Topic 1 · page 6 of 10

Gas volumes

At the same temperature and pressure, a mole of any gas takes up the same volume. Use 24.0 dm³ at room conditions, and pV = nRT in SI units for anything else.

15 min read1.8(i)1.8(ii)1.8(iii)

Read first: Masses from equations, Solutions and parts per million

Equal volumes, equal numbers 1.8(i)

Gas molecules are tiny compared with the spaces between them, so the volume of a gas depends on how many molecules there are, not on what they are. This gives Avogadro's law:

So for gases, the volume ratio is the mole ratio. You can read volumes straight from the equation without going through masses at all.

Worked exampleBurning methane

CHX4(g)+2 OX2(g)→COX2(g)+2 HX2O(l) \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}

20 cm³ of methane needs 2 × 20 = 40 cm³ of oxygen and makes 20 cm³ of carbon dioxide, all measured at the same temperature and pressure.

Molar volume 1.8(ii)

n=V24.0V=n×24.0(V in dm3, at RTP) n = \frac{V}{24.0} \qquad V = n \times 24.0 \qquad (V \text{ in dm}^3,\ \text{at RTP})

Worked exampleCarbon dioxide from limestone

What volume of CO₂, measured at RTP, forms when 5.00 g of calcium carbonate reacts with excess hydrochloric acid?

CaCOX3(s)+2 HCl(aq)→CaClX2(aq)+HX2O(l)+COX2(g) \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)}
  1. n(CaCO₃) = 5.00 ÷ 100.1 = 0.04995 mol
  2. Ratio 1 : 1, so n(CO₂) = 0.04995 mol
  3. V = 0.04995 × 24.0 = 1.20 dm³ (1200 cm³)

Worked exampleGas left over after a reaction

20 cm³ of methane is burned in 100 cm³ of oxygen. What is the total volume of gas afterwards, at room temperature and pressure?

  1. Methane needs 40 cm³ of oxygen. There is 100 cm³, so oxygen is in excess and 60 cm³ is left over.
  2. 20 cm³ of CO₂ forms. The water condenses to a liquid.
  3. Total gas: 60 + 20 = 80 cm³

The ideal gas equation 1.8(iii)

The molar volume only works at the conditions it is quoted for. For any other temperature or pressure, use the ideal gas equation:

pV=nRT pV = nRT

Every quantity must be in SI units. This is where most marks are lost.

Symbol Quantity SI unit Converting
p pressure Pa kPa × 10³ = Pa
V volume m³ cm³ × 10⁻⁶ = m³, dm³ × 10⁻³ = m³
n amount mol
R gas constant 8.31 J mol⁻¹ K⁻¹ given on the data sheet
T temperature K °C + 273 = K

Worked exampleVolume of a gas at a stated temperature and pressure

Find the volume of 0.500 mol of gas at 100 kPa and 298 K.

  1. p = 100 kPa = 100 000 Pa
  2. V = nRT ÷ p = (0.500 × 8.31 × 298) ÷ 100 000 = 0.0124 m³
  3. 0.0124 m³ × 1000 = 12.4 dm³

Molar mass of a volatile liquid

A volatile liquid evaporates easily. If you vaporise a weighed sample and measure the volume of vapour, pV = nRT gives the amount, and the molar mass follows:

M=mn=mRTpV M = \frac{m}{n} = \frac{mRT}{pV}

Worked exampleIdentifying a volatile liquid

0.150 g of a liquid is vaporised at 100 °C and 101 kPa. The vapour occupies 79.4 cm³. Find the molar mass.

  1. Convert: T = 100 + 273 = 373 K. p = 101 000 Pa. V = 79.4 × 10⁻⁶ m³.
  2. n = pV ÷ RT = (101 000 × 79.4 × 10⁻⁶) ÷ (8.31 × 373) = 2.59 × 10⁻³ mol
  3. M = 0.150 ÷ 2.59 × 10⁻³ = 58.0 g mol⁻¹, which matches propanone, C₃H₆O.

Check yourself

  1. What volume of oxygen is needed to burn 50 cm³ of propane completely, and what volume of CO₂ forms? (C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, all gas volumes at RTP)
  2. What volume of hydrogen at RTP forms when 0.486 g of magnesium reacts with excess hydrochloric acid?
  3. What mass of calcium carbonate reacts with excess acid to make 2.40 dm³ of CO₂ at RTP?
  4. At what temperature, in K and in °C, does 1.00 mol of gas occupy 25.0 dm³ at 101 kPa?
  5. 0.200 g of a volatile liquid gives 61.0 cm³ of vapour at 373 K and 101 kPa. Calculate its molar mass.
  6. A student uses p = 101 and V = 61.0 in question 5 without converting. By what factor is the calculated n wrong?

Answers cover these until you have tried the questions

  1. Oxygen: 5 × 50 = 250 cm³. CO₂: 3 × 50 = 150 cm³.
  2. n(Mg) = 0.486 ÷ 24.3 = 0.0200 mol. Mg + 2HCl → MgCl₂ + H₂, ratio 1 : 1, so n(H₂) = 0.0200 mol. V = 0.0200 × 24.0 = 0.480 dm³ (480 cm³).
  3. n(CO₂) = 2.40 ÷ 24.0 = 0.100 mol = n(CaCO₃). m = 0.100 × 100.1 = 10.0 g
  4. T = pV ÷ nR = (101 000 × 0.0250) ÷ (1.00 × 8.31) = 304 K, which is 304 − 273 = 31 °C.
  5. n = (101 000 × 61.0 × 10⁻⁶) ÷ (8.31 × 373) = 1.99 × 10⁻³ mol. M = 0.200 ÷ 1.99 × 10⁻³ = 101 g mol⁻¹
  6. Pressure is 1000 times too small (×10⁻³) and volume is 10⁶ times too large. Together, n is 10³ = 1000 times too large.