Mini Chemistry
Part of Mini EducationSwitch subject

Ideal Gas Model and pV = nRT

Learn and apply Ideal Gas Model and pV = nRT in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
On this page

Ideal Gas Model and pV = nRT: Orientation

Most gas calculations are method-mark questions. If you do the same 3 checks every time (K, units for R, volume conversion), pV = nRT becomes almost automatic.

Definitions (Must Know)

A. Ideal gas model

The ideal gas model assumes:

  • gas particles have negligible volume
  • there are no intermolecular forces between particles except during collisions
  • particles are in constant random motion
  • collisions between particles and with the container walls are perfectly elastic
  • the average kinetic energy of the particles is proportional to absolute temperature
Ideal gas particle model assumptionsContainer diagram with small particles and random motion arrows beside the five kinetic-theory assumptions for an ideal gas.Wall collisions produce gas pressure and are elastic.Five assumptions to quote1) Particle volume is negligible.2) No forces act except during collisions.3) Particles move constantly and randomly.4) All collisions are perfectly elastic.5) Average KE ∝ absolute temperature.Particle size and path lengths are not to scale.
The model links random particle motion and elastic wall collisions to five assumptions. Low pressure is a condition that makes the negligible-volume and no-force assumptions more realistic; it is not an extra assumption.

B. Ideal gas equation

For an ideal gas: pV = nRT

  • p: pressure
  • V: volume
  • n: amount of gas (mol)
  • T: absolute temperature (K)
  • R: molar gas constant (given in the Data Booklet)

C. Absolute temperature (K)

Convert Celsius to kelvin. Use the syllabus convention + 273 unless more precision is requested: T(K) = T(°C) + 273

Detailed Explanations

A. Unit discipline (do this before substituting)

  • 1 dm³ = 1.00 × 10⁻³ m³
  • 1 cm³ = 1.00 × 10⁻⁶ m³ (so 1000 cm³ = 1 dm³)
  • 1 kPa = 1000 Pa

Useful shortcut: if you use p in kPa and V in dm³, then pV is numerically in J (because kPa·dm³ = Pa·m³), so using R = 8.31 remains consistent.

The 8.31 Shortcut (Why kPa·dm³ Works)

1 kPa · 1 dm³ = (1000 Pa)(1.00 × 10⁻³ m³) = 1 Pa m³ = 1 J.

So using p in kPa and V in dm³ still matches R = 8.31.

B. A repeatable workflow (method marks)

  1. Write pV = nRT and rearrange for the unknown.
  2. Convert units (especially T, p, V).
  3. Substitute with units shown.
  4. Round at the end and check the magnitude.

Mini example of Step 2:

  • 27°C → 300 K
  • 250 cm³ → 0.250 dm³

C. Determining relative molecular mass

For a gas sample of mass m:

n = pV/RT and M = m/n = mRT/pV

When m is in grams, M is obtained in g mol⁻¹. Report the same numerical value as Mᵣ, which has no unit.

Worked Examples

Modelled example 1

Calculate amount in SI units

Core

Problem

A gas has pressure 1.00 × 10⁵ Pa and volume 2.40 × 10⁻² m³ at 300 K. Calculate n.
Study the worked solution

  1. Check units

    Method

    Keep Pa, m³ and K with R = 8.31 J mol⁻¹K⁻¹.

    Reason

    1 J = 1 Pa m³, so the units are consistent.

    Working

    n = pV/(RT).

  2. Substitute

    Method

    Evaluate without changing the given SI units.

    Reason

    The rearrangement makes amount the subject.

    Working

    n = ((1.00 × 10⁵)(2.40 × 10⁻²))/(8.31)(300) = 0.962 mol.

Guided practice 2

Convert laboratory gas data

About 6 min

Problem

A gas occupies 250 cm³ at 27°C and 100 kPa. Calculate n.

Convert then calculate

Hints

Hint 1: temperature
Use T = 27 + 273.
Hint 2: volume
With kPa, convert 250 cm³ to 0.250 dm³.
View solution step by step

  1. Convert

    Method

    Use 300 K and 0.250 dm³.

    Reason

    Absolute temperature is required and kPa pairs with dm³ for the stated numerical R.

    Working

    T = 300 K; V = 0.250 dm³.

