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Lesson 4 introduced pV = nRT and the basic rearrangements used to find p, V, n, or T for an ideal gas. This lesson deepens that treatment to the full A-Level level. We synthesise the empirical gas laws of Boyle (1662, pV constant at fixed T and n), Charles (1787, V proportional to T at fixed p and n), Gay-Lussac (1809, p proportional to T at fixed V and n), and Avogadro (1811, equal volumes of any gases at the same T and p contain equal numbers of molecules). Émile Clapeyron unified these in 1834 into a single equation of state, pV = nRT, with R as the universal gas constant. We will derive the molar volumes at RTP (24.0 dm³) and STP (22.7 dm³), use the density form ρ = pM/RT, handle gases collected over water with Dalton's law of partial pressures, apply the combined gas law p₁V₁/T₁ = p₂V₂/T₂, tackle synoptic gas-plus-stoichiometry problems, and finally examine quantitatively how and why real gases deviate from ideality. The compressibility factor Z and the van der Waals equation are not strictly examinable at A-Level but underpin the entropy treatment of §3.1.8 and the Kp work of §3.1.10, so a working familiarity here pays dividends downstream.
Spec mapping (AQA 7405): The principal anchor is §3.1.2.5 (the ideal gas equation and gas-volume calculations). The material connects forward to §3.1.4 (calorimetry and combustion enthalpy, where gas-volume calculations supply n for ΔH determinations), §3.1.10 (equilibrium constants Kp, where partial pressures are computed from total pressure and mole fractions), and §3.1.5 (kinetics — gas-phase collision theory rests on the same kinetic-molecular assumptions as the ideal-gas model). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 covers the four assumptions of an ideal gas, the numerical value and units of R (8.314 J K⁻¹ mol⁻¹), and Dalton's law of partial pressures. AO2 examines numerical use of pV = nRT, ρ = pM/RT, partial-pressure subtraction, and the combined gas law p₁V₁/T₁ = p₂V₂/T₂. AO3 (the most heavily weighted at A-Level) targets multi-step synoptic problems — typically gas-plus-stoichiometry chains where gas-law output feeds reactant or product mass calculations — and evaluative questions on the limits of ideal-gas behaviour, including reasoned predictions of which real gas will deviate more under specified conditions.
pV = nRT
| Symbol | Quantity | SI Unit |
|---|---|---|
| p | Pressure | Pa (pascals) |
| V | Volume | m³ |
| n | Amount of substance | mol |
| R | Gas constant | 8.314 J K⁻¹ mol⁻¹ |
| T | Temperature | K (kelvin) |
An ideal gas obeys pV = nRT exactly, and is defined by four model assumptions:
Real gases deviate from ideal behaviour at high pressures (particles are forced close together; their own volume is no longer negligible, and intermolecular forces become significant) and low temperatures (particles move slowly and reside long enough near each other for attractive forces to influence trajectories). Ideal behaviour is approached at high temperatures and low pressures — the regime where the four assumptions hold well.
Key Point: A real gas behaves most ideally when its temperature is far above its boiling point and its pressure is well below the critical pressure. Helium and neon at room temperature and 1 atm are excellent approximations to ideal; CO₂ at 30 °C and 60 atm is poor.
The molar volume Vₘ is the volume occupied by one mole of an ideal gas at specified conditions. It is obtained directly from pV = nRT by setting n = 1:
Vₘ = RT/p
Three sets of "standard" conditions appear in chemistry, and it is important to keep them distinct.
Room temperature and pressure as defined by AQA: T = 298 K, p = 100 kPa = 1.00 × 10⁵ Pa.
Vₘ = RT/p = (8.314 × 298) / 100 000 = 2477.6 / 100 000 = 0.02478 m³ = 24.8 dm³
The AQA data sheet rounds this to 24.0 dm³ mol⁻¹ for use in mole-from-volume calculations at RTP, reflecting an averaged "room temperature" closer to 293 K rather than the strict 298 K. Use 24.0 dm³ unless told otherwise — that is the convention rewarded in AQA mark schemes.
Standard temperature and pressure: T = 273.15 K, p = 100 kPa.
Vₘ = (8.314 × 273.15) / 100 000 = 2270.9 / 100 000 = 0.02271 m³ = 22.7 dm³
The older definition uses T = 273.15 K, p = 101.325 kPa = 1 atm:
Vₘ = (8.314 × 273.15) / 101 325 = 2270.9 / 101 325 = 0.02241 m³ = 22.414 dm³
The famous "22.4 dm³" figure encountered in older textbooks refers to this 1-atm STP. AQA examinations use 100 kPa as standard pressure.
