Edexcel A-Level Chemistry Paper 2, June 2017: Question 4
6 marks · Medium difficulty · Calculations
Calculate the mass of nitrogen gas in a car tyre using the ideal gas equation and explain atomic radius trends between nitrogen and oxygen.
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Question text
4 Traditionally, high-flying aircraft and Formula 1 racing cars have had their tyres
inflated with nitrogen gas instead of air. Recently, this practice has been extended to
some other cars.
(a) A car tyre is filled with nitrogen gas to a volume of 8.98 dm3 and a pressure of
207 kPa at 20 °C.
(i) Using the Ideal Gas Equation, calculate the mass of nitrogen gas, in grams,
present in the car tyre under these conditions. Give your answer to an
appropriate number of significant figures.
(3)
(ii) During a car journey, the tyres become warm. Use the Ideal Gas Equation to
deduce the effect that this has on the pressure in the tyres.
(1)
(b) One reason for the use of nitrogen gas in car tyres is that less gas is lost from the
tyres during use because nitrogen molecules are larger than oxygen molecules.
A suggested explanation for this is that nitrogen atoms are larger than oxygen atoms.
Explain why a nitrogen atom is larger than an oxygen atom.
(2)
(Total for Question 4 = 6 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
4(a)(i) Example of calculation: (3)
conversion of pressure, volume and temperature to correct 207kPa = 207 000 Pa
units (1) 8.98 dm3 = 0.00898 m3,
20°C = 293 K
rearrangement of ideal gas equation so n=PV ÷ RT and n= 207 000 x 0.00898 =
calculation of n (1) 8.31 x 293
= 0.7634…
conversion of answer into mass to 2/3 SF (1) = 0.7634…. x 28 = 21.37647..
= 21.4 / 21 (g)
Correct answer with no working
scores 3
TE on both parts of the calculation
Question
Answer Additional Guidance Mark
Number
4(a)(ii) The temperature increase will result in an increase in Allow p ∝ T (1)
pressure because p is (directly) proportional to T Reference to p=nRT/V
(at constant volume and moles of gas)
Question
Answer Additional Guidance Mark
Number
4(b) An explanation that makes reference to the following points: Reference to molecule scores (0) (2)
fewer protons (in nitrogen) (1) Accept reverse arguments in
terms of oxygen
Allow
weaker (effective) nuclear charge
Allow smaller atomic number
result in a weaker nuclear attraction because shielding is the Do not award if incorrect numbers
same/electrons are in the same (sub)shell (in oxygen)/same of protons stated or if ions
number of electron shells (1) referred to
Do not award ‘charge density’
Ignore references to electron
repulsion and electronegativity
(Total for Question 4 = 6 marks)
How to answer it
Ideal Gas Equations & Periodic Trends Study Guide
This question assesses your ability to apply the Ideal Gas Equation (PV = nRT) with strict unit conversions, evaluate proportional relationships in gases, and explain periodic atomic radius trends using nuclear charge and electron shielding.
Question 4(a)(i) - Ideal Gas Calculation
Using the Ideal Gas Equation to find mass (3 marks)
✅ Correct Answer
- Moles: n = 0.7634... mol
- Mass: 21.4 g or 21 g
💡 Key Knowledge
- Ideal Gas Equation: PV = nRT
- Standard SI units are mandatory: Pressure in Pascals (Pa), Volume in cubic metres (m³), Temperature in Kelvin (K).
- Molar mass of N₂ = 28.0 g mol⁻¹
🧠 Exam Technique
- Step 1: Convert units carefully ( kPa to Pa × 10³, dm³ to m³ × 10⁻³, °C to K + 273).
- Step 2: Rearrange to n = PV / RT and calculate moles.
- Step 3: Multiply moles by molar mass and round to 2 or 3 significant figures.
❌ Common Errors
- Forgetting to convert dm³ to m³ (losing a factor of 10⁻³).
- Using Celsius instead of Kelvin ( 20°C used as 20 instead of 293 K).
- Using atomic nitrogen mass ( 14.0 ) instead of molecular nitrogen ( 28.0 ).
Question 4(a)(ii) - Proportionality in Gases
Deducing the effect of temperature on pressure (1 mark)
✅ Correct Answer
The pressure in the tyres increases.
💡 Key Knowledge
- Pressure is directly proportional to temperature at constant volume and moles ( p ∝ T or p = nRT / V ).
- As a car drives, friction warms the tyre, increasing the kinetic energy and velocity of gas molecules.
🧠 Exam Technique
Keep your answer direct and concise. State clearly that pressure increases and reference either p ∝ T or the ideal gas equation rearranged for pressure.
❌ Common Errors
Stating that volume changes instead. Remember, a car tyre has a relatively fixed rigid volume, so volume cannot expand freely like a balloon.
Question 4(b) - Periodic Trends
Explaining atomic radius differences across atoms (2 marks)
✅ Correct Answer
- Nitrogen has fewer protons than oxygen (lower nuclear charge).
- This results in a weaker nuclear attraction on outer electrons, whilst shielding remains the same (same number of electron shells).
💡 Key Knowledge
- Atomic radius depends on two main factors across a period: nuclear charge (proton number) and shielding.
- Nitrogen (atomic number 7) vs Oxygen (atomic number 8) both occupy the n = 2 principal quantum shell, meaning inner electron shielding is identical.
🧠 Exam Technique
Always structure period trend explanations using the two-part formula: 1) Compare nuclear charge / proton number. 2) State shielding is constant, leading to stronger/weaker electrostatic attraction on outer electrons.
❌ Common Errors
- Critical trap: Referring to nitrogen and oxygen as molecules instead of atoms scores 0 marks.
- Vagueness like "oxygen has more electrons" without mentioning protons or nuclear attraction.
- Mentioning "charge density" or "electronegativity" which are irrelevant to basic atomic radius size explanations here.
Topics
Physical Chemistry · Inorganic Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 1: Atomic Structure and the Periodic Table
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.