OCR A-Level Chemistry AS Breadth in chemistry (01), June 2019: Question 6

1 mark · Medium difficulty · Multiple Choice

Calculate the volume of 0.0100 mol of nitrogen gas at 350 °C and 200 kPa using the ideal gas equation.

Practise this question

Question

Multiple choice question asking for the volume of 0.0100 mol of N2 at 350 °C and 200 kPa, with four options A (145 cm3), B (259 cm3), C (145 dm3), and D (259 dm3).
Question text

6 What is the volume of 0.0100 mol of N2 at 350 °C and 200 kPa?

A 145 cm3

B 259 cm3

C 145 dm3

D 259 dm3

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer for question 6 is B.

6 B 1 AO2.2

How to answer it

Calculating Gas Volume Using the Ideal Gas Equation

What this question tests

This question assesses your ability to recall, rearrange, and apply the ideal gas equation ( pV = nRT ), requiring rigorous unit conversions for pressure, temperature, and volume, alongside proper handling of significant figures.

Question 6: Multiple Choice Solution

Exam Part: Single-choice selection (1 mark)

✅ Correct Answer: B

The correct option is B (259 cm³). Working through the ideal gas equation yields a volume of 2.59 × 10⁻⁴ m³ , which converts directly to 259 cm³ .

💡 Key Knowledge

  • The Ideal Gas Equation: pV = nRT
  • Standard SI Units must be used in the formula:
    • p in Pascals (Pa)
    • V in cubic metres (m³)
    • n in moles (mol)
    • R = 8.31 J mol⁻¹ K⁻¹
    • T in Kelvin (K)

🧠 Exam Technique

Always convert your units before rearranging or inputting numbers into your calculator to prevent careless substitution errors. Pay close attention to whether the final multiple-choice options require cubic centimetres ( cm³ ) or cubic decimetres ( dm³ ).

❌ Common Errors

  • Temperature trap: Forgetting to add 273 to convert Celsius to Kelvin (using 350 instead of 623 K).
  • Pressure trap: Failing to multiply kilopascals by 1000 to convert to Pascals.
  • Volume unit mix-up: Forgetting that m³ converts to cm³ by multiplying by 10⁶ , leading to option C or D.

📐 Step-by-Step Calculation

  1. Rearrange the equation for Volume ( V ):
    V = nRT / p
  2. Convert all given values into SI units:
    • n = 0.0100 mol
    • T = 350 + 273 = 623 K
    • p = 200 kPa = 200 × 10³ Pa (or 200,000 Pa)
    • R = 8.31 J mol⁻¹ K⁻¹
  3. Substitute values into the expression:
    V = (0.0100 × 8.31 × 623) / 200,000
    V = 51.7737 / 200,000 = 0.00025886 m³
  4. Convert volume from m³ to cm³:
    0.00025886 m³ × 10⁶ = 258.86 cm³
  5. Round to appropriate significant figures:
    Given data ( 0.0100 mol , 350 °C , 200 kPa ) are given to 3 significant figures, so round to 3 s.f. to get 259 cm³.
Examiner Note: This question successfully discriminates between students who memorise unit conversions mechanically and those who understand dimensional consistency in calculations. Top-scoring candidates easily bypassed distractors like D by tracking their unit conversions from m³ to cm³ carefully.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.