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Gas Calculations (pV = nRT)

Learn and apply Gas Calculations (pV = nRT) in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Gas Calculations with pV = nRT: Orientation

In stoichiometry, pV = nRT is just a tool to get moles of a gas. Once you have moles, the rest is the usual chain: balanced equation → mole ratio → convert to what the question asks.

Definitions (Must Know)

A. Ideal gas equation

The ideal gas equation is:

pV = nRT

B. Molar gas constant, R

The molar gas constant, R, is given in the Data Booklet (commonly 8.31 J mol⁻¹K⁻¹).

C. Absolute temperature (K)

Temperature must be in kelvin: T(K) = T(°C) + 273.15

Detailed Explanations

A. Choosing R (unit match)

Two common consistent sets (same “8.31” value):

Use this RIf you use these units
R = 8.31 J mol⁻¹ K⁻¹p in Pa, V in m³, T in K
R = 8.31 kPa dm³ mol⁻¹ K⁻¹p in kPa, V in dm³, T in K

If you want the full unit-logic explanation, see Ideal Gas Model and pV = nRT. For mixed gases or “collected over water” setups, route through Dalton’s Law and Partial Pressures before substitution.

B. Workflow (gas stoichiometry)

  1. Convert T to K.
  2. Convert p and V to match your R.
  3. Solve for n: n = pV/RT.
  4. If a reaction is involved, use the balanced equation to convert moles of one species to another.
  5. Convert to what the question asks (mass / volume / concentration).

Mini example:

  • At 298 K, p = 100 kPa, V = 2.00 dm³: n = pV/RT = (100 × 2.00)/(8.31 × 298) = 0.0807 mol

C. Finding molar mass, M, of a gas using pV = nRT

  1. Use pV = nRT to calculate moles: n = pV/RT.
  2. Use M = m/n, so: M = mRT/pV

Worked Examples

Modelled example 1

Calculate Amount from Gas Conditions

Core

Problem

A gas occupies 2.00 dm³ at 298 K and 100 kPa. Find the number of moles. Use R = 8.31 kPa dm³ mol⁻¹ K⁻¹.

Study the worked solution

  1. Check the unit system

    Method

    Keep pressure in kilopascals, volume in cubic decimetres and temperature in kelvin.

    Reason

    Those units match the stated value and units of R.

    Working

    p = 100 kPa, V = 2.00 dm³ and T = 298 K

  2. Rearrange for amount

    Method

    Make n the subject of pV = nRT.

    Reason

    Amount is the requested unknown.

    Working

    n = pV/RT

  3. Substitute and calculate

    Working

    n = (100)(2.00)/(8.31)(298) = 0.0807 mol

Guided practice 2

Determine Molar Mass from Gas Data

About 7 min

Problem

A gas sample has mass 0.440 g and occupies 248 cm³ at 298 K and 100 kPa. Find its molar mass. Use R = 8.31 kPa dm³ mol⁻¹ K⁻¹.

Try this before viewing the solution

Hints

Hint 1: match the volume unit
Convert the gas volume from cubic centimetres to cubic decimetres before using the stated R.
Hint 2: connect amount to mass
First calculate n = pV/(RT), then use M = m/n.
View solution step by step

  1. Convert the volume

    Method

    Divide the cubic-centimetre value by 1000.

    Reason

    The stated R uses cubic decimetres.

    Working

    248 cm³ = 0.248 dm³

  2. Calculate the gas amount

    Reason

    The gas equation gives the amount before mass can be converted to molar mass.

    Working

    n = pV/RT = (100)(0.248)/(8.31)(298) = 0.0100 mol

  3. Calculate molar mass

    Method

    Divide the sample mass by its amount.

    Reason

    Molar mass is mass per mole.

    Working

    M = m/n = 0.440/0.0100 = 44.0 g mol⁻¹

Common misconception 3

Correct an Incorrect Rearrangement

Find and correct the mistake

Learner attempt

0.500 mol of a gas is at 298 K and 100 kPa. A learner tries to find its volume using V = np/(RT). Identify the first error and calculate the correct volume.

