(a) Assume that the Earth's atmosphere is pure nitrogen in thermodynamic equilibrium at a temperature of 300 K. Calculate the…

(a) To calculate the height at which the atmospheric density is half its sea-level value, we assume an isothermal atmosphere in thermodynamic equilibrium. The density ρ at height h is given by the barometric formula: ρ = ρ₀ exp(-Mgh/RT) where M is the molar mass, g is gravitational acceleration, R is the gas constant, and T is the temperature. We are given T = 300 K, M = 28 g/mol = 0.028 kg/mol, and g = 9.8 m/s². Setting ρ = ρ₀ / 2: 1/2 = exp(-Mgh/RT) Taking the natural logarithm of both sides: ln(2) = Mgh/RT implies h = RT/Mg ln(2) Substituting the values: h = (8.314 × 300)/(0.028 × 9.8) × 0.693 h = 2494.2/0.2744 × 0.693 ≈ 9089.6 × 0.693 ≈ 6300 m The height is approximately 6.30 km. This result is physically consistent with the scale height of the atmosphere.

(b) The total entropy change is the sum of the entropy change of the body and the reservoir. For the body at constant pressure, the change is: Δ S_body = Cₚ ln((T_f)/(Tᵢ)) The reservoir at constant temperature T_f gives up heat Q = Cₚ(T_f - Tᵢ), so its entropy change is: Δ Sᵣₑₛ = -Q/(T_f) = -(Cₚ(T_f - Tᵢ))/(T_f) The total entropy change is: Δ Sₜₒₜₐₗ = Cₚ ln((T_f)/(Tᵢ)) - (Cₚ(T_f - Tᵢ))/(T_f) Let x = (T_f - Tᵢ)/(T_f). Then (T_f)/(Tᵢ) = 1/(1-x). The expression becomes: (Δ Sₜₒₜₐₗ)/(Cₚ) = ln(1/(1-x)) - x = -ln(1-x) - x Using the Taylor expansion -ln(1-x) = x + (x²)/2 + (x³)/3 + …, we get: (Δ Sₜₒₜₐₗ)/(Cₚ) = (x²)/2 + (x³)/3 + … For small |x| the leading term x²/2 is positive. More generally, let f(x) = −ln(1−x) − x for x below 1. Then f(0) = 0 and f′(x) = x/(1−x), which is negative for x below 0 and positive for x above 0, so f has its minimum 0 at x = 0. Hence ΔS_total is positive for either sign of (T_f − Tᵢ)/T_f, and zero only when T_f = Tᵢ, as the second law requires.

(c) Consider two point charges q moving with velocity v⃗. The magnetic field produced by one charge at the location of the other, separated by distance d, is given by the Biot-Savart law for a moving charge: B⃗ = (μ₀)/(4π) (q (v⃗ × r̂))/(d²) The magnetic force on the second charge is F⃗ = q v⃗ × B⃗. Since the line joining the charges is perpendicular to v⃗, the magnitude of the force is: F = q v B = q v ( (μ₀)/(4π) qv/(d²) ) = (μ₀)/(4π) (q² v²)/(d²) The direction of the force is determined by the cross product. For two charges moving in the same direction, the magnetic force is attractive, acting toward each other. This is analogous to the attraction between parallel currents. The electric repulsion q²/(4πε₀d²) is larger by c²/v², so the net force is still repulsive: F_mag/F_el = v²/c².

(d) Let I₁ be the current from the 1.5 V cell, I₂ from the 2 V cell, and I₃ through the 5 Ω resistor. Applying Kirchhoff's Current Law (KCL) at the node: I₁ + I₂ = I₃ Applying Kirchhoff's Voltage Law (KVL) to the first loop (1.5 V cell and 5 Ω resistor): 1.5 = I₁(1) + I₃(5) Applying KVL to the second loop (2 V cell and 5 Ω resistor): 2 = I₂(2) + I₃(5) Substituting I₃ = I₁ + I₂:

1. 1.5 = I₁ + 5(I₁ + I₂) = 6I₁ + 5I₂
2. 2 = 2I₂ + 5(I₁ + I₂) = 5I₁ + 7I₂ Solving this system: From (1), I₁ = (1.5 - 5I₂)/6. Substituting into (2): 2 = 5((1.5 - 5I₂)/6) + 7I₂ implies 12 = 7.5 - 25I₂ + 42I₂ implies 17I₂ = 4.5 I₂ = 4.5/17 ≈ 0.265 A I₁ = (1.5 - 5(0.265))/6 = 0.175/6 ≈ 0.029 A I₃ = 0.029 + 0.265 = 0.294 A

(e) With the switch K, the 4 V source charges the circuit, and when K is opened the inductor (with its 4 Ω resistance) and the capacitor form a closed series R-L-C loop. Equivalently, for sinusoidal excitation the R-L branch is in parallel with C. The resonant frequency ω₀ is determined by the condition where the imaginary part of the admittance is zero. The undamped natural frequency is ω₀ = 1/(√LC); for this configuration the actual resonance condition is ω_r² = 1/(LC) − R²/L². Given L = 1 mH = 10⁻³ H and C = 2 mF = 2 × 10⁻³ F: ω₀ = 1/(√(10⁻³ × 2 × 10⁻³)) = 1/(√(2 × 10⁻⁶)) = 1000/(√2) ≈ 707 rad/s The Q-factor for a parallel RLC circuit (where R is in series with L) is given by Q = (ω₀ L)/R. Q = (707 × 10⁻³)/4 = 0.707/4 ≈ 0.18 Equivalently, Q = (1/R)√(L/C) = (1/4)√(0.5) ≈ 0.177. The circuit does not have a resonant frequency. For the R-L-C loop, critical damping needs R = 2√(L/C) = 1.41 Ω, and here R = 4 Ω is larger, so the loop is overdamped (Q below 1/2): the capacitor discharges without oscillation. For sinusoidal drive, R² = 16 Ω² exceeds L/C = 0.5 Ω², so 1/(LC) − R²/L² is negative and the susceptance never vanishes.