Govt. Exams
Entrance Exams
Efficiency η = W/Q_h = 0.4. Work done W = 0.4 × 1000 = 400 J. Heat rejected Q_c = Q_h - W = 1000 - 400 = 600 J
At equilibrium, ΔG = 0. When ΔG < 0, reaction is spontaneous; when ΔG > 0, reaction is non-spontaneous.
Carnot efficiency = 1 - (T_cold/T_hot) = 1 - (300/400) = 1 - 0.75 = 0.25 = 25%
Internal energy depends only on initial and final states, not on the path. Heat and work are path functions.
By first law of thermodynamics: ΔU = q - w = 500 - 200 = 300 J (where w is work done by the system)
Efficiency η = W/Q_in = (Q_in - Q_out)/Q_in = (500 - 300)/500 = 200/500 = 0.40 = 40%
In an adiabatic process, Q = 0 (no heat transfer). From first law: ΔU = Q - W = 0 - W = -W, so ΔU = -W.
For an isothermal reversible process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁). Since V₂ > V₁, work is positive (work done by the gas).
Heat and work are path functions because their values depend on the path followed during the process. Internal energy, enthalpy, and Gibbs free energy are state functions.
Using first law: ΔU = Q - W, so Q = ΔU + W = -50 + 30 = -20 J. Heat is released by the system.
