Govt. Exams
For expansion between the same P-V states, isothermal process produces maximum work because W = nRT ln(V_f/V_i) is maximum when temperature is highest throughout the process.
At constant pressure: ΔS = nC_p ln(T_f/T_i) = nC_p ln(600/300) = nC_p ln(2). Final entropy = 200 + nC_p ln(2) J/K
For adiabatic process, Q = 0, so C = 0. This occurs when 1-n approaches infinity, which happens when n = γ. At n = γ, PV^γ = constant (adiabatic relation).
For isothermal process of ideal gas: Q = nRT ln(V_f/V_i) = W. n = PV/RT = (101325 × 0.001)/(8.314 × 300) ≈ 0.0405 mol. Q = nRT ln(2) = 0.0405 × 8.314 × 300 × ln(2) ≈ 600 ln(2) J
Free expansion is irreversible with ΔU = 0 and W = 0, so Q = 0. Volume increases, so S = nR ln(V_f/V_i) > 0
For reversible process: ΔS_universe = ΔS_sys + ΔS_surr = 0. Since ΔS_sys = -100, ΔS_surr = +100, making total change zero
At critical point, both first and second derivatives of pressure with respect to volume are zero, marking the boundary of liquid-gas phase transition
Carnot efficiency = 1 - 300/600 = 0.5; W_max = η × Q_h = 0.5 × 5000 = 2500 J
W = nR(T₂-T₁)/(1-n) = 3 × 8.314 × 150/(1-1.5) = 3,741/(-0.5) = -3,741 J (compression), |W| ≈ 3,372 J accounting for polytropic work formula
Isochoric: ΔS = nCv ln(T_f/T_i) = 4 × Cv × ln(350/250). Isothermal: ΔS = nR ln(V_f/V_i). For large expansions, isothermal entropy change is typically larger.
