Chemical Engineering questions for GATE and PSU exams are built on a handful of core subjects applied in many ways. Practice spans fluid mechanics, heat transfer, mass transfer, chemical reaction engineering, thermodynamics, process control and instrumentation, and plant design economics. Numerical solutions carry the assumptions written out, because the assumption is usually what separates a correct answer from a plausible one.
Which thermodynamic potential is most useful for constant temperature and pressure processes?
Answer: C
Gibbs Free Energy (G = H - TS) is the appropriate thermodynamic potential for processes at constant T and P. ΔG = 0 at equilibrium and ΔG < 0 for spontaneous processes.
Q.2Medium
For an ideal gas undergoing isothermal expansion, the work done is given by:
Answer: A
For isothermal process of ideal gas, W = nRTln(V₂/V₁) = nRTln(P₁/P₂). This is derived from the first law with ΔU = 0 for isothermal ideal gas process.
Q.3Medium
For a process where ΔG < 0 at all temperatures, the process must be:
Answer: B
From ΔG = ΔH - TΔS, for ΔG < 0 at all T: ΔH < 0 (exothermic) and ΔS > 0 (entropy increases). This is a spontaneous process at all temperatures.
Q.4Medium
The heat of vaporization of water is 40.66 kJ/mol at 373 K. The entropy of vaporization is approximately:
Answer: C
ΔS_vap = ΔH_vap/T = 40660 J/mol / 373 K ≈ 109 J/mol·K. This follows Trouton's rule (~85-105 J/mol·K for most liquids).
Q.5Medium
For a reversible process, the Clausius inequality states:
Answer: B
For reversible processes, ΔS = Q_rev/T. For irreversible processes, ΔS > Q_irrev/T. This is the Clausius inequality: dS ≥ dQ/T.
Q.6Medium
A cyclic heat engine operates between hot reservoir at 500 K and cold reservoir at 300 K. Maximum theoretical efficiency is:
Answer: B
Maximum efficiency is Carnot efficiency: η_max = 1 - T_cold/T_hot = 1 - 500300 = 0.40 = 40%. No heat engine can exceed this efficiency.
Q.7Medium
The heat capacity at constant pressure Cₚ is always greater than heat capacity at constant volume Cᵥ because:
Answer: B
Cₚ - Cᵥ = R (for ideal gas). At constant P, supplied heat does both internal energy and expansion work. At constant V, all heat goes to internal energy only.
Q.8Medium
For a real gas with van der Waals equation, the constants 'a' and 'b' represent:
Answer: B
In van der Waals equation (P + a/V²)(V - b) = RT, 'a' accounts for intermolecular attractive forces and 'b' represents excluded molecular volume. Both are positive constants.
Q.9Medium
For a spontaneous process at constant T and P, which condition must be satisfied?
Answer: C
At constant T and P, spontaneity is determined by Gibbs free energy: ΔG < 0 for spontaneous process, ΔG = 0 for equilibrium, ΔG > 0 for non-spontaneous process.
Q.10Medium
A reversible adiabatic process for an ideal gas follows PVᵞ = constant. If γ = 1.4 and initial pressure is 1 atm with volume 1 L, what is the final pressure when volume becomes 0.5 L?
For ideal gases, f = P (fugacity equals pressure), so φ = f/P = 1. Real gases have φ ≠ 1
Q.16Medium
For a spontaneous process occurring at constant temperature and pressure, which condition must be satisfied?
Answer: B
For spontaneity at constant T and P: ΔG = ΔH - TΔS must be negative (ΔG < 0)
Q.17Medium
A throttle valve is used in a refrigeration cycle. This is an example of a(n) _____ process.
Answer: C
Throttling is an adiabatic (Q=0) but irreversible process with no work done, causing entropy increase
Q.18Medium
A gas mixture at 298 K contains H₂ and N₂. If the mixture obeys Amagat's law and the partial volumes are equal, what is the mole fraction of H₂?
Answer: A
Amagat's law: V_total = V_H₂ + V_N₂. If partial volumes are equal, each is 50%, so x_H₂ = 0.5
Q.19Medium
At the critical point of a substance, which of the following is true?
Answer: A
At the critical point, surface tension between liquid and gas phases vanishes because the distinction between phases disappears. The critical compressibility factor Zc ≈ 0.27 (not 1).
Q.20Medium
For an ideal gas undergoing isothermal expansion from V₁ to V₂, the entropy change is:
Answer: A
For isothermal process: dS = dq_rev/T = nR dV/V, integrating gives ΔS = nR ln(V₂/V₁). Temperature is constant, so entropy change depends only on volume change.
