Entrance Exams
Govt. Exams
For isothermal process: ΔS = Q/T = 5000/300 = 16.67 J/K. Change in entropy is positive as heat is absorbed.
By Dalton's law: P_N2 = x_N2 × P_total = 0.40 × 5 = 2 bar
At the triple point, solid, liquid, and vapor phases are in thermodynamic equilibrium. For water, this occurs at 273.16 K and 611.657 Pa.
T1 = 298 K. Isentropic: T2s = T1(P2/P1)^((γ-1)/γ) = 298 × 8^(0.4/1.4) ≈ 298 × 1.741 ≈ 520 K. Actual: T2 = T1 + (T2s - T1)/η_is = 298 + (520-298)/0.8 ≈ 465 K
For adiabatic process: T2/T1 = (P2/P1)^((γ-1)/γ) or T2 = T1 × r^(-(γ-1)) = 850 × 8^(-0.4/1.4) ≈ 850 × 0.485 ≈ 412 K
The Clausius statement of second law: For isolated systems, dS_universe ≥ 0. dS = 0 for reversible, dS > 0 for irreversible processes.
At 80 kPa, saturated vapor properties are h_g ≈ 2660.3 kJ/kg and s_g ≈ 7.31 kJ/kg·K. Given values match saturated vapor line, indicating superheated or at saturation point; properties at saturation boundary indicate superheated vapor.
μ_JT = (∂T/∂P)_H. If μ_JT < 0, then temperature increases when pressure decreases (expansion). Negative μ_JT occurs above inversion temperature.
COP = Q_h / W = 5. Therefore, W = Q_h / COP = 1000 / 5 = 200 J
Work W = ∫P dV. In isochoric process, dV = 0, therefore W = 0 regardless of pressure or temperature changes.