(a) (i) Calculate the coefficient of viscosity of air at temperatures : (I) 298 K and (II) 0 K. Assume that the collision…

(a)(i) Viscosity of air from kinetic theory

For a gas of hard spheres, momentum is carried between layers by molecules that travel about one mean free path λ:

• η = (1/3) ρ v̄ λ
• ρ = N m (N = number density)
• λ = 1/(√2 σ N), where σ = πd² is the collision cross-section

Combining, η = m v̄/(3√2 σ), with v̄ = √(8RT/πM). Note that η is independent of pressure and increases as √T.

Data:

• σ = 0.28 nm² = 0.28 × 10⁻¹⁸ m² = 2.8 × 10⁻¹⁹ m²
• M = 29 g mol⁻¹ = 0.029 kg mol⁻¹
• m = M/N_A = 0.029/6.022 × 10²³ = 4.816 × 10⁻²⁶ kg

(I) At 298 K

v̄ = √(8 × 8.314 × 298/(π × 0.029)) = 466.4 m s⁻¹

η = (4.816 × 10⁻²⁶ × 466.4)/(3 × 1.414 × 2.8 × 10⁻¹⁹) = 1.89 × 10⁻⁵ kg m⁻¹ s⁻¹ (Pa s)

This agrees well with the measured value, about 1.85 × 10⁻⁵ Pa s.

(II) At 0 K

• v̄ ∝ √T, so v̄ = 0, and the formula gives η = 0.
• This is only a formal limit. The model assumes a dilute gas, but air liquefies near 80 K, and at 0 K there is no gas at all. The result is therefore physically meaningless.
• If "0 °C" (273.15 K) is intended: η = 1.89 × 10⁻⁵ × √(273.15/298) = 1.81 × 10⁻⁵ Pa s (measured ≈ 1.72 × 10⁻⁵ Pa s).

(a)(ii) Boyle temperatures of Ar, CH₄ and C₆H₆

• For a van der Waals gas, the second virial coefficient is B = b − a/RT. At the Boyle temperature B = 0, so T_B = a/(Rb).
• T_B is high when attraction (a) is strong relative to molecular size (b).

The attractions here are all London dispersion forces, which grow with polarisability, that is, with the number of electrons and the size and diffuseness of the electron cloud:

• Ar is a small, compact atom (18 electrons);
• CH₄ has 10 electrons, but they are spread over a larger molecule, so its polarisability is somewhat higher than argon's;
• C₆H₆ has 42 electrons, including a delocalised π cloud, so it is highly polarisable.

Order: Ar < CH₄ < C₆H₆

van der Waals estimates, using a in L² bar mol⁻² and b in L mol⁻¹:

• Ar: 1.355/(0.08314 × 0.0320) ≈ 509 K (experimental 411.5 K)
• CH₄: 2.283/(0.08314 × 0.0428) ≈ 642 K (experimental 510 K)
• C₆H₆: 18.24/(0.08314 × 0.1154) ≈ 1900 K

The van der Waals model overestimates T_B, but the order is unaffected.

(b)(i) Ethanol or dimethyl ether: which has the greater surface tension?

Ethanol has the greater surface tension. At 20 °C it is about 22 mN m⁻¹, while liquefied dimethyl ether near room temperature is roughly half that.

• Same molar mass: the two are isomers (C₂H₆O, M = 46 g mol⁻¹). The difference therefore cannot come from molar mass or from dispersion forces, which are similar. It comes from the type of intermolecular force.
• Ethanol (CH₃CH₂–OH): its O–H group is both a hydrogen-bond donor and acceptor, so molecules are linked by an extended hydrogen-bonded network. A molecule at the surface loses part of this network, so creating new surface costs more energy.
• Dimethyl ether (CH₃–O–CH₃): it has no O–H bond. It can accept but not donate a hydrogen bond, so only dipole–dipole and dispersion forces hold its molecules together.
• Consistent evidence: the same difference in cohesion shows in boiling points (ethanol 78 °C; dimethyl ether −24 °C) and in enthalpies of vaporisation.

(b)(ii) Pressure difference across the surface of a water droplet (r = 150 nm)

By the Laplace equation for a spherical surface, Δp = p_in − p_out = 2γ/r.

• γ(water, 25 °C) = 0.0720 N m⁻¹
• r = 150 nm = 1.50 × 10⁻⁷ m

Δp = (2 × 0.0720)/(1.50 × 10⁻⁷) = 9.6 × 10⁵ Pa ≈ 9.5 atm

Units check: N m⁻¹ ÷ m = N m⁻² = Pa.

