Questions on gases and the ideal gas equation

Question 1

Describe how to find the relative molecular mass of a volatile liquid using a gas syringe. Include how the result is calculated and how the method keeps it accurate.

Accepted Answer

Indicative content

hold a gas syringe in a heated jacket at a steady temperature well above the boiling point of the liquid, and record its starting reading
fill a hypodermic syringe with the liquid and weigh it on a balance reading to three decimal places before injecting
inject a few tenths of a cubic centimetre through the self-sealing cap, then reweigh the hypodermic syringe so the mass injected is found by difference
record the gas syringe reading once it is steady, the jacket temperature in kelvin and the atmospheric pressure in pascals
convert the increase in volume to cubic metres, use n = pV ÷ RT to find the amount of vapour, and divide the mass injected by that amount to give Mr
running well above the boiling point keeps the vapour close to ideal, and weighing rather than reading a volume of liquid avoids needing its density

Levels

Level 3 (5–6 marks) the heated gas syringe, weighing the hypodermic syringe before and after injecting, recording volume, temperature in kelvin and pressure in pascals, the use of pV = nRT and the division of mass by amount are all present, and the answer explains why running well above the boiling point and weighing rather than measuring volume keep the result accurate
Level 2 (3–4 marks) the gas syringe method and pV = nRT are right; the mass is not found by weighing by difference, a unit is left unconverted, or nothing is said about what keeps the result accurate
Level 1 (1–2 marks) the liquid is vaporised in a gas syringe or pV = nRT is quoted, usually with the volume left in cm³ or the temperature in °C

Question 2

Calculate the relative molecular mass of a volatile liquid from the gas syringe results given in the table. R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

M1 V = 62.0 × 10⁻⁶ = 6.20 × 10⁻⁵ m³ and p = 1.01 × 10⁵ Pa M1 n = pV ÷ RT = (1.01 × 10⁵ × 6.20 × 10⁻⁵) ÷ (8.31 × 363) = 2.076 × 10⁻³ mol M1 M = m ÷ n = 0.150 ÷ 2.076 × 10⁻³ A1 Mr = 72.3, to three significant figures, with no unit

Question 3

Calculate the molecular formula of a gaseous hydrocarbon from the combustion volumes given in the table. The hydrocarbon burns completely in excess oxygen, and all volumes are at room temperature and pressure.

Accepted Answer

M1 volume of carbon dioxide = 110 − 30.0 = 80.0 cm³, and the 30.0 cm³ left is unreacted oxygen M1 volume of oxygen used = 150 − 30.0 = 120 cm³ M1 hydrocarbon : O₂ : CO₂ = 20.0 : 120 : 80.0 = 1 : 6 : 4, so x = 4 and x + y/4 = 6, giving y = 8 A1 the molecular formula is C₄H₈

Question 4

Deduce the molar gas volume from a student's results, given in the table, for calcium carbonate reacting with excess ethanoic acid, and suggest why the value is below 24.0 dm³ mol⁻¹. Mr of CaCO₃ is 100.1.

Accepted Answer

M1 amounts of CaCO₃, and so of CO₂, from n = m ÷ 100.1: 5.99 × 10⁻⁴, 1.20 × 10⁻³, 1.80 × 10⁻³ and 2.40 × 10⁻³ mol M1 gradient of volume against amount, for example (55.3 − 13.6) ÷ (2.398 × 10⁻³ − 5.99 × 10⁻⁴) = 2.32 × 10⁴ cm³ mol⁻¹ A1 molar gas volume = 23.2 dm³ mol⁻¹; allow 22.7 to 23.4 dm³ mol⁻¹ from a single result or a line of best fit B1 carbon dioxide is slightly soluble in the aqueous acid, or some escapes before the bung is fitted, so less gas is collected than is made

Question 5

Give each of these values in the unit used in pV = nRT: 350 cm³, 2.50 dm³ and −23 °C.

Accepted Answer

B1 350 cm³ = 350 × 10⁻⁶ m³ = 3.50 × 10⁻⁴ m³ B1 2.50 dm³ = 2.50 × 10⁻³ m³ B1 −23 °C = −23 + 273 = 250 K

Question 6

Calculate the amount, in moles, of gas in a 1.20 dm³ flask at 101 kPa and 20 °C. R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

M1 conversions: V = 1.20 × 10⁻³ m³, p = 1.01 × 10⁵ Pa, T = 293 K M1 n = pV ÷ RT = (1.01 × 10⁵ × 1.20 × 10⁻³) ÷ (8.31 × 293) A1 n = 0.0498 mol, to three significant figures

Question 7

Calculate the pressure, in kPa, of 0.500 mol of gas in a 10.0 dm³ container at 300 K. R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

M1 V = 10.0 × 10⁻³ = 1.00 × 10⁻² m³ M1 p = nRT ÷ V = 0.500 × 8.31 × 300 ÷ (1.00 × 10⁻²) = 1.25 × 10⁵ Pa A1 p = 125 kPa, to three significant figures

Question 8

Calculate the volume of carbon dioxide, in dm³ at room temperature and pressure, formed when 2.50 g of calcium carbonate reacts with excess hydrochloric acid. The molar gas volume at RTP is 24.0 dm³ mol⁻¹; Mr of CaCO₃ is 100.1.

