(a) (i) The melting point of tin is 232°C, its latent heat of fusion is 14 cal/g and the specific heat of solid and molten tin are 0·055 and 0·064 cal/g °C respectively. Calculate the change in entropy when 1·0 gm of tin is heated from 100°C to 300°C

Question

(a) (i) The melting point of tin is 232°C, its latent heat of fusion is 14 cal/g and the specific heat of solid and molten tin are 0·055 and 0·064 cal/g °C respectively. Calculate the change in entropy when 1·0 gm of tin is heated from 100°C to 300°C.

(ii) Calculate the efficiency of an engine having compression ratio 13·8 and expansion ratio 6 and working on diesel cycle. Given γ = 1·4. (10+5 marks)

(b) (i) Write the expression for the Fermi-Dirac distribution. Plot the Fermi-Dirac distribution at T = 0 and for T₁ > T₂ > 0. Now from the plot propose two alternative definitions of the Fermi level.

(ii) Calculate the probability of an electron occupying an energy level 0·02 eV above the Fermi level at T = 300 K. (15+5 marks)

(c) Given an infinite line charge of charge density 2 nCm⁻¹ parallel to the y-axis and passing through the point (3, 0, 4) m and an infinite sheet of charge of charge density 4 nCm⁻² parallel to the x-y plane and passing through the point (0, 0, 6) m. Calculate the electric field intensity at the point (10, 10, 10) m. Assume free space. (15 marks)

UPSC Answer Check · Accepted Answer

(a)(i) Entropy is a state function, so evaluate ΔS along a reversible path: heat solid to melting point, melt it, heat liquid to 300°C. T₁ = 100 + 273.15 = 373.15 K, T_m = 232 + 273.15 = 505.15 K, T₂ = 300 + 273.15 = 573.15 K. For m = 1.0 g: ΔS₁ = ∫_T₁^T_m m c_s dT/T = m c_s ln(T_m/T₁) = 1.0 × 0.055 × ln(505.15/373.15) = 0.01666 cal/K. ΔS₂ = m L/T_m = 1.0 × 14/505.15 = 0.02771 cal/K. ΔS₃ = ∫_T_m^T₂ m c_l dT/T = m c_l ln(T₂/T_m) = 1.0 × 0.064 × ln(573.15/505.15) = 0.00808 cal/K. Total entropy change: ΔS = 0.01666 + 0.02771 + 0.00808 = 0.05245 cal/K = 0.05245 × 4.184 = 0.2195 J/K. Condition: pressure constant and phase changes reversible. Final answer: ΔS ≈ 0.0525 cal/K = 0.220 J/K. (a)(ii) For an ideal air-standard Diesel cycle, let compression ratio r = V₁/V₂ = 13.8 and expansion ratio R = V₄/V₃ = 6. Then cut-off ratio ρ = V₃/V₂ = r/R = 13.8/6 = 2.3. Efficiency: η = 1 − [1/(γ r^(γ−1))] × [(ρ^γ − 1)/(ρ − 1)]. Given γ = 1.4: r^(γ−1) = 13.8^0.4 = 2.8573, ρ^γ = 2.3^1.4 = 3.2094. Thus: η = 1 − [(3.2094 − 1)/(1.4 × 2.8573 × (2.3 − 1))] = 1 − [2.2094/(1.4 × 2.8573 × 1.3)] = 1 − 0.42486 = 0.57514. So η ≈ 57.5%. (b)(i) The Fermi-Dirac distribution is: f(E) = 1/[exp((E − E_F)/(k_B T)) + 1], where f(E) is the probability that a state of energy E is occupied. Plot description: take E on the horizontal axis and f(E) on the vertical axis. At T = 0, the curve is a step: f(E) = 1 for E < E_F, f(E) = 0 for E > E_F, and conventionally f(E_F) = 1/2. For T > 0, the step becomes smooth. At E = E_F, f(E_F) = 1/2 for every finite T > 0. For T₁ > T₂ > 0, the curve for T₁ is more spread out; it lies above the T₂ curve for E > E_F and below it for E < E_F. The rounding width is of order k_B T. Two alternative definitions of the Fermi level from the plot: The energy at which the occupation probability is exactly 1/2 at any finite temperature T > 0. The limiting energy at T → 0 at which the occupation drops from 1 to 0, i.e. the energy separating occupied and unoccupied states. (b)(ii) Given ΔE = E − E_F = 0.02 eV, T = 300 K. k_B T = 8.617 × 10⁻⁵ × 300 = 0.02585 eV. Then: f = 1/[exp(ΔE/k_B T) + 1] = 1/[exp(0.02/0.02585) + 1] = 1/[exp(0.7736) + 1] = 1/(2.1676 + 1) = 0.3157. So the probability is f ≈ 0.316 = 31.6%. (c) Use superposition of the fields due to the infinite line charge and the infinite sheet charge. For the line charge: λ = 2 nC/m = 2 × 10⁻⁹ C/m. It is parallel to the y-axis and passes through (3, 0, 4) m. At P = (10, 10, 10) m, the perpendicular displacement from the line in the x-z plane is: r = (10 − 3)î + (10 − 4)k̂ = 7î + 6k̂. Distance d = √(7² + 6²) = √85 m. By Gauss’s law for an infinite line charge: E_line = λ/(2π ε₀ d) r̂ = λ/(2π ε₀ d²) r = [2 × 10⁻⁹/(2π ε₀ × 85)](7î + 6k̂). Using 1/(4π ε₀) = 8.9876 × 10⁹ N m²/C², 1/(2π ε₀) = 1.7975 × 10¹⁰. Coefficient = (2 × 10⁻⁹ × 1.7975 × 10¹⁰)/85 = 0.42294. Thus: E_line = 0.42294(7î + 6k̂) = 2.9606î + 2.5377k̂ N/C. For the sheet: σ = 4 nC/m² = 4 × 10⁻⁹ C/m², parallel to the x-y plane at z = 6 m. The point P has z = 10 m, so it lies above the sheet. The field points away from the positive sheet, i.e. in +k̂ direction: E_sheet = σ/(2ε₀) k̂ = [4 × 10⁻⁹/(2 × 8.854 × 10⁻¹²)] k̂ = 225.88 k̂ N/C. Total field: E = E_line + E_sheet = 2.9606î + 0ĵ + (2.5377 + 225.88)k̂ = 2.96î + 228.42k̂ N/C. Magnitude: |E| = √(2.96² + 228.42²) = 228.44 N/C.

(a) (i) The melting point of tin is 232°C, its latent heat of fusion is 14 cal/g and the specific heat of solid and molten tin…

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