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Avogadro’s law and the ideal gas law
Gay-Lussac's law of combining volumes, Avogadro's law, molar volume at 25 °C and 0 °C and 100 kPa, the ideal gas law PV = nRT, molar mass and density of gases, gas stoichiometry, when real gases are not ideal, and finding the molar mass of butane and magnesium from gas volumes.
A litre of hydrogen and a litre of carbon dioxide look identical: colourless, invisible, filling their containers. One weighs 22 times as much as the other, yet at the same temperature and pressure they contain exactly the same number of molecules. That surprising fact, first suggested by Amedeo Avogadro in 1811, means that for gases, measuring a volume is as good as counting particles.
Gay-Lussac's law of combining volumes
In 1808 Gay-Lussac found that when gases react, the volumes of reacting gases and of any gaseous products are in simple whole-number ratios, provided all volumes are measured at the same temperature and pressure. Two volumes of hydrogen react with one volume of oxygen to give two volumes of steam.
Avogadro's law and molar volume
Avogadro explained combining volumes by proposing that equal volumes of all gases, at the same temperature and pressure, contain equal numbers of particles. It follows that, at constant temperature and pressure, the volume of a gas is directly proportional to the amount of gas:
$$V \propto n \qquad \frac{V_1}{n_1} = \frac{V_2}{n_2}$$
This works because gas particles are so far apart that their own size is irrelevant: the volume depends on how many particles there are, not what they are.
The volume of one mole of any (ideal) gas is the molar volume, Vm. The HSC data sheet gives two values, both at 100 kPa:
| Conditions | Molar volume |
|---|---|
| 0 °C (273.15 K) and 100 kPa | 22.71 L mol⁻¹ |
| 25 °C (298.15 K) and 100 kPa | 24.79 L mol⁻¹ |
$$n = \frac{V}{V_m}$$
Older books (and some textbooks) use 22.4 L mol⁻¹ at "STP" of 0 °C and 101.3 kPa (1 atm). Since 1982 standard pressure has been 100 kPa, which gives 22.71 L. In HSC answers, use the data sheet values and state the conditions.
The ideal gas law: PV = nRT
Boyle's, Charles's, Gay-Lussac's and Avogadro's laws combine into one equation:
$$PV = nRT$$
where R is the universal gas constant, 8.314 J K⁻¹ mol⁻¹. With R in these units, use P in kPa, V in L, n in mol and T in K (kPa × L = J, so the units work). If you use P in Pa, V must be in m³. Check: for 1 mol at 298.15 K and 100 kPa, V = nRT ÷ P = 1 × 8.314 × 298.15 ÷ 100 = 24.79 L, the molar volume.
A 12.0 L scuba cylinder is filled with air to 20.0 MPa at 20 °C. How many moles of air does it hold, and what volume would that air occupy at 25 °C and 100 kPa?
P = 20.0 MPa = 2.00 × 10⁴ kPa; T = 293.15 K.
$n = \dfrac{PV}{RT} = \dfrac{2.00\times10^4 \times 12.0}{8.314 \times 293.15} = 98.5\ \text{mol}$
V = n × Vm = 98.5 × 24.79 = 2.44 × 10³ L. A cylinder the size of a large drink bottle holds enough air to fill a small room, which is why compressed gas cylinders are treated with great care.
Molar mass and density of a gas
Substituting n = m ÷ M into PV = nRT gives a way to find the molar mass of a gas from its mass and volume:
$$M = \frac{mRT}{PV} \qquad \text{density}\ \rho = \frac{m}{V} = \frac{PM}{RT}$$
A 2.00 g sample of a gas occupies 1.12 L at 25 °C and 100 kPa. Calculate its molar mass and suggest what it could be.
n = 1.12 ÷ 24.79 = 0.04518 mol (molar volume works because the conditions are 25 °C and 100 kPa).
M = 2.00 ÷ 0.04518 = 44.3 g mol⁻¹: carbon dioxide (44.01) or propane (44.09). A chemical test, such as limewater, would tell them apart.
Denser gases have larger molar masses. At 25 °C and 100 kPa, CO₂ has a density of 44.01 ÷ 24.79 = 1.78 g L⁻¹ compared with about 1.17 g L⁻¹ for air, which is why carbon dioxide collects in low-lying places such as cellars and wine vats.
Stoichiometry with gases
For gases the mole map gains its last route: n = V ÷ Vm (at the data sheet conditions) or n = PV ÷ RT (at any conditions). When all substances are gases at the same T and P, you can skip moles completely and use volume ratios.
What volume of CO₂, measured at 25 °C and 100 kPa, is produced when 10.0 g of propane burns completely? C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
n(C₃H₈) = 10.0 ÷ 44.09 = 0.2268 mol; n(CO₂) = 3 × 0.2268 = 0.6804 mol.
V = n × Vm = 0.6804 × 24.79 = 16.9 L.
What volume of oxygen is needed to burn 20.0 L of propane completely, with both gases at the same temperature and pressure?
