Gas volumes and the ideal gas equationEdexcel International A Level Chemistry: Revision notes
Section 1
Molar volume and gas volumes from equations
The molar volume is the volume occupied by one mole of any gas at a stated temperature and pressure. At room temperature and pressure (rtp, 20 °C and 1 atm) it is 24.0 dm3 mol-1. Volume (dm3) = amount (mol) x 24.0.
To use an equation for gas volumes: (1) balance the equation; (2) find the amount of the known substance; (3) use the ratio to find the amount of the gas; (4) multiply by 24.0 dm3 mol-1. Example: 0.0500 mol of propane burns in C3H8 + 5O2 → 3CO2 + 4H2O, needing 0.250 mol O2, or 6.00 dm3. Gas volumes are in the same ratio as the equation coefficients at the same temperature and pressure.
Water is a liquid at rtp, so it is not included when working out gas volumes.
Section 2
The ideal gas equation
The ideal gas equation is pV = nRT, where p is the pressure in Pa, V is the volume in m3, n is the amount in mol, R is the gas constant, 8.31 J K-1 mol-1, and T is the temperature in K. It applies to gases and the vapours of volatile liquids.
Unit conversions are the main trap: 1 dm3 = 10^-3 m3; 1 cm3 = 10^-6 m3; 1 kPa = 10^3 Pa; K = °C + 273. Rearrangements: n = pV / RT, V = nRT / p, p = nRT / V, T = pV / nR.
Do not forget to convert cm3 and dm3 to m3, and °C to K. Most lost marks are unit slips, not algebra.
Section 3
Worked example: pressure of a gas
0.250 mol of methane is in a 2.50 dm3 cylinder at 300 K. V = 2.50 x 10^-3 m3. p = nRT / V = (0.250 x 8.31 x 300) / (2.50 x 10^-3) = 2.49 x 10^5 Pa. At constant n and V, pressure is proportional to the kelvin temperature, so cooling to 280 K gives p = 2.49 x 10^5 x 280 / 300 = 2.33 x 10^5 Pa.
At constant n and T, pressure is inversely proportional to volume (halving the volume doubles the pressure), and at constant n and p, volume is proportional to T.
Section 4
Finding molar mass from gas data
Volatile liquids can be vaporised and treated as ideal gases. Measure the mass m of the sample, the volume, the temperature and the pressure. Find n = pV / RT, then M = m / n.
Example: 0.205 g of a vapour occupies 85.0 cm3 at 373 K and 101 kPa. n = (101 000 x 85.0 x 10^-6) / (8.31 x 373) = 2.77 x 10^-3 mol, so M = 0.205 / 2.77 x 10^-3 = 74.0 g mol-1. Compare with candidates to deduce the compound.
Section 5
Core Practical 1: molar volume of a gas
A standard method reacts a known mass of magnesium ribbon with an excess of dilute hydrochloric acid and collects the hydrogen in a gas syringe: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g). (1) Weigh the magnesium accurately; (2) measure the acid into a conical flask; (3) connect the gas syringe; (4) add the magnesium and stopper the flask quickly; (5) record the final volume when the syringe stops moving, plus the room temperature and pressure.
The amount of hydrogen equals the amount of magnesium. Molar volume = volume of hydrogen / amount of magnesium. Typical results are a little low because of leaks, gas escaping before the bung is fitted, or an oxide layer on the magnesium. Using excess acid ensures that all the magnesium reacts.
Must Know
- Molar volume is 24.0 dm3 mol-1 at rtp; volume = amount x 24.0
- pV = nRT: p in Pa, V in m3, T in K, R = 8.31
- Convert cm3 and dm3 to m3 (x 10^-6 and x 10^-3), kPa to Pa, °C to K
- M = m / n, with n from pV / RT, identifies a volatile liquid
- Core Practical 1: use excess acid, weigh the metal, record the volume, T and p
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Gas volumes and the ideal gas equation
- A student reacts 0.120 g of magnesium ribbon with an excess of dilute hydrochloric acid and collects the hydrogen in a gas syringe at room temperature and pressure (rtp). The molar volume of a gas at rtp is 24.0 dm³ mol⁻¹. The relative atomic mass of magnesium is 24.3.The student collects only 108 cm³ of hydrogen. Suggest two reasons why the measured volume is less than the calculated volume.2 marks
- A gas cylinder of fixed volume 2.50 dm³ contains 4.00 g of methane, CH₄ (molar mass 16.0 g mol⁻¹), at 300 K. Take R = 8.31 J K⁻¹ mol⁻¹ and the molar volume at rtp as 24.0 dm³ mol⁻¹.Calculate the volume the methane would occupy at rtp and use this to state whether the gas in the cylinder is at a higher or lower pressure than atmospheric pressure.2 marks
- A volatile liquid X is one of propanone (C₃H₆O), diethyl ether (C₄H₁₀O) or hexane (C₆H₁₄). A sample of mass 0.205 g is vaporised completely in a gas syringe and occupies 85.0 cm³ at 373 K and a pressure of 101 kPa. Take R = 8.31 J K⁻¹ mol⁻¹. Relative atomic masses: H = 1.0, C = 12.0, O = 16.0.Calculate the amount, in mol, of X in the gas syringe.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).