The pressure inside the droplet is about 9.5 atm above the outside pressure, so the total inside pressure is about 10.5 atm. This excess pressure is why tiny droplets have a higher vapour pressure (Kelvin equation) and evaporate faster than large ones.

(c)(i) Change in Helmholtz energy for a reversible isothermal compression

• dA = −S dT − p dV, so at constant T, dA = −p dV and ΔA = w_rev (the reversible work done on the gas).
• For n mol of ideal gas, p = nRT/V, which gives:

ΔA = −∫ p dV = −nRT ln(V₂/V₁) = nRT ln(V₁/V₂)

Equivalently, ΔU = 0 for an isothermal ideal gas, so ΔA = −TΔS, with ΔS = nR ln(V₂/V₁).

Substitution:

• n = 1, T = 298 K, V₁ = 100.0 L, V₂ = 22.4 L
• RT = 8.314 × 298 = 2477.6 J mol⁻¹
• ln(100.0/22.4) = ln 4.464 = 1.496
• ΔA = 2477.6 × 1.496 = +3.71 × 10³ J = +3.71 kJ

Interpretation:

• ΔA is positive because work must be done on the gas to compress it. The compressed state has lower entropy and so a higher Helmholtz energy.
• For an isothermal ideal gas Δ(pV) = 0, so here ΔG = ΔA = +3.71 kJ as well.

(c)(ii) Why a tyre gets hot when pumped, and whether it can be inflated without a rise in temperature

• Why it heats: each pump stroke compresses air quickly. There is almost no time for heat to escape, so the compression is nearly adiabatic (q ≈ 0). By the first law, ΔU = q + w ≈ w > 0: the work done on the air raises its internal energy, and hence its temperature.
• Size of the effect: for a reversible adiabatic compression of air (γ = 1.4) from 1 atm to 3 atm absolute, T₂ = T₁(p₂/p₁)^((γ−1)/γ) = 298 × 3^0.286 ≈ 408 K inside the pump cylinder.
• Why the tyre itself warms less: the hot air mixes with the air already in the tyre, and heat passes to the pump barrel and rubber. Friction of the piston adds a little heat, but it is secondary.
• Can it be inflated without a temperature rise? Yes, in principle, if the compression is made isothermal. Pumping very slowly, in small strokes with pauses, lets the heat of compression flow to the surroundings at every stage (q = −w, ΔU = 0 for an ideal gas). A strictly isothermal process is an idealisation needing infinitely slow compression. In practice, slow pumping with a well-cooled pump and hose keeps the rise negligible. Letting a hot tyre cool back to ambient temperature brings it to the same final state, though its pressure then falls slightly.

(c)(iii) Pressure of O₂ over NiO at 25 °C

NiO(s) ⇌ Ni(s) + ½ O₂(g)

• The activities of the pure solids are 1, so K = (p_O₂/p°)^½.
• At equilibrium ΔG = 0, and ΔG = ΔG° + RT ln Q gives ΔG° = −RT ln K.

½ ln p_O₂ = −ΔG°/RT = −212 000/(8.314 × 298) = −85.57

ln p_O₂ = −171.14, so p_O₂ = e^(−171.14) ≈ 4.8 × 10⁻⁷⁵ atm

(log₁₀ p_O₂ = −74.32, taking p° ≈ 1 atm.)

Significance: this equilibrium ("dissociation") pressure is immeasurably small. NiO is therefore thermodynamically very stable at 25 °C and will not decompose in air, where p_O₂ = 0.21 atm is vastly greater. Nickel is instead oxidised. To reduce NiO, the O₂ pressure must be lowered below this value (for example with H₂ or C), or the temperature must be raised.

(c)(iv) Final temperature after Joule–Thomson expansion

• The expansion through a porous plug occurs at constant enthalpy, and μ_JT = (∂T/∂p)_H.
• Treating μ_JT as constant, ΔT = μ_JT Δp.

Working:

• Δp = 0.95 − 200.00 = −199.05 atm
• ΔT = 0.150 K atm⁻¹ × (−199.05 atm) = −29.9 K
• T_final = 292.15 K − 29.86 K = 262.3 K ≈ −10.9 °C

Because μ_JT > 0, the gas is below its inversion temperature and cools on expansion. This is the principle of Linde liquefaction. Above the inversion temperature (as for H₂ and He at room temperature) μ_JT < 0 and the gas would warm instead.