(a) A ternary system consists of three components (A, B and C) in equilibrium with two phases. Determine the number of degrees of…

(a) Gibbs phase rule is F = C − P + 2. For a ternary system, C = 3. With two phases, P = 2. Therefore F = 3 − 2 + 2 = 3 degrees of freedom.

Thus three intensive variables must be fixed to specify the state. They may be taken as temperature T, pressure P, and one composition variable, e.g. mole fraction of A in one phase. The composition of the other phase is then fixed by the equilibrium tie-line.

If T and P are both fixed, F = 1: one composition variable can still vary, and the two-phase tie-line shifts through the two-phase region. If P is fixed, F = 2; if T is fixed, F = 2. A temperature change alters mutual solubilities and moves the tie-lines; at some temperature the two phases may become identical or one phase may disappear. A pressure change affects equilibrium mainly through the difference in molar volumes of the two phases. High pressure favours the phase of smaller molar volume. In condensed systems the pressure effect is usually small unless pressures are very high. At fixed T and P, varying one phase composition changes the other phase composition along the tie-line.

(b) Van der Waals modified the ideal gas equation because the ideal gas assumptions fail for real gases:

• Molecules are not point particles; they have finite volume. The available volume is reduced from V to V − b per mole.
• Intermolecular attractive forces reduce the pressure exerted on the walls. The pressure correction is proportional to the square of density, a/V².

Hence for one mole, (P + a/V²)(V − b) = RT. For n moles, (P + a n²/V²)(V − n b) = nRT. Here a is the attraction parameter and b is the excluded-volume parameter.

The critical constants are the temperature T_c, pressure P_c and molar volume V_c at the critical point, where liquid and gas phases become identical and the isotherm has a horizontal inflection. Thus at the critical point, (∂P/∂V)_T = 0 and (∂²P/∂V²)_T = 0.

For one mole, P = RT/(V − b) − a/V². First derivative: −RT/(V − b)² + 2a/V³ = 0, so RT = 2a(V − b)²/V³. Second derivative: 2RT/(V − b)³ − 6a/V⁴ = 0, so RT = 3a(V − b)³/V⁴. Equating these gives 2a(V − b)²/V³ = 3a(V − b)³/V⁴, hence 2V = 3(V − b), so V_c = 3b.

Then RT_c = 2a(2b)²/(27b³) = 8a/(27b), so T_c = 8a/(27Rb). Also P_c = RT_c/(V_c − b) − a/V_c² = (8a/(27b))/(2b) − a/(9b²) = 4a/(27b²) − 3a/(27b²) = a/(27b²).

The critical compressibility factor is Z_c = P_c V_c/(R T_c) = 3/8.

(c) Let the applied uniform field be E₀ along the z-axis, so E₀ = E e_z. The sphere is an isolated neutral conductor of radius a.

(i) Induced surface charge density Outside the sphere the potential satisfies Laplace’s equation, ∇²Φ = 0. Far from the sphere, Φ → −E z = −E r cosθ. On the conducting sphere, Φ must be constant; choose Φ(a, θ) = 0.

The axisymmetric solution with the correct far field is Φ(r, θ) = (−E r + B/r²) cosθ. At r = a, Φ = 0, so −E a + B/a² = 0, hence B = E a³. Therefore Φ(r, θ) = −E r cosθ + E a³ cosθ/r².

The surface charge density is σ = ε₀ E_r(a, θ). Now E_r = −∂Φ/∂r = E cosθ + 2E a³ cosθ/r³. At r = a, E_r = 3E cosθ. Thus σ(θ) = 3ε₀ E cosθ = 3ε₀ E₀·e_r on r = a. The net induced charge is zero because ∫σ dA = 0.

(ii) Electric field at point P Using E = −∇Φ, the spherical components outside the sphere are E_r = −∂Φ/∂r = E cosθ(1 + 2a³/r³), E_θ = −(1/r)∂Φ/∂θ = −E sinθ(1 − a³/r³), E_φ = 0. Therefore for r ≥ a, E(r, θ) = E cosθ(1 + 2a³/r³) e_r − E sinθ(1 − a³/r³) e_θ. Inside the conducting sphere, r < a, E = 0.

Equivalently, in vector form, E = E₀ − (a³/r³)E₀ + 3a³(E₀·r)r/r⁵, r ≥ a, where r is the position vector and r = |r|. At the surface r = a, this reduces to E = 3E cosθ e_r, which is normal to the sphere, as required for a conductor. Far away, E → E₀.