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Ideal gas equation lab answers for A Level chemistry (free PDF download)

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The method and the results are on the ideal gas equation lab. Do the analysis yourself first, then check the answers below.

Quick answer: The amount of substance, in mol, links mass, particle number, solution volume and gas volume. These answers go with the Chemistry 11-12 lab on ideal gas equation: the graph, the calculations, the uncertainties and the conclusion explained, for all 8 analysis questions.

Ideal gas equation lab answers for A Level chemistry (free PDF download)

Here are the ideal gas equation lab answers for chemistry grades 11-12: graph, calculations, uncertainties and conclusion. First do the lab yourself, then compare, and go back to the lesson if needed.

The model answers use R = 8.31 J K⁻¹ mol⁻¹ and the partial pressure of butane after subtracting the water vapour pressure. Rounding differences in the last figure are acceptable.

Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.1.2

Ideal gas equation lab answers: the analysis explained

Question 1 Pressure of the butane

p = 101.3 − 2.5 = 98.8 kPa = 9.88 × 10⁴ Pa.

Question 2 Amount of butane in trial 2

V = 100 cm³ = 1.00 × 10⁻⁴ m³. n = pV ÷ RT = (9.88 × 10⁴ × 1.00 × 10⁻⁴) ÷ (8.31 × 294) = 9.88 ÷ 2443 = 4.04 × 10⁻³ mol.

Question 3 Molar mass from trial 2

M = 0.24 ÷ 0.00404 = 59.3 g mol⁻¹.

Question 4 Mean molar mass

Trial 1: n = 3.24 × 10⁻³ mol, M = 58.7. Trial 3: n = 4.85 × 10⁻³ mol, M = 57.7. Mean = (58.7 + 59.3 + 57.7) ÷ 3 = 58.6 g mol⁻¹.

Question 5 Accuracy

True M = 4 × 12.0 + 10 × 1.0 = 58.0 g mol⁻¹. Difference = (58.6 − 58.0) ÷ 58.0 × 100 = 1.0 %.

Question 6 Uncertainty in the mass loss

Two readings, each ±0.01 g, give ±0.02 g on the loss. Percentage uncertainty = 0.02 ÷ 0.24 × 100 = 8.3 %.

Question 7 Systematic error

Some butane dissolves, so the measured volume of gas is too small. The amount n is too small, and M = m ÷ n is too large. The calculated molar mass is too high.

Question 8 Improvement

Collect a larger volume and mass (for example 250 cm³ with a bigger mass loss) so that the fixed ±0.02 g uncertainty becomes a smaller percentage, or use a balance reading to 0.001 g.

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