Chemical Engineering - MCQ Practice Questions

First Law: ΔU = Q - W. Q = +500 J (absorbed), W = +200 J (work by system). ΔU = 500 - 200 = 300 J. Positive indicates internal energy increases.

For a spontaneous process at constant T and P, ΔG must be negative. ΔG = 0 indicates equilibrium, and ΔG > 0 indicates non-spontaneous process.

Clausius-Clapeyron equation (d ln P/dT = ΔH_vap/RT²) applies specifically to phase equilibria and shows direct relationship between vapor pressure and temperature.

For a reversible adiabatic process, dq = 0, therefore ΔS = ∫dq_rev/T = 0. Entropy remains constant during reversible adiabatic processes.

At constant pressure, q_p = ΔH (change in enthalpy). This is the definition of enthalpy and is a key relationship in engineering thermodynamics.

Raoult's law states P_i = P_i°x_i for ideal solutions. Total pressure P = P₁°x₁ + P₂°x₂. This assumes ideal mixing behavior.

At constant volume, ΔV = 0, so w = -P∫dV = 0. Therefore, ΔU = q + w = q + 0 = q. All heat goes into internal energy change.

For isothermal expansion of an ideal gas, ΔS = nR ln(V_f/V_i) = nR ln($\frac{5}{2}$) > 0. Volume increases, so entropy increases.

In (P + a/V²)(V - b) = RT, 'a' corrects for intermolecular forces reducing pressure, while 'b' corrects for molecular volume.

Clausius-Clapeyron equation: ln(P₂/P₁) = -(ΔH_vap/R)(1/T₂ - 1/T₁) describes phase equilibrium.

At phase equilibrium, μ_liquid = μ_vapor = μ_solid. This equality determines equilibrium conditions.

ΔG < 0 indicates spontaneous, irreversible process under constant T and P conditions.

First law: ΔU = Q - W = 500 - 300 = 200 J (using convention W = work by system).