The molar volume of liquid water is approximately:
Vₘ(H₂O, l) = M/ρ = 18.0 / 1.00 = 18.0 cm³ mol⁻¹ = 0.0180 dm³ mol⁻¹
Compared with 24.0 dm³ mol⁻¹ for an ideal gas at RTP, this is a factor of 24.0 / 0.0180 ≈ 1333. One mole of a substance in the gas phase occupies roughly 1300 times more space than in the condensed phase — a useful order-of-magnitude check whenever a calculation produces a suspect gas-volume figure.
Exam Tip: Unless told otherwise, use 24.0 dm³ mol⁻¹ as the molar volume at RTP. If a question states "at 273 K and 100 kPa", use 22.7 dm³ instead — or fall back on pV = nRT, which never fails when the inputs are clean.
Combining n = m/M with pV = nRT gives pV = (m/M)RT, which rearranges to:
M = mRT / (pV)
This is the workhorse equation for identifying volatile compounds by measuring the mass of a known volume of vapour at controlled p and T — historically the Victor Meyer method (a sealed-tube vaporiser feeding a gas-collection cylinder). At A-Level, the technique appears either as a pure pV = nRT calculation (Example 1) or paired with a density measurement (Example 2).
0.250 g of a gaseous compound occupies 98.0 cm³ at 100 kPa and 373 K. Calculate its molar mass.
Convert to SI: p = 100 000 Pa; V = 98.0 × 10⁻⁶ m³ = 9.80 × 10⁻⁵ m³; m = 0.250 g; T = 373 K; R = 8.314 J K⁻¹ mol⁻¹.
M = mRT / (pV) = (0.250 × 8.314 × 373) / (100 000 × 9.80 × 10⁻⁵) = 775.3 / 9.80 = 79.1 g mol⁻¹
A molar mass near 79 is consistent with methanol-d₃ (CD₃OH, M = 35) or benzene (C₆H₆, M = 78.1) — benzene is the strong candidate at 373 K (its boiling point is 353 K, so it vaporises readily).
A gas has a density of 1.78 g dm⁻³ at 298 K and 100 kPa. Calculate its molar mass and suggest its identity.
In 1 dm³ (= 1.00 × 10⁻³ m³), mass = 1.78 g. Then:
M = mRT / (pV) = (1.78 × 8.314 × 298) / (100 000 × 1.00 × 10⁻³) = 4408.5 / 100.0 = 44.1 g mol⁻¹
A molar mass of 44.1 g mol⁻¹ matches carbon dioxide (CO₂) (M = 44.0) and propane (C₃H₈) (M = 44.1) — to within experimental tolerance, they are indistinguishable on density alone. Discrimination requires a chemical test (lime-water turbidity for CO₂; combustibility for propane) or an instrumental method (IR absorption at 2349 cm⁻¹ for CO₂'s asymmetric stretch; characteristic C-H stretches near 2960 cm⁻¹ for propane).
The density of a gas follows from rearranging pV = nRT with n = m/M:
pV = (m/M) RT ⟹ m/V = pM/RT ⟹ ρ = pM / RT
with ρ in kg m⁻³ if M is in kg mol⁻¹, p in Pa, T in K (or in g m⁻³ if M is in g mol⁻¹, dividing by 1000 to convert to g dm⁻³).
At fixed T and p, density is directly proportional to molar mass — a result that justifies the rule "heavier gases sink". At fixed M and T, density is proportional to pressure (compress and you condense the gas towards a liquid); at fixed M and p, density is inversely proportional to temperature (hot air rises because it is less dense).
Calculate the density of oxygen gas (O₂) at 298 K and 101.3 kPa. M(O₂) = 32.0 g mol⁻¹.
ρ = pM / RT = (101 300 × 32.0) / (8.314 × 298) = 3 241 600 / 2477.6 = 1308 g m⁻³ = 1.31 g dm⁻³
For comparison, the density of liquid water is 1000 g dm⁻³ — gaseous O₂ at RTP is ~760 times less dense than the liquid that fish breathe through.
Compare the densities of N₂, CO₂, H₂, and NH₃ at RTP and predict the appropriate laboratory delivery method for each.