  2. Calculate

    Method

    Substitute into n = pV/(RT).

    Reason

    All quantities now use a consistent set.

    Working

    n = (100)(0.250)/(8.31)(300) = 1.00 × 10⁻² mol.

Common misconception 3

Do not substitute Celsius

Find and correct the mistake

Learner attempt

A learner substitutes T = 25 for a gas at 25°C in pV = nRT. Locate and correct the first error.

Diagnose

Required scale

View solution step by step

  1. Identify the model requirement

    Method

    Replace Celsius by absolute temperature.

    Reason

    Kinetic energy is proportional to thermodynamic temperature, not the Celsius reading.

    Working

    T = 25 + 273 = 298 K.

  2. Correct the substitution

    Method

    Use 298 in the denominator or numerator as required.

    Reason

    Using 25 would make the calculated gas quantity physically invalid.

    Working

    Use pV = nR(298).

Examiner practice 4

Determine relative molecular mass

4 marks

Examination question

A 0.352 g gas sample occupies 125 cm³ at 298 K and 100 kPa. Determine Mᵣ. [4 marks]

Show all working

View solution step by step

  1. Convert volume

    1 mark

    Method

    Convert to 0.125 dm³.

    Reason

    This matches pressure in kPa and the laboratory-unit form of R.

    Working

    V = 0.125 dm³.

  2. Use gas data

    2 marks

    Method

    Apply M = mRT/(pV).

    Reason

    Combining n = pV/RT with M = m/n gives the required relationship.

    Working

    M = (0.352)(8.31)(298)/(100)(0.125) = 69.7 g mol⁻¹.

  3. State relative mass

    1 mark

    Method

    Report the numerical relative molecular mass.

    Reason

    Mᵣ is dimensionless although its value matches molar mass in g mol⁻¹.

    Working

    Mᵣ = 69.7.

Challenge 5

Find gas volume

Minimal support

Problem

Calculate the volume occupied by 0.0500 mol of an ideal gas at 310 K and 95.0 kPa. Give the answer in dm³.

Rearrange independently

Hints

Hint 1: subject
Rearrange to V = nRT/p.
Hint 2: unit-set
kPa with R = 8.31 gives dm³.
View solution step by step

  1. Select the unit route

    Method

    Use kPa, K and R = 8.31 kPa dm³ mol⁻¹K⁻¹.

    Reason

    The requested volume then emerges directly in dm³.

    Working

    V = nRT/p.

  2. Evaluate

    Method

    Substitute the new known and unknown arrangement.

    Reason

    Volume, rather than amount, is now the subject.

    Working

    V = (0.0500)(8.31)(310)/95.0 = 1.36 dm³.

Mind Stretchers

Mind stretcher 1Extension

An impure sample of magnesium has mass 0.250 g. It reacts with excess hydrochloric acid and produces 240 cm³ of H₂ at 298 K and 100 kPa in a gas syringe (dry gas). Calculate the percentage of Mg in the sample.

Show Hint

Find dry hydrogen moles first, then use the equation ratio to recover the magnesium mass before calculating a percentage.

Show Answer

Mark scheme:

  • Equation: Mg + 2HCl → MgCl₂ + H₂
  • V = 240 cm³ = 0.240 dm³.
  • n(H₂) = pV/RT = (100 × 0.240)/(8.31 × 298) = 9.69 × 10⁻³ mol.
  • From the equation, n(Mg) = n(H₂) = 9.69 × 10⁻³ mol.
  • Mass of Mg = nAᵣ = (9.69 × 10⁻³)(24.3) = 0.235 g.
  • Percentage of Mg = (0.235/0.250) × 100 = 94.0%.

Mind stretcher 2Extension

A gas has density 1.25 g dm⁻³ at 298 K and 100 kPa. Calculate Mᵣ.

Show Hint

Choose a convenient volume. Density gives its mass, while the gas equation gives its amount.

Show Answer

Mark scheme:

  • Density ρ = m/V, so m = ρ V.
  • Substitute into M = mRT/pV: M = (ρ RT)/p.
  • M = (1.25)(8.31)(298)/100 = 30.9 g mol⁻¹.
  • Mᵣ ≈ 31.0 (3 s.f.).