1.5.3Additional lessons
The Ideal Gas Equation — IB Chemistry HL Revision Notes & Topical Questions
5 min read 6 parts Lesson 20 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 6The Equation and Its Variables
The ideal gas equation PV = nRT combines all the gas laws into one relationship for an ideal gas.
P (pressure)
- P (pressure) Must be in pascals, Pa, when R = 8.31; convert kPa by multiplying by 10³ and bar by 10⁵.
V (volume)
- V (volume) Must be in cubic metres, m³; 1 dm³ = 10⁻³ m³ and 1 cm³ = 10⁻⁶ m³ when converting.
n (amount)
- n (amount) The number of moles of gas, found from mass divided by molar mass or from a balanced equation.
R (gas constant)
- R (gas constant) 8.31 J K⁻¹ mol⁻¹, the constant that links the other four variables together.
T (temperature)
- T (temperature) Must be in kelvin; add 273 (or 273.15) to the Celsius temperature before using it.
Part 2 of 6Special Cases of the Gas Laws
Special Cases of the Gas Laws
Boyle's law
- Boyle's law At constant n and T, pressure is inversely proportional to volume, so P₁V₁ = P₂V₂.
Charles's law
- Charles's law At constant n and P, volume is directly proportional to the kelvin temperature.
Pressure law
- Pressure law At constant n and V, pressure is directly proportional to the kelvin temperature.
Avogadro's law
- Avogadro's law At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules.
| Condition | Fixed | Relationship |
|---|---|---|
| Boyle | n, T | P ∝ 1/V |
| Charles | n, P | V ∝ T |
| Pressure law | n, V | P ∝ T |
Part 3 of 6Using the Equation
1
Write down every quantity given and identify the one that must be calculated.
2
Convert pressure to pascals, volume to cubic metres and temperature to kelvin before substituting anything.
3
Rearrange the equation so the unknown stands alone, for example V = nRT/P or n = PV/RT.
4
Substitute the values with R = 8.31 and calculate, then convert the answer to a convenient unit such as dm³.
5
To find molar mass, calculate n first and then divide the measured mass by n.
Part 4 of 6Molar Volume at STP
STP conditions
- STP conditions Standard temperature and pressure are 273 K and 1.00 × 10⁵ Pa as used in this course.
Molar volume
- Molar volume One mole of any ideal gas occupies 22.7 dm³ at STP, found from V = RT/P.
Use in calculations
- Use in calculations Gas volume at STP equals moles multiplied by 22.7 dm³ mol⁻¹, which avoids using the full equation.
Different conditions
- Different conditions At room temperature the molar volume is larger, so the 22.7 value cannot be reused there.
Part 5 of 6Worked Example
Question
- A 0.640 g sample of a gas occupies 250 cm³ at 100 kPa and 300 K.
- Find its molar mass.
- V = 250 cm³ = 2.50 × 10⁻⁴ m³ and P = 1.00 × 10⁵ Pa.
- n = PV/RT = (1.00 × 10⁵ × 2.50 × 10⁻⁴)/(8.31 × 300) = 1.00 × 10⁻² mol.
- M = mass/n = 0.640 g / 1.00 × 10⁻² mol.
Answer M ≈ 64 g mol⁻¹, consistent with sulfur dioxide, SO₂.
Part 6 of 6Exam Checklist & Tips
1.5.3 – Key Points for the Exam
- Ideal gas A hypothetical gas whose particles have no volume and no attractions and obey PV = nRT exactly.
- R The gas constant, 8.31 J K⁻¹ mol⁻¹.
- STP 273 K and 1.00 × 10⁵ Pa.
- Molar volume The volume occupied by one mole of gas, 22.7 dm³ mol⁻¹ at STP.
Common Mistakes
- Leaving the volume in dm³ or cm³ and the pressure in kPa, which gives an answer wrong by a power of ten.
- Using the temperature in °C instead of kelvin.
- Using 22.7 dm³ mol⁻¹ for gases measured at room temperature rather than at STP.
- Forgetting to find moles first when the question asks for molar mass.
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