Gas calculations
Gas calculations
The ideal gas equation and gas molar mass
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The ideal gas equation connects pressure, volume, temperature and amount. Write pV = nRT, then rearrange for the quantity you need.
- Use pressure in Pa, volume in m³ and temperature in K when using R = 8.31 J mol⁻¹ K⁻¹.
- Convert before substituting: add 273 to temperature in °C for these school calculations, and use 1 dm³ = 10⁻³ m³ or 1 cm³ = 10⁻⁶ m³.
- Calculate the amount in moles. If mass is supplied, use Mr = m/n to find relative molecular mass, using mass in grams.
The ideal-gas model is a better approximation at low pressure and high temperature. The particles then occupy a smaller fraction of the volume, and attractions have less effect.
R gas constant, 8.31 J mol⁻¹ K⁻¹, unless the question supplies a different value; T absolute temperature of the gas, in K; V gas volume, in m³ in the ideal gas equation; n amount of the specified substance or particles, in mol; p gas pressure, in Pa in the ideal gas equation.
| Conversion | Operation |
|---|---|
| kPa → Pa | × 1000 |
| cm3 → m3 | ÷ 1 000 000 |
| dm3 → m3 | ÷ 1000 |
| °C → K | Add 273, or 273.15 if the question precision requires it |
Convert before substituting
Convert 250 cm3, 100 kPa and 27 °C to SI units.
Worked answer
Amount of gas
Find the number of moles in 250 cm3 at 100 kPa and 300 K. R = 8.31 J mol−1 K−1.
Worked answer
n = (100 000 × 2.50 × 10−4)/(8.31 × 300) = 0.010028… mol
Gas volume
Find the volume of 0.0200 mol gas at 298 K and 101 kPa. R = 8.31 J mol−1 K−1.
Give your answer in m³ and cm³.
Worked answer
V = (0.0200 × 8.31 × 298)/101 000 = 4.90372… × 10−4 m3
Mr relative molecular or formula mass; no unit; R gas constant, 8.31 J mol⁻¹ K⁻¹, unless the question supplies a different value; T absolute temperature of the gas, in K; V gas volume, in m³ in the ideal gas equation; m mass of the specified sample or substance, in g; n amount of the specified substance or particles, in mol; p gas pressure, in Pa in the ideal gas equation.
Relative molecular mass of a vapour
A 0.180 g vapour occupies 100 cm3 at 100 kPa and 400 K. R = 8.31 J mol−1 K−1.
Calculate the relative molecular mass, Mᵣ, of the vapour.
Worked answer
n = (100 000 × 1.00 × 10−4)/(8.31 × 400) = 0.00300842… mol
Mr = 0.180/0.00300842…
Relative molecular mass from gas density
A gas has density 1.20 g dm−3 at 100 kPa and 300 K.
Calculate the relative molecular mass, Mᵣ, of the gas. Use R = 8.31 J mol⁻¹ K⁻¹.
Worked answer
Choose 1.00 dm3, which is 0.00100 m3 and has mass 1.20 g.
n = (100 000 × 0.00100)/(8.31 × 300) = 0.0401123… mol
Mr = 1.20/0.0401123…
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
A 1.05 g sample vaporises completely. Its vapour occupies 425 cm³ at 102 kPa and 25 °C. Assume ideal behaviour and use R = 8.31 J mol⁻¹ K⁻¹.
- Calculate the amount in moles of vapour. (3 marks)
- Calculate its relative molecular mass. (2 marks)
- State one further experimental condition, other than the assumptions already given, required for this result to be valid. (1 mark)
- A gas has density 2.00 g dm⁻³ at 101 kPa and 298 K. Calculate its relative molecular mass. (3 marks)
- Explain why ideal-gas behaviour is a better approximation at low pressure and high temperature. (2 marks)
Total: 11 marks
Show answer & marking points
(a) p = 102000 Pa, V = 4.25 × 10⁻⁴ m³ and T(K) = temperature(°C) + 273
T = 25 + 273 = 298 K. [1] n = pV/RT [1] = 0.017505… mol = 0.0175 mol. [1] Accept the use of 298.15 K with consistent working.
