(a) One real gas equation is van der Waals equation (P + (n² a)/(V²))(V - nb) = nRT Which parameter in the above equation…

(a) van der Waals parameters (5 marks)

The parameter a reflects the intermolecular force of attraction: the term n²a/V² is the correction to the pressure, because attraction reduces the force with which molecules strike the wall. The parameter b reflects repulsion: it is the excluded volume, about four times the actual volume of the molecules per mole, and nb is subtracted from V. In SI units, a is in Pa m⁶ mol⁻² (equivalently N m⁴ mol⁻²) and b is in m³ mol⁻¹.

(b) Liquids A and B (5 marks)

A high surface tension means strong intermolecular attraction. Since γ_A > γ_B:

• (i) Vapour pressure: B is larger, because its molecules escape more easily from the weaker attraction.
• (ii) Cohesive force: A is larger, since surface tension measures cohesion.
• (iii) Boiling point: A is higher, because more energy is needed to overcome the attraction.
• (iv) Capillary action: A shows the larger rise, since h = 2γ cos θ/(rρg), for the same tube, density and contact angle.
• (v) Wetting property: B wets better, as its weaker cohesion lets it spread over the solid; A tends to form drops.

(c) Two metal blocks in thermal contact (5 marks)

Heat lost by the hot block equals heat gained by the cold block, and the heat capacities are equal, so the final temperature is the mean: T_f = (300 + 400)/2 = 350 K.

For each block ΔS = C ln(T_f/T_i).

ΔS = 15 ln(350/300) + 15 ln(350/400) = 15 ln(122500/120000) = 15 × 0.0206 ≈ +0.31 J K⁻¹.

It is positive, as required for an irreversible process in an isolated system.

(d) Dissociation of N₂O₄ (5 marks)

Start with 1 mol N₂O₄ and degree of dissociation α = 0.20: N₂O₄ = 0.80 mol, NO₂ = 0.40 mol, total 1.20 mol. The mole fractions are 0.80/1.20 = 2/3 and 0.40/1.20 = 1/3, and at P = 1 atm these equal the partial pressures.

K_p = p²(NO₂)/p(N₂O₄) = (1/3)²/(2/3) = 1/6 ≈ 0.167, which also equals 4α²P/(1 − α²) = 0.16/0.96.

ΔG° = −RT ln K_p = −8.314 × 300 × ln(0.1667) = −2494 × (−1.792) ≈ +4.47 × 10³ J mol⁻¹ ≈ +4.5 kJ mol⁻¹.

The positive value agrees with K_p < 1.

(e) Components and phases for NH₄Cl (5 marks)

The equilibrium is NH₄Cl(s) ⇌ NH₃(g) + HCl(g). Components C = S − R − r, where S is the number of species, R the number of independent reactions, and r the extra composition restrictions.

(i) Heated in a closed container, there are three species, one equilibrium, and NH₃ and HCl are in equal amounts because both come only from the salt: C = 3 − 1 − 1 = 1. The phases are two, solid NH₄Cl and one gas mixture (P = 2). Degrees of freedom F = C − P + 2 = 1.

(ii) With additional NH₃, the equal-amount restriction no longer holds: C = 3 − 1 = 2. The phases are still two (solid and gas), so F = 2 − 2 + 2 = 2.

(f) Silver concentration cell (5 marks)

Both electrodes are silver in AgNO₃, so this is a concentration cell without transference (with a salt bridge) and E° = 0. The electrode in the more concentrated solution, C₂, is the cathode and the dilute one, C₁, is the anode. For the Ag⁺/Ag couple, n = 1:

ΔE = (RT/F) ln(C₂/C₁) = 0.05916 log₁₀(1 × 10⁻²/5 × 10⁻⁴) = 0.05916 × log₁₀ 20 = 0.05916 × 1.301 ≈ 0.077 V.

So the cell potential is about 77 mV at 25 °C, treating the concentrations as activities.

(g) Why a potentiometer is used (5 marks)

A potentiometer measures the potential when the cell delivers zero current, by balancing it against a known opposing voltage. This is needed because the cell has internal resistance, so if a current i were drawn, the terminal voltage would be E − iR, lower than the true EMF. A current would also cause polarisation and change the electrode concentrations, and the process would no longer be reversible. Only at null (zero current) is the potential the equilibrium (reversible) EMF, which can be related to ΔG = −nFE. An ordinary voltmeter draws current and reads less than the EMF.

(h) Activation energy of food spoilage (5 marks)

From the Arrhenius equation, ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂), with T₁ = 281.15 K, T₂ = 303.15 K and k₂/k₁ = 40.

ln 40 = 3.689, and (1/281.15 − 1/303.15) = 3.5568 × 10⁻³ − 3.2987 × 10⁻³ = 2.58 × 10⁻⁴ K⁻¹.

E_a = 8.314 × 3.689/(2.58 × 10⁻⁴) ≈ 1.19 × 10⁵ J mol⁻¹ ≈ 119 kJ mol⁻¹.

(i) Colour of a red flower in green light (5 marks)

A flower appears red because its pigments absorb the other wavelengths, chiefly green and blue, and reflect or transmit red. In green light, the only light present is green, which the pigment absorbs, and there is no red light to reflect. Almost nothing reaches the eye from the petals, so the flower appears black (or very dark), while the green leaves look green.

(j) Adsorption at low temperature (5 marks)

Adsorption is spontaneous when ΔG = ΔH − TΔS < 0. When a gas is held on a surface its freedom of movement falls, so ΔS < 0 and −TΔS is positive; for ΔG to be negative ΔH must be negative and larger in size than TΔS. Adsorption is therefore exothermic, since bonds form between the gas and the surface. As temperature falls, the unfavourable −TΔS term shrinks, so ΔG becomes more negative and adsorption more favourable. This agrees with Le Chatelier's principle: lowering temperature favours the exothermic process, and physical adsorption in particular decreases as temperature rises.