Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.3.4 to 1.3.5

Avogadro's Law and the Combined Boyle-Charles Law.

Blow a little more air into a balloon and it swells. Is there a rule for how much? There is, and it is so simple that a striking fact falls out of it: at standard conditions 22.4 L of any gas holds exactly the same number of molecules.

1. Volume and moles

Look at the experiment in Figure 1.9. Two piston cylinders, A and B. A holds 0.10 mol, which is 4.4 g, of dry ice (solid CO₂); B holds 0.20 mol, which is 8.8 g. Heat them and both turn to gas. The pressure and the temperature are the same in both, yet the volume in B is twice the volume in A.

Twice the gas, twice the volume. So at constant pressure and temperature the volume of a gas is proportional to its number of moles: V ∝ n (P, T constant). Put in a constant K and V = Kn, that is V/n = K.

This is Avogadro's law: at constant temperature and pressure, equal volumes of any ideal gas contain the same number of particles, that is, the same number of moles. Hydrogen or CO₂, it is the count of particles that sets the volume.

Fill a balloon a little at a time, in moles. Watch the volume and the count of molecules together.

2. Standard conditions and molar volume

Volume also changes with temperature and pressure, so a standard state is fixed for comparison. The standard conditions of a gas (STP) are a temperature of 0°C or 273.15 K and a pressure of 1 atm, 760 mm Hg or 760 torr. The standard molar volume there is 22.4141 L, or 22.4 L in short. Use 22.4 L in calculations.

The number of molecules in one mole volume of a gas is the Avogadro number (N_A), and N_A = 6.022×10²³.

One mole of molecules is one gram-molecular mass and holds 6.022×10²³ molecules: 44 g CO₂ = 1 mol, 18 g H₂O = 1 mol. One mole of atoms is one gram-atomic mass and holds 6.022×10²³ atoms: 23 g Na, for example.

Molar volume is the volume of one mole of gas under standard conditions. A hotter gas swells, so the figure moves with the condition: 22.4141 L at STP, 24.040 L at 20°C and 1 atm, and 24.789 L at 25°C or SATP. Match the pairs below.

3. From moles to molecules, and molecules to atoms

The book's calculations share one mould. Problem 1.16: how many molecules in 16 g of O₂? One mole of O₂ weighs 32 g, so 16 g is half a mole: 6.022×10²³ × 16/32 = 3.011×10²³. Problem 1.17: 1.0 g of hydrogen holds 6.022×10²³ atoms, because it is one gram-atom. Problem 1.18: the mass of one Na atom (Na = 23) = 23/(6.022×10²³) = 3.82×10⁻²³ g. Divide the mass of a mole by N_A and you have the mass of one.

Problem 1.19: if 55.6 mg of a graphite pencil is used over 500 signatures, the carbon atoms per signature = 6.022×10²³ × 55.6/(12 × 1000 × 500) = 5.580386×10¹⁸. The 1000 turns mg into g, the 12 gives moles of carbon, the 500 splits it per signature. Problem 1.20: molecules in 1 mL of oxygen at STP = 6.022×10²³/(22.4×10³) = 2.688×10¹⁹, because 22.4 L = 22.4×10³ mL. The value holds for every gas, since Avogadro's law does not care which gas it is.

When the question asks for atoms, multiply the number of molecules by the atomicity: 2 for diatomic O₂, 3 for triatomic O₃, 4 for the four-atom NH₃. So 0.5 mol of NH₃ holds 0.5 × 6.022×10²³ × 4 = 1.2044×10²⁴ atoms. Drop each molecule into its multiplier box.

Three exercise calculations. 1.25(a): 90 g of water is 90/18 = 5 mol, that is 5 × 6.022×10²³ = 3.011×10²⁴ molecules. 1.25(f): 2.2 g of CO₂ at standard conditions occupies (2.2/44) × 22.4 = 1.12 L. 1.26(a): 10 mL of NH₃ at standard conditions has mass (10/22414) × 17 = 7.5845×10⁻³ g. Run them through the converter below, or type your own numbers. Nothing here goes anywhere. The notebook is yours.

4. Boyle and Charles in one equation

Boyle's law says V ∝ 1/P (n, T constant) and Charles's law says V ∝ T (n, P constant). Join them and V ∝ T/P, that is V = KT/P or PV/T = K (n constant). This is the combined law.

For one gas in two different states the K is the same, so the combined equation is P₁V₁/T₁ = P₂V₂/T₂.

Application: 2 L of gas at 1 atm and 27°C is taken to 2 atm and 127°C. What is the new volume? Temperatures in kelvin: T₁ = 300 K, T₂ = 400 K. V₂ = P₁V₁T₂/(T₁P₂) = 1 × 2 × 400/(300 × 2) = 1.33 L. Doubling the pressure tries to halve the volume, while heating from 300 to 400 K pushes it back up; between the two, 2 L settles at 1.33 L. One more: 500 mL of gas at STP taken to 740 mmHg and 25°C gives V₂ = 760 × 500 × 298/(273 × 740) = 560.5 mL, about 0.56 L.

5. Check yourself

Three questions, one at a time.

Work the second one out.

The last question.

The whole note on seven cards.

Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.3.4 to 1.3.5.