(a) A gas has only two particles a and b. Show with the help of diagrams how these two particles can be arranged in three energy…

(a) Arrangements of two particles in three states Two particles are placed in three single-particle states 1, 2, 3. The count depends on whether the particles are distinguishable and whether a state may hold more than one particle.

(i) Maxwell–Boltzmann: particles are distinguishable (labelled a, b) and any number may occupy a state. Number of arrangements = gⁿ = 3² = 9.

(ii) Fermi–Dirac: particles are identical fermions (half-integral spin), indistinguishable, and obey the Pauli exclusion principle, so at most one particle per state. Number of arrangements = ᵍCₙ = ³C₂ = 3.

(iii) Bose–Einstein: particles are identical bosons (integral spin), indistinguishable, with no restriction on occupancy. Number of arrangements = ⁽ᵍ⁺ⁿ⁻¹⁾Cₙ = ⁴C₂ = 6.

Result: MB 9, FD 3, BE 6. Relative to MB, the doubly occupied states carry more weight in BE (3 of 6 against 3 of 9: bunching) and are forbidden in FD (exclusion). When the number of states g is much larger than the number of particles, multiple occupancy becomes rare and both quantum counts approach gⁿ/n!, the corrected MB result.

(b) Boiling pressure of water at 150 °C Clausius–Clapeyron equation: dP/dT = L/[T(v₂ − v₁)] Data (CGS): L = 540 cal g⁻¹ = 540 × 4.2 × 10⁷ = 2.268 × 10¹⁰ erg g⁻¹; v₂ − v₁ = 1676 cm³ g⁻¹; known point: water boils at T = 373 K under P₀ = 1 atm = 10⁶ dyne cm⁻². dP/dT = 2.268 × 10¹⁰/(373 × 1676) = 2.268 × 10¹⁰/6.251 × 10⁵ = 3.63 × 10⁴ dyne cm⁻² K⁻¹ Treating dP/dT as constant for ΔT = 150 − 100 = 50 K: ΔP = 3.63 × 10⁴ × 50 = 1.81 × 10⁶ dyne cm⁻² = 1.81 atm P = P₀ + ΔP = 1.00 × 10⁶ + 1.81 × 10⁶ = 2.81 × 10⁶ dyne cm⁻² ≈ 2.81 atm (This linear estimate assumes L and Δv stay constant over 100–150 °C; the given Δv refers to 100 °C.)

(c) Ring around a solenoid with decreasing current Inside a long solenoid B = μ₀nIₛ (along the axis); outside, B ≈ 0. So the flux through the ring (radius b > a) is only that through the solenoid cross-section: Φ = μ₀nIₛ · πa²

(i) Induced current (Faraday's law, ring self-inductance neglected): ε = −dΦ/dt = −μ₀nπa² dIₛ/dt Iᵣ = ε/R = −(μ₀nπa²/R) dIₛ/dt As Iₛ is reduced (dIₛ/dt < 0), Iᵣ is positive, i.e. it flows in the same sense as Iₛ, opposing the fall in flux (Lenz's law).

(ii) Joule power in the ring: P = Iᵣ²R = (μ₀nπa²)²(dIₛ/dt)²/R = μ₀²n²π²a⁴(dIₛ/dt)²/R

(iii) Induced electric field: by symmetry E is azimuthal and depends only on the distance s from the axis. Using ∮E·dl = −dΦ/dt on a circle of radius s: Outside (s > a): E(2πs) = −μ₀nπa² dIₛ/dt ⇒ E = −(μ₀na²/2s)(dIₛ/dt) φ̂ Inside (s < a): E(2πs) = −μ₀nπs² dIₛ/dt ⇒ E = −(μ₀ns/2)(dIₛ/dt) φ̂ Just outside the surface (s = a), E = −(μ₀na/2)(dIₛ/dt). The field is non-zero outside even though B = 0 there, and at the ring E(b)·2πb = ε, as required.