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A Level · Cambridge (CIE) · 9701

Calculate gas amounts with pV = nRT

Rearrange the ideal gas equation and keep pressure, volume, temperature and the gas constant in compatible units.

States of matter · Cambridge 9701 AS Chemistry · 4.1

AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Keep pressure, volume, amount and absolute temperature consistent in gas calculations. Written practice is self-checked

Before you begin

  • Rearrange a formula and convert between common metric volume units.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which temperature scale is used in pV = nRT?

What you will learn

  • State and rearrange pV = nRT
  • Use pressure, volume and temperature to calculate gas amount
  • Convert temperature to kelvin and volume to compatible units

Use absolute temperature and compatible units

The ideal gas equation is pV = nRT, where p is pressure, V is volume, n is amount in mol, R is the molar gas constant and T is absolute temperature in kelvin.

With R = 8.31 J mol⁻¹ K⁻¹, use p in pascals and V in cubic metres so pV is in joules. Convert Celsius to kelvin with T/K = t/°C + 273. Avoid mixing kPa with m³ or cm³ without an explicit conversion.

Rearrange before substituting values

For amount, n = pV/(RT). For a fixed amount of ideal gas, pV/T is constant between two states, provided the amount does not change.

Estimate whether the result is plausible: a higher pressure or volume at the same temperature corresponds to more gas particles; a higher absolute temperature increases the pV product for a fixed amount.

Pause and explain

With p in Pa, V in m³ and R in J mol⁻¹ K⁻¹, which expression gives n?

Put the idea to work

Worked example

Find n for a gas at 100 000 Pa, volume 2.493 × 10⁻³ m³ and temperature 300 K. Use R = 8.31 J mol⁻¹ K⁻¹.

Show the worked solution
  1. n = pV/(RT).
  2. pV = 100 000 × 0.002493 = 249.3 J.
  3. RT = 8.31 × 300 = 2493 J mol⁻¹; divide.

Answer n = 0.100 mol.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Calculate n for p = 2.00 × 10⁵ Pa, V = 1.247 × 10⁻³ m³ and T = 300 K. Use R = 8.31 J mol⁻¹ K⁻¹.

Give me a hint

Use n = pV/(RT) and keep the SI units together.

Compare my reasoning
  1. pV = 2.00 × 10⁵ × 1.247 × 10⁻³ = 249.4 J.
  2. RT = 8.31 × 300 = 2493 J mol⁻¹.
  3. Divide pV by RT.

n ≈ 0.100 mol.

Look for these in your work

  • Equation rearranged
  • Units compatible
  • Amount calculated
2 · Independent

Convert 27 °C to kelvin and explain why it must be converted before using the gas equation.

Give me a hint

Add the Celsius temperature to 273 for this syllabus calculation.

Compare my reasoning
  1. T = 27 + 273.
  2. T = 300 K.
  3. The equation uses absolute temperature, not Celsius.

27 °C = 300 K. Kelvin is measured from absolute zero, so it is the temperature scale used in pV = nRT.

Look for these in your work

  • Conversion correct
  • Kelvin unit stated
  • Absolute temperature explained
3 · Transfer

A fixed amount of ideal gas is heated from 300 K to 450 K at constant volume. Its initial pressure is 80.0 kPa. Find the final pressure.

Give me a hint

At fixed n and V, pressure is proportional to absolute temperature.

Compare my reasoning
  1. p₂/p₁ = T₂/T₁.
  2. p₂ = 80.0 × 450/300.
  3. Keep the pressure unit unchanged.

p₂ = 120 kPa.

Look for these in your work

  • Temperature ratio used
  • Kelvin temperatures used
  • Pressure unit retained

Common mistakes

  • Using Celsius directly
  • Mixing cm³ and m³
  • Using a pressure unit inconsistent with R
  • Rearranging as n = RT/(pV)
Recall without your notes

What can you explain now?

Which temperature scale is used in pV = nRT?

Compare with the explanation

Kelvin

The equation uses absolute temperature.

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Make it stick

What units make pV and nRT consistent when R is 8.31 J mol⁻¹ K⁻¹?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Chemistry

Course scope and assessment

Four hundred and forty-two original lessons map the published learning outcomes across AS Chemistry Topics 1–22 and A Level Topics 23–37, with guided, independent and transfer practice and explained answers. Every numbered subject-content topic and listed outcome tag now has a mapped lesson sequence. Topic 32 maps all 12 hydroxy-compound outcome tags in 11 lessons; Topic 33 maps all 15 carboxylic-acid and derivative outcome tags in 16 lessons; Topic 34 maps all 22 nitrogen-compound outcome tags in 20 lessons; Topic 35 maps all 12 polymerisation outcome tags in 16 lessons; Topic 36 maps all four organic-synthesis outcome tags in 14 lessons; Topic 37 maps all 22 chromatography and NMR outcome tags in 18 lessons. Cambridge's 2028–2030 Version 1 syllabus states that there are no significant changes which affect teaching from the preceding edition; the routes are aligned to that edition. This is 37 of the 37 numbered subject-content topics. Full cumulative exam assessment, Paper 3 laboratory competence, Paper 5 investigation practice and human subject-teacher review remain outstanding. Written practice is self-checked and does not certify exam readiness.

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