(b) Mᵣ = m/n
Mᵣ = 1.05/0.017505… [1] = 60.0, with no unit. [1]
(c) Any one: no vapour is lost; volume contains only sample vapour; temperature and pressure are measured correctly. [1] Do not credit complete vaporisation or ideal behaviour, which are stated in the question.
(d) Consider 1.00 dm³ gas: mass 2.00 g and volume 1.00 × 10⁻³ m³. [1] n = pV/(RT)
n = (101000 × 0.00100)/(8.31 × 298) = 0.040785… mol. [1] Mᵣ = 2.00/n = 49.0. [1]
(e) At low pressure the particles are far apart, so their own volume is a small fraction of the gas volume. [1] At high temperature their kinetic energy is larger relative to the effect of attractions. [1]
Gas volumes at the same temperature and pressure
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For gases at the same temperature and pressure, volume is proportional to amount in moles. This lets you use balanced-equation coefficients as gas-volume ratios.
Check both conditions before using the shortcut. It does not turn the equation coefficients into volume ratios for liquids or solids.
If a molar gas volume Vₘ is supplied, use n = V/Vₘ with matching volume units. For example, with Vₘ = 24.0 dm³ mol⁻¹, first convert 120 cm³ to 0.120 dm³. Then n = 0.120/24.0 = 0.00500 mol.
Carbon monoxide combustion
2CO(g) + O2(g) → 2CO2(g). Mix 60.0 cm3 CO with 50.0 cm3 O2 at the same temperature and pressure.
Identify the limiting reactant and calculate the volume of each gas present after complete reaction, measured at the original temperature and pressure.
Worked answer
2CO + O₂ → 2CO₂. At the same temperature and pressure, gas volumes are proportional to moles. Divide each starting reactant volume by its coefficient; the smaller value identifies the limiting reactant.
| Quantity | CO | O₂ | CO₂ |
|---|---|---|---|
| Initial volume / cm³ | 60.0 | 50.0 | 0 |
| Coefficient | 2 | 1 | 2 |
| Volume ÷ coefficient / cm³ | 60.0/2 = 30.0 (smaller) | 50.0/1 = 50.0 | Product: not compared |
| Volume used or formed / cm³ | 30.0 × 2 = 60.0 used | 30.0 × 1 = 30.0 used | 30.0 × 2 = 60.0 formed |
| Final volume / cm³ | 60.0 − 60.0 = 0 | 50.0 − 30.0 = 20.0 | 60.0 |
CO is limiting. The final mixture contains 20.0 cm³ O₂ and 60.0 cm³ CO₂; no CO remains.
Ammonia volume
N2 + 3H2 → 2NH3. What volume of NH3 is formed from 90.0 cm3 H2 with excess N2 at the same conditions?
Worked answer
V(NH3) = 90.0 × 2/3
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
80.0 cm³ N₂ and 180 cm³ H₂ react completely according to N₂(g) + 3H₂(g) → 2NH₃(g). All gas volumes are measured at the same temperature and pressure.
- Determine the limiting reactant and calculate the total final gas volume. (4 marks)
- Explain why equation coefficients can be used as gas-volume ratios here. (1 mark)
- A separate gas sample occupies 360.0 cm³ where the supplied molar gas volume is 24.0 dm³ mol⁻¹. Calculate the number of moles of gas in this sample. (2 marks)
Total: 7 marks
Show answer & marking points
(a) N₂ + 3H₂ → 2NH₃. Compare starting volume ÷ coefficient, since all volumes are at the same temperature and pressure.