Since ρ = pM/RT, at fixed T and p, density tracks molar mass:
| Gas | M (g mol⁻¹) | ρ at RTP (g dm⁻³) | ρ relative to air (≈1.20) |
|---|---|---|---|
| H₂ | 2.0 | 0.0819 | 0.069 (much less dense) |
| NH₃ | 17.0 | 0.696 | 0.580 (less dense) |
| Air (avg) | 29.0 | 1.19 | 1.00 |
| N₂ | 28.0 | 1.15 | 0.964 (≈ air) |
| CO₂ | 44.0 | 1.80 | 1.51 (denser) |
| Cl₂ | 71.0 | 2.90 | 2.43 (much denser) |
Practical consequences: CO₂ and Cl₂ (denser than air) are collected by downward delivery (i.e. upward displacement of air, gas tube pointing down, gas sinks to fill the receiver). H₂ and NH₃ (less dense than air) are collected by upward delivery (gas tube pointing up, gas rises and displaces air downwards). Gases close in density to air (N₂, O₂) are best collected over water if soluble considerations permit — gas density alone is too close to discriminate by displacement.
The industrial significance of ρ = pM/RT is substantial: gas-density measurements are still used in process control (orifice-plate flow meters infer mass flow from pressure-drop and gas density) and in oceanography (dissolved-gas analysis converts measured concentrations into partial pressures via the ideal-gas law).
When a gas is generated by reaction and collected by displacement of water in an inverted measuring cylinder, the collected gas is necessarily saturated with water vapour. The cylinder therefore contains a binary mixture: the target gas plus water vapour at its equilibrium vapour pressure at the experimental temperature.
Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases equals the sum of the partial pressures of each component, where the partial pressure of a component is the pressure it would exert if it occupied the same volume alone at the same temperature:
p_total = p₁ + p₂ + p₃ + ...
For a gas collected over water:
p_total = p_gas + p_water ⟹ p_gas = p_total − p_water
The vapour pressure of water depends sensitively on temperature (rising roughly doubling every 10-15 °C):
| Temperature (°C) | Vapour pressure of water (kPa) |
|---|---|
| 15 | 1.71 |
| 20 | 2.34 |
| 25 | 3.17 |
| 30 | 4.24 |
| 40 | 7.38 |
| 50 | 12.34 |
| 60 | 19.93 |
| 100 | 101.3 |
At 100 °C the vapour pressure of water equals 1 atm — which is precisely the definition of the normal boiling point.
Hydrogen gas is collected over water at 25 °C and atmospheric pressure 101.3 kPa. 120 cm³ of wet gas is collected. The vapour pressure of water at 25 °C is 3.17 kPa. Calculate the moles of dry hydrogen.
p(H₂) = 101.3 − 3.17 = 98.13 kPa = 98 130 Pa
V = 120 cm³ = 1.20 × 10⁻⁴ m³; T = 298 K
n = pV / RT = (98 130 × 1.20 × 10⁻⁴) / (8.314 × 298) = 11.776 / 2477.6 = 4.75 × 10⁻³ mol
If the water-vapour correction had been omitted, the calculated n would have been (101 300 × 1.20 × 10⁻⁴)/(8.314 × 298) = 4.91 × 10⁻³ mol — a 3.2% overestimate.
A 250 cm³ volume of oxygen is collected over water at 30 °C and 100 kPa. Vapour pressure of water at 30 °C is 4.24 kPa. Calculate (i) moles of dry O₂; (ii) the percentage error if the vapour pressure correction were ignored.
(i) p(O₂) = 100.0 − 4.24 = 95.76 kPa = 95 760 Pa; V = 2.50 × 10⁻⁴ m³; T = 303 K.
n(O₂) = (95 760 × 2.50 × 10⁻⁴) / (8.314 × 303) = 23.94 / 2519.1 = 9.50 × 10⁻³ mol
(ii) Uncorrected: n = (100 000 × 2.50 × 10⁻⁴)/(8.314 × 303) = 25.00/2519.1 = 9.92 × 10⁻³ mol. Percentage error = (9.92 − 9.50)/9.50 × 100% = 4.4%.
The correction is small at 15-25 °C (~2-3%) but grows rapidly with temperature: at 50 °C the vapour-pressure correction is already 12.3% of atmospheric, and ignoring it would be a serious experimental blunder. Above 60 °C, gas collection over water is rarely worthwhile — collection over mercury or a drying tube becomes preferable.
For a fixed amount of gas (n constant), pV/T = nR is itself a constant. Comparing two states:
p₁V₁ / T₁ = p₂V₂ / T₂
The combined gas law follows from pV = nRT by holding n constant — it is not an independent law but a direct algebraic consequence. It is most useful when converting gas-volume measurements taken at one set of (p, T) conditions to those at another, without needing to know the absolute amount n.
A gas occupies 500 cm³ at 20 °C and 100 kPa. What volume does it occupy at 80 °C and 200 kPa?
p₁ = 100 kPa, V₁ = 500 cm³, T₁ = 293 K; p₂ = 200 kPa, V₂ = ?, T₂ = 353 K.