Volume ratio = mole ratio: O₂ : C₃H₈ = 5 : 1, so V(O₂) = 5 × 20.0 = 100 L. Air is only about 21% oxygen, so about 480 L of air is needed.
0.250 g of magnesium reacts with excess hydrochloric acid. What volume of hydrogen is produced at 30 °C and 98.5 kPa?
n(Mg) = 0.250 ÷ 24.31 = 0.01028 mol = n(H₂) (1 : 1).
$V = \dfrac{nRT}{P} = \dfrac{0.01028 \times 8.314 \times 303.15}{98.5} = 0.263\ \text{L} = 263\ \text{mL}$
When are real gases not ideal?
The ideal gas law relies on the assumptions of the kinetic particle theory (2H). Two of them fail for real gases:
- Particles do have volume. At very high pressure the particles are squeezed so close that their own volume is a significant part of the total, so the gas can't be compressed as much as PV = nRT predicts.
- Particles do attract each other. At low temperature particles move slowly enough for intermolecular forces to pull them together, lowering the pressure. Eventually the gas condenses, which an ideal gas never would.
Real gases behave most ideally at high temperature and low pressure, and gases with small, non-polar particles (He, H₂, N₂) are closer to ideal than polar or larger ones (NH₃, H₂O, CO₂). At room conditions most gases are within about 1% of ideal, which is why the molar volume works so well.
Investigations: molar mass from gas volumes
- Collect 100 mL of butane at 25 °C and 100.0 kPa. With no corrections the molar mass comes out at about 56.3 g mol⁻¹, roughly 3% low.
- Turn on Subtract water vapour pressure. Why does the result now match butane, C₄H₁₀ (58.12 g mol⁻¹)?
- Turn off Equalise water levels. Is the gas now above or below air pressure, and which way does the answer move?
- Turn off Dry the lighter before weighing. Why does a few hundredths of a gram of water matter so much?
- Change the room temperature to 15 °C and 30 °C. Does the corrected molar mass change? Does the size of the water vapour correction?
Test yourself
What volume is occupied by 0.350 mol of nitrogen gas at 0 °C and 100 kPa?
0.350 × 22.71 = 7.95 L.
How many moles of gas are in a 5.00 L container at 150 kPa and 50 °C?
n = 150 × 5.00 ÷ (8.314 × 323.15) = 0.279 mol.
N₂(g) + 3H₂(g) → 2NH₃(g). What volume of ammonia forms from 60 mL of hydrogen with excess nitrogen, all at the same T and P?
Volume ratio NH₃ : H₂ = 2 : 3, so 60 × 2/3 = 40 mL.
Under which conditions does a real gas behave most like an ideal gas?
Particles are far apart (low P) and move too fast for attractions to matter (high T).
0.640 g of a gas occupies 496 mL at 25 °C and 100 kPa. What is its molar mass?
n = 0.496 ÷ 24.79 = 0.0200 mol; M = 0.640 ÷ 0.0200 = 32.0 g mol⁻¹ (O₂).
Label the apparatus for finding the molar mass of butane.
Butane experiment: sort each error by its effect.
Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of . At 25 °C and 100 kPa one mole of gas occupies L. In PV = nRT, R is 8.314 J K⁻¹ mol⁻¹, so P is in when V is in litres. Real gases deviate from ideal behaviour at high and low temperature.
A student evaluating the butane experiment collected 152 mL of gas over water at 24 °C and an air pressure of 100.8 kPa. The lighter lost 0.351 g. The vapour pressure of water at 24 °C is 3.0 kPa. (a) Calculate the molar mass of the gas, with and without allowing for water vapour. (b) Evaluate the method, referring to your results and at least two other sources of error, and suggest one improvement.
(a) T = 297.15 K, V = 0.152 L. Without correction: n = 100.8 × 0.152 ÷ (8.314 × 297.15) = 6.20 × 10⁻³ mol; M = 0.351 ÷ 6.20 × 10⁻³ = 56.6 g mol⁻¹. With correction, P(butane) = 100.8 − 3.0 = 97.8 kPa: n = 97.8 × 0.152 ÷ (8.314 × 297.15) = 6.02 × 10⁻³ mol; M = 0.351 ÷ 6.02 × 10⁻³ = 58.3 g mol⁻¹.
(b) Butane, C₄H₁₀, is 58.12 g mol⁻¹. The corrected result is only 0.4% high, while ignoring water vapour gives a result 2.6% low, so the correction is essential and, with it, the method is accurate. Other errors: water remaining on the lighter would make the mass lost too small and M too low; if the water level inside the cylinder were above the level outside, the gas would be below air pressure and M would again be too low; some butane dissolves in water, so the volume is too small and M too high. Commercial lighter fuel also contains some propane and isobutane, so the gas is not pure butane. The method relies on single measurements; repeating the experiment three or more times and averaging would improve reliability, and collecting the gas in a gas syringe would avoid the water vapour and level problems altogether.