and state the main sources of error.

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    Apparatus: a sealed flask of dry air of fixed volume connected by a short capillary to a Bourdon gauge or pressure sensor, immersed in a water bath with a thermometer and stirrer. Method: heat the bath in steps from about 10 °C to 90 °C, stir, allow 2–3 minutes for thermal equilibrium, and record θ and p at each step; the volume is constant and the mass of gas fixed. Analysis: plot p (y) against θ (x) in °C; the graph is a straight line, and extrapolating it to p = 0 gives the intercept on the temperature axis, which should be near −273 °C. Errors: the gas in the connecting tube is not at bath temperature (systematic), thermometer and gauge reading uncertainties (random), slow thermal equilibrium, and the very long extrapolation, which magnifies any error in the gradient.

73. Exam technique/trapSL & HL

    Explain how uncertainties are handled in a Boyle's law investigation, including error bars, gradients and systematic versus random errors.

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    Absolute uncertainty is the ± value on a reading, e.g. V = 25.0 ± 0.5 cm³; fractional uncertainty is Δx/x, here 2 %. For a product or quotient, fractional uncertainties add: the uncertainty in pV is the sum of those in p and V, so a constant pV within combined uncertainty supports the law. When plotting p against 1/V, the fractional uncertainty in 1/V equals that in V, so error bars on the x-axis grow at small volumes. Draw maximum and minimum gradient lines through all error bars; the gradient uncertainty is (m_max − m_min)/2. Random errors scatter points either side of the line and are reduced by repeating and averaging; a systematic error such as dead volume or a zero-offset gauge shifts every point and shows as a non-zero intercept, not as scatter.

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