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A Level · Cambridge (CIE) · 9702

Use the ideal-gas equation of state

Recognise the ideal-gas proportionality, use both equation-of-state forms and relate R to the Boltzmann constant.

Section 15 · Ideal gases

Ideal gases · Cambridge International AS & A Level Physics 9702 · 15.2

A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 15. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.

Before you begin

  • Use pressure, volume, temperature in kelvins, energy and scientific notation.

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

Quick readiness check

For a fixed amount of ideal gas, which temperature belongs in pV ∝ T?

What you will learn

  • Recognise an ideal gas from pV ∝ T for fixed amount of gas
  • Use pV = nRT and pV = NkT
  • Use k = R/NA

An ideal gas has a simple equation of state

For a fixed amount of gas, an ideal gas obeys pV ∝ T, where T is thermodynamic temperature in kelvins. The proportionality fails if the amount changes, so keep n fixed when using this form as a relationship between states.

The equation of state is pV = nRT, where p is pressure in pascals, V is volume in cubic metres, n is amount in moles, R is the molar gas constant, and T is in kelvins.

Count particles or use moles consistently

The same equation can be written pV = NkT, with N the number of molecules and k the Boltzmann constant. Use either n with R or N with k; do not mix the two forms.

The constants are related by k = R/NA. With R = 8.31 J K⁻¹ mol⁻¹ and NA = 6.02 × 10²³ mol⁻¹, k is about 1.38 × 10⁻²³ J K⁻¹.

pV = nRT = NkT · k = R/NA

Pause and explain

Which equation matches particle count N?

Put the idea to work

Worked example

A gas at 300 K has pressure 1.00 × 10⁵ Pa and volume 0.0249 m³. Estimate the amount of gas.

Show the worked solution
  1. Use n = pV/(RT).
  2. Substitute the SI values and R = 8.31 J K⁻¹ mol⁻¹.
  3. Round to a sensible number of significant figures.

Answer n = (1.00 × 10⁵ × 0.0249)/(8.31 × 300) ≈ 1.00 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

A gas has p = 1.00 × 10⁵ Pa, V = 8.31 × 10⁻³ m³ and T = 300 K. Find n using R = 8.31 J K⁻¹ mol⁻¹.

Give me a hint

Use n = pV/(RT) and retain SI units.

Compare my reasoning
  1. n = pV/(RT).
  2. pV = 831 J and RT = 2493 J mol⁻¹.
  3. Divide and state moles.

n ≈ 0.333 mol.

Look for these in your work

  • Equation rearranged
  • pV and RT evaluated
  • Amount unit stated
2 · Independent

Estimate k from R = 8.31 J K⁻¹ mol⁻¹ and NA = 6.02 × 10²³ mol⁻¹.

Give me a hint

Use k = R/NA; the mole units cancel.

Compare my reasoning
  1. k = 8.31/(6.02 × 10²³).
  2. Evaluate the power of ten.
  3. Give J K⁻¹.

k ≈ 1.38 × 10⁻²³ J K⁻¹.

Look for these in your work

  • Constant relation used
  • Mole units cancel
  • Boltzmann unit stated
3 · Transfer

A fixed sample of ideal gas is heated from 300 K to 450 K at constant volume. Compare its final and initial pressures.

Give me a hint

At fixed volume and fixed amount, p is proportional to absolute temperature.

Compare my reasoning
  1. p₂/p₁ = T₂/T₁.
  2. p₂/p₁ = 450/300 = 1.5.
  3. The final pressure is 1.5 times the initial pressure.

The pressure increases by a factor of 1.5.

Look for these in your work

  • Fixed state variables identified
  • Kelvin ratio used
  • Pressure ratio interpreted

Common mistakes

  • Using Celsius temperature in the gas equation
  • Changing the amount while applying pV ∝ T
  • Combining N with R or n with k
Recall without your notes

What can you explain now?

For a fixed amount of ideal gas, which temperature belongs in pV ∝ T?

Compare with the explanation

Thermodynamic temperature in kelvins

The proportionality uses absolute temperature.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

When should you use pV = nRT, and how is it related to pV = NkT?

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 Physics

Course scope and assessment

One hundred and sixty-two original lessons cover all 12 published subject-content outcomes in Topic 1 (sections 1.1–1.4), all nine outcomes in Topic 2 (section 2.1), all 13 outcomes in Topic 3 (sections 3.1–3.3), all 13 outcomes in Topic 4 (sections 4.1–4.3), all 11 outcomes in Topic 5 (sections 5.1–5.2), all 10 outcomes in Topic 6 (sections 6.1–6.2), all 16 outcomes in Topic 7 (sections 7.1–7.5), all 12 outcomes in Topic 8 (sections 8.1–8.4), all 15 outcomes in Topic 9 (sections 9.1–9.3), all 16 outcomes in Topic 10 (sections 10.1–10.3), all 18 outcomes in Topic 11 (sections 11.1–11.2), all seven outcomes in Topic 12 (sections 12.1–12.2), all 12 outcomes in Topic 13 (sections 13.1–13.4), all eight outcomes in Topic 14 (sections 14.1–14.3), and all nine outcomes in Topic 15 (sections 15.1–15.3), plus all four outcomes in Topic 16 (sections 16.1–16.2), and all ten outcomes in Topic 17 (sections 17.1–17.3), plus all 12 outcomes in Topic 18 (sections 18.1–18.5), all nine outcomes in Topic 19 (sections 19.1–19.3), all 19 outcomes in Topic 20 (sections 20.1–20.5), and all eight outcomes in Topic 21 (sections 21.1–21.2), and all 17 outcomes in Topic 22 (sections 22.1–22.4), and all 13 outcomes in Topic 23 (sections 23.1–23.2), and all 16 outcomes in Topic 24 (sections 24.1–24.3), and all 11 outcomes in Topic 25 (sections 25.1–25.3), of the 2025–2027 syllabus. Each lesson has guided, independent and application practice with explained answers. All 25 syllabus topics now have mapped teaching and practice; the 2028–2030 changes outside Topics 23–25 still need reconciliation. Online study cannot demonstrate supervised Paper 3 laboratory competence. Paper 5 planning and analysis, full timed cumulative assessment and human subject-teacher review remain outstanding.

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