| Quantity | N₂ | H₂ | NH₃ |
|---|---|---|---|
| Initial volume / cm³ | 80.0 | 180 | 0 |
| Coefficient | 1 | 3 | 2 |
| Volume ÷ coefficient / cm³ | 80.0/1 = 80.0 | 180/3 = 60.0 (smaller) | Product: not compared |
| Volume used or formed / cm³ | 60.0 × 1 = 60.0 used | 60.0 × 3 = 180 used | 60.0 × 2 = 120 formed |
| Final volume / cm³ | 80.0 − 60.0 = 20.0 [1] | 0 | 120 [1] |
H₂ is limiting because its volume ÷ coefficient is smaller. [1] Total final gas volume = 20.0 + 120 = 140 cm³. [1]
(b) At matching temperature and pressure, gas volume is proportional to amount, so the mole ratios are volume ratios. [1]
(c) 360.0 cm³ = 0.3600 dm³. [1] n = V/Vₘ
n = 0.3600/24.0 = 0.0150 mol. [1]
Combine solution amounts, limiting reactants and gases
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A reaction involving a solid, a solution and a gas still uses the same mole-ratio bridge. Each starting quantity just has a different route to moles.
- Use n = m/Mr for the solid and n = cV for the solution, with V in dm³.
- If both reactant quantities are supplied, check which is limiting.
- Use the limiting amount and balanced equation to find moles of gas.
- Convert the gas amount to volume using pV = nRT or a supplied molar gas volume.
Do not calculate a gas volume from a reactant that is actually in excess. The reactant that runs out sets the maximum amount of gas.
Magnesium with hydrochloric acid
0.300 g Mg reacts with 20.0 cm3 of 1.00 mol dm−3 HCl. Mg + 2HCl → MgCl2 + H2. Ar(Mg) = 24.3. Find the maximum H2 volume at 100 kPa and 298 K; R = 8.31 J mol−1 K−1.
Worked answer
Mg + 2HCl → MgCl₂ + H₂. Compare the starting moles divided by each coefficient.
| Quantity | Mg | HCl |
|---|---|---|
| Mass / g | 0.300 | Not needed |
| Aᵣ or Mᵣ (no unit) | 24.3 | Not needed |
| Concentration, c / mol dm⁻³ | Solid | 1.00 |
| Volume, V / dm³ | Solid | 20.0/1000 = 0.0200 |
| Amount, n / mol | n = m/Aᵣ 0.300/24.3 = 0.012345… | n = cV 1.00 × 0.0200 = 0.0200 |
| Coefficient | 1 | 2 |
| n ÷ coefficient / mol | 0.012345…/1 | 0.0200/2 = 0.0100 (smaller) |
HCl is limiting. HCl : H₂ = 2 : 1, so n(H₂) = 0.0200/2 = 0.0100 mol.
V = nRT/p = (0.0100 × 8.31 × 298)/100000 = 2.47638 × 10⁻⁴ m³ = 248 cm³ H₂.
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
0.500 g CaCO₃ reacts with 25.0 cm³ of 0.300 mol dm⁻³ HCl. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. Use Mᵣ(CaCO₃) = 100.1 and R = 8.31 J mol⁻¹ K⁻¹.
- Determine the limiting reactant. (3 marks)
- Calculate the maximum CO₂ volume at 100 kPa and 298 K, in cm³. (3 marks)
Total: 6 marks
Show answer & marking points
(a) Compare the starting moles divided by the coefficient.
| Quantity | CaCO₃ | HCl |
|---|---|---|
| Mass / g | 0.500 | Not needed |
| Aᵣ or Mᵣ (no unit) | 100.1 | Not needed |
| Concentration, c / mol dm⁻³ | Solid | 0.300 |
| Volume, V / dm³ | Solid | 25.0/1000 = 0.0250 |
| Amount, n / mol | n = m/Mᵣ 0.500/100.1 = 0.004995… [1] | n = cV 0.300 × 0.0250 = 0.00750 [1] |
| Coefficient | 1 | 2 |
| n ÷ coefficient / mol | 0.004995…/1 | 0.00750/2 = 0.00375 (smaller) |
HCl has the smaller n ÷ coefficient value, so HCl is limiting. [1]
(b) n(CO₂) = n(known substance) × coefficient(target)/coefficient(known substance)
n(CO₂) = 0.00750/2 = 0.00375 mol. [1] V = nRT/p
V = (0.00375 × 8.31 × 298)/100000 = 9.286425 × 10⁻⁵ m³. [1] V = 92.9 cm³. [1]