V₂ = p₁V₁T₂ / (T₁p₂) = (100 × 500 × 353) / (293 × 200) = 17 650 000 / 58 600 = 301 cm³
Sanity check: heating (293 → 353 K) raises V by factor 1.20; doubling pressure halves V; combined factor = 1.20/2 = 0.602; predicted V₂ = 500 × 0.602 = 301 cm³. ✓
A balloon contains 750 cm³ of helium at 28 °C and 98.5 kPa. What would its volume be at STP (273.15 K, 100 kPa)?
p₁ = 98.5 kPa, V₁ = 750 cm³, T₁ = 301 K; p₂ = 100 kPa, V₂ = ?, T₂ = 273.15 K.
V₂ = p₁V₁T₂ / (T₁p₂) = (98.5 × 750 × 273.15) / (301 × 100) = 20 178 956 / 30 100 = 670 cm³
The volume contracts modestly: cooling reduces V by ~9.3%; small pressure increase reduces V by another 1.5%; net contraction ~10.6%. Converting measurements to STP is the historical convention for reporting gas-collection experiments, allowing direct comparison across laboratories operating at different ambient conditions.
Exam Tip: In the combined gas law you may use any consistent units for p and V (they cancel between the two sides), but T must always be in kelvin. Forgetting the °C → K conversion is the single most common error.
A-Level Paper 2 and Paper 3 commonly chain a gas-law calculation onto a stoichiometric step: gas-law output (n of a gaseous product) feeds a mole-ratio step that determines an unknown reactant mass, or vice versa. The skill is in recognising the bridge and keeping unit conversions clean.
2.00 g of calcium carbonate reacts with excess hydrochloric acid. Calculate the volume of carbon dioxide produced at 35 °C and 98.0 kPa.
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
Step 1 (stoichiometry): n(CaCO₃) = 2.00 / 100.1 = 0.01999 mol.
Step 2 (mole ratio): 1:1 CaCO₃ → CO₂, so n(CO₂) = 0.01999 mol.
Step 3 (gas law): p = 98 000 Pa, T = 308 K, R = 8.314 J K⁻¹ mol⁻¹.
V = nRT / p = (0.01999 × 8.314 × 308) / 98 000 = 51.20 / 98 000 = 5.22 × 10⁻⁴ m³ = 0.522 dm³ (522 cm³).
The "gas + stoichiometry" chain — solid-state mass → moles → mole ratio → gaseous moles → ideal-gas volume — is the dominant template at A-Level and recurs in carbonate decomposition, metal-acid reactions, alcohol vaporisation, and combustion analysis.
0.184 g of ethanol (C₂H₅OH) is vaporised at 100 °C and 100 kPa. Calculate the volume of the ethanol vapour.
M(C₂H₅OH) = 46.0 g mol⁻¹; n = 0.184 / 46.0 = 4.00 × 10⁻³ mol; p = 100 000 Pa; T = 373 K.
V = nRT / p = (4.00 × 10⁻³ × 8.314 × 373) / 100 000 = 12.40 / 100 000 = 1.24 × 10⁻⁴ m³ = 0.124 dm³ (124 cm³).
Once the small ethanol sample is fully vaporised at 100 °C, it behaves as a near-ideal gas (well above the 78 °C boiling point); the answer is the volume the vapour would occupy in a rigid container at exactly 100 °C and 100 kPa.
A sample of zinc reacts with excess sulfuric acid. The hydrogen gas produced occupies 240 cm³ at 20 °C and 100 kPa. Calculate the mass of zinc that reacted.
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
Step 1 (gas law for product H₂): p = 100 000 Pa; V = 2.40 × 10⁻⁴ m³; T = 293 K.
n(H₂) = pV / RT = (100 000 × 2.40 × 10⁻⁴) / (8.314 × 293) = 24.0 / 2436.0 = 9.85 × 10⁻³ mol.
Step 2 (mole ratio): 1:1 Zn → H₂, so n(Zn) = 9.85 × 10⁻³ mol.
Step 3 (mass): m(Zn) = 9.85 × 10⁻³ × 65.4 = 0.644 g.
This chain — gas volume → moles via pV = nRT → reactant mass via mole ratio — is the reverse of Example 6. Both must be fluent.
The ideal-gas equation is an excellent approximation at moderate p and T, but real gases deviate measurably. The compressibility factor Z quantifies the deviation:
Z = pV / nRT
For an ideal gas Z = 1 at all conditions. For a real gas:
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