(a) The volume of a certain gas is found to be 5.0×10⁻⁴ m³ mol⁻¹ at 273 K and 3.0×10⁶ Pa. This gas obeys the van der Waals'…

(a) van der Waals gas: constant b, compression factor and interactions

Given: Vₘ = 5.0 × 10⁻⁴ m³ mol⁻¹, T = 273 K, P = 3.0 × 10⁶ Pa, a = 0.50 m⁶ Pa mol⁻², R = 8.314 J K⁻¹ mol⁻¹.

(i) The van der Waals equation for one mole is (P + a/Vₘ²)(Vₘ − b) = RT.

a/Vₘ² = 0.50 / (5.0 × 10⁻⁴)² = 2.0 × 10⁶ Pa, so P + a/Vₘ² = 5.0 × 10⁶ Pa.

RT = 8.314 × 273 = 2269.7 Pa m³ mol⁻¹, so Vₘ − b = 2269.7 / (5.0 × 10⁶) = 4.539 × 10⁻⁴ m³ mol⁻¹.

b = 5.0 × 10⁻⁴ − 4.539 × 10⁻⁴ = 4.6 × 10⁻⁵ m³ mol⁻¹ (46 cm³ mol⁻¹).

(ii) Z = PVₘ/RT = (3.0 × 10⁶ × 5.0 × 10⁻⁴) / 2269.7 = 1500 / 2269.7 = 0.66.

(iii) Z < 1 shows that the gas is more compressible than an ideal gas. The attraction term a/Vₘ² (2.0 × 10⁶ Pa) is large compared with P (3.0 × 10⁶ Pa), so attractive forces dominate over repulsion at this temperature and pressure. Repulsion (molecular size) still exists, as b is positive, but it is the smaller effect here.

(b) Pressure difference needed to drink through a slanted straw

The pressure difference needed equals the hydrostatic pressure of the vertical height h of the water column. For a 15 cm straw at 45° (assuming the water has to be lifted the vertical height of the straw), h = 15 cm × sin 45° = 10.6 cm = 0.106 m.

ΔP = ρgh = 1000 kg m⁻³ × 9.8 m s⁻² × 0.106 m = 1.04 × 10³ Pa.

Since 1 torr = 133.3 Pa, ΔP = 1.04 × 10³ / 133.3 = 7.8 torr. It is the vertical height that matters, not the slanted length.

(c) Diameter of the capillary

A bubble blown in a liquid has one liquid surface, so the excess pressure is ΔP = 2γ/r.

ΔP = 2.16 torr × 133.3 Pa/torr = 288 Pa, γ = 0.072 N m⁻¹.

r = 2γ/ΔP = 2 × 0.072 / 288 = 5.0 × 10⁻⁴ m = 0.05 cm.

Diameter = 2r = 0.10 cm (1.0 mm).

(d) Silicate structures: pyroxenes, amphiboles and phyllosilicates

The basic unit of all silicates is the SiO₄⁴⁻ tetrahedron, with Si at the centre and O at the four corners; they join by sharing corner oxygens.

• Pyroxenes (single chain): each tetrahedron shares two corners. Empirical formula of the anion: (SiO₃)ₙ²ⁿ⁻, e.g. enstatite MgSiO₃, diopside CaMg(SiO₃)₂.
• Amphiboles (double chain): two single chains are linked by sharing further corners (alternately two and three corners per tetrahedron). Empirical formula of the anion: (Si₄O₁₁)ₙ⁶ⁿ⁻, e.g. tremolite Ca₂Mg₅(Si₄O₁₁)₂(OH)₂.
• Phyllosilicates (sheet): each tetrahedron shares three corners, forming a two-dimensional hexagonal sheet. Empirical formula of the anion: (Si₂O₅)ₙ²ⁿ⁻, e.g. talc, mica and clay minerals.