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The molar volume of hydrogen and the gas constant

Weigh a strip of magnesium, let it react with hydrochloric acid inside a gas burette standing mouth-down in water, and collect the hydrogen over water. Level the water, subtract its vapour pressure, and work out the gas constant R and the molar volume of hydrogen at 0 °C and 100 kPa from your reading — then see how much each forgotten correction costs.

Difficulty ●○○ ≈ 15 min

Aim Measure the gas constant R and the molar volume from the hydrogen a reaction gives, and show what Dalton’s law and levelling the water contribute.
Mg + 2HCl → MgCl₂ + H₂ · collected over water · 50 mL gas burette

Magnesium and acid: the molar volume of hydrogen

Ribbon not yet inserted

The gas burette holds 10 mL of 3 M HCl topped up with water and stands mouth-down in a tall cylinder of water; the magnesium ribbon, caged in copper wire, hangs near the mouth. The denser acid sinks to the ribbon.

01020304050
Magnesium ribbon 0.0371g
Gas reading 0.00mL
Water temperature 17.6°C
Barometer 101.11kPa
Inside level above outside: — cm
Subtract in the calculation

When the reaction is over, R and the molar volume are worked out here from your reading.

Computed: the hydrogen equals the magnesium in moles (all of it reacts). The gas in the tube is hydrogen saturated with water vapour at the water’s temperature; its pressure is the barometric pressure less the water column ρgh between the levels (ρ = 998.2 kg/m³, g = 9.80665 m/s²), and hydrogen’s partial pressure is that less the vapour pressure of water (IAPWS-IF97). The tube’s cross-section is 0.80 cm² (1.25 cm per mL); as you raise or lower it, (p − ρgh)V = nRT is solved for the gas volume. Hydrogen is treated as ideal (Z − 1 ≈ 6 × 10⁻⁴ at 1 bar and 300 K). Readings are rounded to 0.05 mL. The ribbon’s mass, the water temperature and the pressure are set by the run number. Not modelled: hydrogen dissolving in the water, magnesium already oxidised on the ribbon, and the small change of water’s density with temperature; the bubbles and the pace of the reaction are illustrative.

Procedure

  1. Choose a run number and note the ribbon’s mass (analytical balance, ±0.1 mg), the water temperature and the barometer.
  2. Insert the ribbon: the acid sinks onto it, hydrogen bubbles rise to the top of the tube and the water level falls. Wait until the ribbon has dissolved and the bubbling stops.
  3. Raise or lower the tube until the level inside matches the water outside, then read the gas volume at the bottom of the meniscus, to 0.05 mL.
  4. Calculate: p(H₂) = p(barometer) − p°(H₂O, T); n(H₂) = n(Mg); R = pV/(nT); then convert the volume to 0 °C and 100 kPa for the molar volume.
  5. Untick “the vapour pressure of water”, then move the tube without correcting for the levels, and watch how far R drifts.

What to notice

  • Raising the tube lifts the water inside, lowers the gas pressure and lets the gas expand; lowering it does the opposite — Boyle’s law at work. Only when the levels match is the gas at the barometric pressure.
  • At 22 °C water’s vapour pressure is about 2.6 kPa: forget to subtract it and R comes out about 2.7 % high (8.54 instead of 8.31).
  • A 15 cm difference between the levels is about 1.5 kPa: left uncorrected it shifts R by about 1.5 % more. With both corrections R agrees with 8.314 to within the rounding of the reading.

In a real lab

  • Hydrochloric acid is corrosive — wear goggles; hydrogen is flammable, so keep flames away.
  • The oxide film on the ribbon dissolves without giving hydrogen and makes the result low: polish the ribbon with sandpaper or steel wool first.
  • A little hydrogen dissolves in the water, and gas warmer than the water reads high: after the reaction, wait a few minutes for the gas to reach the water’s temperature before reading.
  • If the levels cannot be matched, measure the height difference h and subtract ρgh from the barometric pressure (about 98 Pa per centimetre of water).

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