(a) Calculate the number of collisions that oxygen makes per second on 1·00 cm² of the surface of the vessel containing them if…

(a)

The collision frequency on a surface is obtained from the kinetic theory of gases. For an ideal gas in equilibrium, the number of molecules striking unit area per second is Z = (1/4)n v̄ where n = P/(kT) and v̄ = √(8kT/(πm)). Thus Z = P / √(2πmkT).

Given: P = 1.00 × 10⁻⁶ atm = 1.00 × 10⁻⁶ × 1.01325 × 10⁵ Pa = 0.101325 Pa. T = 25°C = 298.15 K. For O₂, m = 32.00 × 10⁻³ kg mol⁻¹ / 6.022 × 10²³ mol⁻¹ = 5.314 × 10⁻²⁶ kg.

Therefore 2πmkT = 2 × π × 5.314 × 10⁻²⁶ × 1.381 × 10⁻²³ × 298.15 = 1.374 × 10⁻⁴⁵. So √(2πmkT) = 3.707 × 10⁻²³ kg m s⁻¹.

Hence Z = 0.101325 / (3.707 × 10⁻²³) = 2.733 × 10²¹ m⁻² s⁻¹.

Area = 1.00 cm² = 1.00 × 10⁻⁴ m². Collisions per second = Z × A = 2.733 × 10²¹ × 1.00 × 10⁻⁴ = 2.73 × 10¹⁷ s⁻¹.

Final answer: 2.73 × 10¹⁷ collisions per second on 1.00 cm².

(b)(i)

The work done creates new surface, so it equals surface tension times increase in surface area. Let one mole of water have molar volume Vm. If N droplets of radius r are formed, volume is conserved: N × (4/3)πr³ = Vm so N = 3Vm/(4πr³).

Total new surface area = N × 4πr² = 3Vm/r. Work W = γ × area = 3γVm/r. Therefore r = 3γVm/W.

Using density of liquid water ≈ 1.00 g cm⁻³ and M = 18.0 g mol⁻¹, Vm = 18.0 cm³ mol⁻¹ = 1.80 × 10⁻⁵ m³ mol⁻¹. γ = 0.072 J m⁻², W = 10.0 J mol⁻¹.

Thus r = 3 × 0.072 × 1.80 × 10⁻⁵ / 10.0 = 3.89 × 10⁻⁷ m.

Final answer: radius of each droplet = 3.89 × 10⁻⁷ m = 0.389 μm.

(b)(ii)

Volume of one droplet = (4/3)πr³. With r = 3.89 × 10⁻⁷ m, r³ = 5.89 × 10⁻²⁰ m³. Volume of droplet = (4/3)π × 5.89 × 10⁻²⁰ = 2.47 × 10⁻¹⁹ m³.

Moles of water in one droplet = volume of droplet / Vm = 2.47 × 10⁻¹⁹ / 1.80 × 10⁻⁵ = 1.37 × 10⁻¹⁴ mol.

Number of molecules = 1.37 × 10⁻¹⁴ × 6.022 × 10²³ = 8.25 × 10⁹.

Final answer: about 8.2 × 10⁹ water molecules per droplet.

(c)(i)

A phase diagram of butane is drawn with pressure on the y-axis and temperature on the x-axis. Mark the following points:

• Triple point: temperature just below −138°C, pressure = 3 × 10⁻⁵ torr = 3.95 × 10⁻⁸ atm.
• Normal melting point: (−138°C, 1 atm).
• Normal boiling point: (0°C, 1 atm).
• Critical point: (152°C, 38 atm).

Draw the sublimation curve from the triple point downwards to lower temperatures and pressures. Draw the vaporisation curve from the triple point through (0°C, 1 atm) to the critical point (152°C, 38 atm). Draw the fusion curve from the triple point through (−138°C, 1 atm) upwards, with a positive slope. The regions are: solid on the left/above the fusion and sublimation curves, liquid between fusion and vaporisation curves, and gas/vapour on the right/below the vaporisation curve. Above the critical point, there is no distinction between liquid and gas.

(c)(ii)

At 140°C, butane is below its critical temperature 152°C. At 1 atm and 140°C, it is a vapour because the normal boiling point is 0°C. When it is compressed isothermally, its pressure rises until it reaches the saturation vapour pressure at 140°C. Since 140°C < 152°C, this saturation pressure lies below the critical pressure 38 atm and above 1 atm. At that pressure, liquid and vapour coexist. Further compression converts vapour into liquid. Therefore, during compression from 1 atm to 40 atm at 140°C, two phases are present for part of the process.

Final answer: yes, liquid and vapour coexist during part of the compression.

(c)(iii)

At 200°C, butane is above its critical temperature 152°C. Above the critical temperature, no liquid-vapour coexistence boundary exists, irrespective of pressure. Compression from 1 atm to 40 atm keeps the substance in a single supercritical fluid phase.

Final answer: no, two phases are not present at any time.

(d)

Heat required to melt 100 g ice at 0°C: Q₁ = mLf = 100 g × 334 J g⁻¹ = 3.34 × 10⁴ J.

Heat required to warm the melted water from 0°C to 40°C: Q₂ = mcΔT = 0.100 kg × 4184 J kg⁻¹ K⁻¹ × 40 K = 1.6736 × 10⁴ J.

Total heat required: Q = Q₁ + Q₂ = 3.34 × 10⁴ + 1.6736 × 10⁴ = 5.0136 × 10⁴ J.

This heat comes from condensation of water vapour. If m grams condense, Q = mLv = m × 2260 J g⁻¹. So m = 5.0136 × 10⁴ / 2260 = 22.18 g.

Final answer: approximately 22.2 g of water condenses in the container.

(e)

Crystalline solids become more defective at higher temperature because defect formation increases configurational entropy. The formation of a defect costs enthalpy, so ΔG = ΔH − TΔS. At absolute zero, ΔH dominates and defects are unfavourable. As temperature increases, the −TΔS term becomes more negative and stabilises defects.

For Schottky defects, the equilibrium fraction is approximately proportional to exp(−ΔH/(2kT)); for Frenkel defects, a similar exponential law holds. Thus the defect concentration rises steeply with temperature. In addition, higher thermal energy increases atomic vibrations, anharmonic displacements, and mobility of defects, allowing more vacancies, interstitials, and dislocation-related defects to appear. Near the melting point, the defect concentration can become appreciable. Therefore, increasing temperature generally makes crystalline solids more defective.