Accepted Answer

M1 n(CaCO₃) = 2.50 ÷ 100.1 = 0.02498 mol M1 one mole of CaCO₃ gives one mole of CO₂, so V = 0.02498 × 24.0 A1 V = 0.599 dm³, to three significant figures

Question 9

Calculate the temperature, in K, at which 0.100 mol of gas occupies 3.00 dm³ at 100 kPa. R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

M1 V = 3.00 × 10⁻³ m³ and p = 1.00 × 10⁵ Pa M1 T = pV ÷ nR = (1.00 × 10⁵ × 3.00 × 10⁻³) ÷ (0.100 × 8.31) A1 T = 361 K, to three significant figures

Question 10

Explain why a student's graph of the volume of a fixed amount of gas at constant pressure against its temperature in °C is a straight line that does not pass through the origin, and why pV = nRT needs the temperature in kelvin.

Accepted Answer

B1 the volume of a gas is proportional to its temperature measured from absolute zero B1 absolute zero is −273 °C, so the line meets the temperature axis at −273 °C, not at 0 °C; allow −273.15 °C B1 the Kelvin scale starts at absolute zero, so only a kelvin temperature is proportional to pV; a Celsius value would give a wrong n, p or V

Question 11

Explain why the gas syringe in the method for finding the Mr of a volatile liquid is kept well above the boiling point of the liquid.

Accepted Answer

B1 all of the injected liquid must vaporise, or the volume measured would be for only part of the mass B1 close to its boiling point a vapour is not ideal, because intermolecular forces pull the molecules together B1 the volume would then be smaller than pV = nRT predicts, and the calculated Mr would be too high

Question 12

Suggest why the mass of liquid injected is found by weighing the hypodermic syringe before and after injection, rather than by reading the volume of liquid on the syringe.

Accepted Answer

B1 the volume injected is a few tenths of a cubic centimetre, too small to read precisely from the syringe scale B1 turning a volume into a mass needs the density, which may be unknown and changes with temperature B1 weighing by difference gives the mass that left the syringe, and liquid left in the needle is not counted

Question 13

Compare finding the amount of a gas from its volume using pV = nRT with using the molar gas volume of 24.0 dm³ mol⁻¹.

Accepted Answer

B1 both give the amount of gas from its volume, and both assume the gas behaves ideally B1 24.0 dm³ mol⁻¹ applies only at room temperature and pressure, whereas pV = nRT works at any temperature and pressure B1 pV = nRT needs every quantity in SI units, whereas the molar volume is used with the volume in dm³

Question 14

Evaluate a student's claim that a gas syringe result of Mr = 81.7 for hexane, whose true Mr is 86.0, is accurate to within 1%.

Accepted Answer

B1 the percentage difference is (86.0 − 81.7) ÷ 86.0 × 100 = 5.0%, so the claim of 1% is not supported B1 a gas syringe read to the nearest 1 cm³ on about 50 cm³ alone carries an uncertainty of about 2%, so 1% is not achievable with this apparatus B1 a low Mr means the amount was overestimated, which points to a systematic error such as a volume reading that is too large, for example from air in the syringe expanding

Question 15

Calculate the volume of oxygen needed to burn 0.500 dm³ of propane completely, and the volume of carbon dioxide formed, all volumes at room temperature and pressure: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).

Accepted Answer

M1 at the same temperature and pressure, gas volumes react in the ratio of the coefficients, 1 : 5 : 3 A1 volume of oxygen = 5 × 0.500 = 2.50 dm³, to three significant figures A1 volume of carbon dioxide = 3 × 0.500 = 1.50 dm³, to three significant figures

Question 16

Explain why ammonia departs further from ideal gas behaviour than helium at the same temperature and pressure.

Accepted Answer

B1 an ideal gas has no forces between its molecules and molecules of negligible volume B1 ammonia molecules are polar and form hydrogen bonds with each other, so the attractions between them are much stronger than the weak induced dipole–dipole forces between helium atoms B1 ammonia molecules are also larger than helium atoms, so their own volume is a bigger share of the gas volume

Question 17

State the ideal gas equation, and state the SI unit of each quantity in it.

Accepted Answer

B1 pV = nRT, where R is 8.31 J K⁻¹ mol⁻¹ B1 p in pascals, V in cubic metres, n in moles and T in kelvin

Question 18

Explain why one mole of hydrogen and one mole of carbon dioxide occupy almost the same volume at the same temperature and pressure.

Accepted Answer

B1 in a gas the molecules are far apart, so their own volume is a tiny fraction of the volume of the gas B1 the volume therefore depends on the number of molecules, the temperature and the pressure, not on the size or mass of the molecules

Question 19

Suggest why the Mr of ethanoic acid measured from its vapour just above its boiling point comes out close to 120 rather than 60.

Accepted Answer

B1 ethanoic acid molecules form hydrogen bonds in pairs, so many exist as dimers in the vapour B1 there are fewer particles than expected, so the calculated amount is smaller and the Mr is larger, up to twice 60

Measurement	Value
Mass of liquid injected	0.150 g
Increase in gas syringe volume	62.0 cm³
Temperature of the jacket	363 K
Pressure	101 kPa

Stage	Volume / cm³
Hydrocarbon	20.0
Oxygen added	150
Gas left after cooling to room temperature	110
Gas left after passing through sodium hydroxide solution	30.0

Mass of CaCO₃ / g	Volume of CO₂ / cm³
0.060	13.6
0.120	27.7
0.180	41.2
0.240	55.3