Choose the correct rearrangement

Correct expression for V

View solution step by step

  1. Isolate volume

    Method

    Divide both sides of pV = nRT by p.

    Reason

    The learner has moved R and T to the wrong side of the expression.

    Working

    V = nRT/p

  2. Calculate the volume

    Working

    V = (0.500)(8.31)(298)/100 = 12.4 dm³

Examiner practice 4

Calculate Carbon Dioxide Volume

4 marks

Problem

1.00 g of calcium carbonate reacts with excess hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) Calculate the volume of CO₂ produced at 298 K and 100 kPa. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate calcium carbonate amount

    1 mark

    Method

    Divide mass by molar mass.

    Reason

    The equation ratio applies to amounts in moles, not directly to masses.

    Working

    n(CaCO₃) = 1.00/100 = 0.0100 mol

  2. Use the equation ratio

    1 mark

    Reason

    The equation has a 1:1 ratio between CaCO₃ and CO₂.

    Working

    n(CO₂) = 0.0100 mol

  3. Apply the ideal gas equation

    1 mark

    Method

    Rearrange to V = nRT/p and use the given conditions.

    Reason

    The carbon dioxide amount, temperature and pressure are now known.

    Working

    V = (0.0100)(8.31)(298)/100 = 0.248 dm³

  4. State a suitable final volume

    1 mark

    Working

    V = 0.248 dm³ = 248 cm³

Challenge 5

Transfer a Gas Sample to New Conditions

Minimal support

Problem

The 2.00 dm³ gas sample from Example 1 is sealed, heated from 298 K to 350 K and compressed to 1.50 dm³. No gas escapes. Calculate its new pressure. Use R = 8.31 kPa dm³ mol⁻¹ K⁻¹.

Try this before viewing the solution

Hints

Hint 1: identify what stays fixed
Because the sample is sealed, its amount remains the 0.0807 mol found in Example 1.
Hint 2: use the new state
Apply p = nRT/V using 350 K and 1.50 dm³.
View solution step by step

  1. Carry the amount into the new state

    Method

    Use the amount already calculated for the sealed sample.

    Reason

    Heating and compression change its conditions, not the number of gas particles.

    Working

    n = 0.0807 mol

  2. Calculate the new pressure

    Method

    Rearrange the ideal gas equation to p = nRT/V.

    Working

    p = (0.0807)(8.31)(350)/1.50 = 157 kPa

Mind Stretchers

Mind stretcher 1Extension

0.500 g of CaCO₃ reacts with 25.0 cm³ of 0.300 mol dm⁻³ HCl: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) (a) Identify the limiting reagent. (b) Calculate the volume of CO₂ formed at 298 K and 100 kPa.

Show Hint

Choose one consistent pressure-volume unit system, convert temperature to kelvin, and rearrange only after writing the equation.

Show Answer

Mark scheme:

  • n(CaCO₃) = 0.500/100 = 0.00500 mol
  • n(HCl) = cV = 0.300 × (25.0/1000) = 0.00750 mol
  • Need 2 mol HCl per 1 mol CaCO₃, so 0.00500 mol CaCO₃ would need 0.0100 mol HCl → HCl is limiting.
  • From the equation: 2 mol HCl gives 1 mol CO₂, so n(CO₂) = 0.00750/2 = 0.00375 mol.
  • V = nRT/p = (0.00375 × 8.31 × 298)/100 = 0.0928 dm³ (= 92.8 cm³).

Mind stretcher 2: Using gas density to determine molar massExtension

Question. A gas has density 1.84 g dm⁻³ at 100 kPa and 298 K. Determine its molar mass using R = 8.31 kPa dm³ mol⁻¹ K⁻¹.

Show Hint

Take a 1.00 dm³ sample, so its mass is 1.84 g.

Show Answer

For 1.00 dm³, n = pV/RT = 100(1.00)/(8.31 × 298) = 0.0404 mol. Hence M = m/n = 1.84/0.0404 = 45.5 g mol⁻¹.