AQA A-Level Chemistry Paper 3, 2017: Question 34

1 mark · Medium difficulty · Multiple Choice

Calculate the pressure of a mixture of oxygen and nitrogen gas when placed into an evacuated flask of known volume at constant temperature.

Practise this question

Question

Question 34: 130 cm cubed of oxygen and 40 cm cubed of nitrogen, each at 298 K and 100 kPa, were placed into an evacuated flask of volume 0.50 dm cubed. What is the pressure of the gas mixture in the flask at 298 K? Options are A: 294 kPa, B: 68.0 kPa, C: 34.0 kPa, D: 13.7 kPa.
Question text

130 cm3 of oxygen and 40 cm3 of nitrogen, each at 298 K and 100 kPa, were placed

into an evacuated flask of volume 0.50 dm3.

What is the pressure of the gas mixture in the flask at 298 K?

[1 mark]

A 294 kPa

B 68.0 kPa

C 34.0 kPa

D 13.7 kPa

Mark scheme

Show the mark scheme Mark scheme showing Question 34 with the correct answer as C.

34 C

How to answer it

Gas Mixtures & Boyle's Law: Total Pressure in an Evacuated Flask

📌 What this question tests

This question assesses your ability to apply the Ideal Gas laws to mixtures of non-reacting gases under constant temperature conditions, specifically:

  • Combining gas volumes at identical temperature and pressure.
  • Converting volume units between cm³ and dm³ .
  • Applying Boyle’s Law ( P₁V₁ = P₂V₂ ) or Dalton's Law of partial pressures efficiently under timed multiple-choice conditions.

Question 34 (Multiple Choice)

Total pressure calculation for mixed gases expanding into a container

✅ Correct Answer

C — 34.0 kPa

The total initial volume is 170 cm³ at 100 kPa . When expanded into a 500 cm³ ( 0.50 dm³ ) flask at constant temperature ( 298 K ), the pressure drops proportionally to 34.0 kPa .

Mark scheme award: [1 mark] for selecting option C.

📐 Step-by-Step Calculation

  1. Total initial gas volume:
    Both gases are at 298 K and 100 kPa .
    V₁ = 130 cm³ + 40 cm³ = 170 cm³
  2. Match volume units:
    Final flask volume, V₂ = 0.50 dm³
    V₂ = 0.50 × 1000 = 500 cm³
  3. Apply Boyle's Law (T is constant at 298 K):
    P₁V₁ = P₂V₂
    100 kPa × 170 cm³ = P₂ × 500 cm³
    P₂ = (100 × 170) / 500 = 17 000 / 500 = 34.0 kPa

💡 Alternative Method: Partial Pressures

You can also calculate the partial pressure of each gas using P₂ = P₁ × (V₁ / V₂) :

  • p(O₂) = 100 kPa × (130 / 500) = 26.0 kPa
  • p(N₂) = 100 kPa × (40 / 500) = 8.0 kPa
  • Total P = p(O₂) + p(N₂) = 26.0 + 8.0 = 34.0 kPa

🧠 Exam Technique & Time Savers

  • Notice constant temperature: Because T remains 298 K throughout, you do not need to use pV = nRT or the molar gas constant ( R = 8.31 J K⁻¹ mol⁻¹ ). Doing so wastes 2 minutes.
  • Sense-check the result: The gas mixture expands from 170 cm³ to 500 cm³ (about a 3-fold expansion). Therefore, pressure must decrease by a factor of 3 (from 100 kPa to ~33 kPa). This immediately eliminates A and B.

❌ Common Errors & Distractor Analysis

  • Distractor A (294 kPa): Occurs from inverting the volume ratio: (500 / 170) × 100 = 294 kPa. Gas expanding into a larger volume cannot experience an increase in pressure!
  • Distractor B (68.0 kPa): Obtained if a student forgets that 0.50 dm³ = 500 cm³ and accidentally halves the calculation, or divides 170 by 250 cm³.
  • Unit Conversion Failure: Leaving 0.50 dm³ without converting to cm³ (or vice versa), resulting in answers with orders of magnitude errors. Always ensure both volumes share the same unit: 1 dm³ = 1000 cm³ .

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.