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.
A system absorbs 500 J of heat and does 200 J of work on surroundings. The change in internal energy is:
Answer: A
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.
Q.122Easy
At constant temperature and pressure, which of the following represents the Gibbs free energy change for a spontaneous process?
Answer: B
For a spontaneous process at constant T and P, ΔG must be negative. ΔG = 0 indicates equilibrium, and ΔG > 0 indicates non-spontaneous process.
Q.123Easy
The Clausius-Clapeyron equation relates vapor pressure to temperature. Which statement is correct?
Answer: A
Clausius-Clapeyron equation (d ln P/dT = ΔH_vap/RT²) applies specifically to phase equilibria and shows direct relationship between vapor pressure and temperature.
Q.124Easy
For an ideal gas undergoing adiabatic compression, the entropy change is:
Answer: C
For a reversible adiabatic process, dq = 0, therefore ΔS = ∫dq_rev/T = 0. Entropy remains constant during reversible adiabatic processes.
Q.125Easy
In a constant pressure process, the heat absorbed by a system equals:
Answer: B
At constant pressure, q_p = ΔH (change in enthalpy). This is the definition of enthalpy and is a key relationship in engineering thermodynamics.
Q.126Easy
For a binary ideal solution, the total vapor pressure at constant T is given by:
Answer: A
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.
Q.127Easy
During a constant volume process, the first law of thermodynamics simplifies to:
Answer: B
At constant volume, ΔV = 0, so w = -P∫dV = 0. Therefore, ΔU = q + w = q + 0 = q. All heat goes into internal energy change.
Q.128Easy
A gas undergoes an isothermal expansion from 2 L to 5 L at 298 K. If the process is reversible, what is the sign of entropy change for an ideal gas?
Answer: A
For isothermal expansion of an ideal gas, ΔS = nR ln(V_f/V_i) = nR ln(25) > 0. Volume increases, so entropy increases.
Q.129Easy
In the van der Waals equation, the term 'a' represents:
Answer: B
In (P + a/V²)(V - b) = RT, 'a' corrects for intermolecular forces reducing pressure, while 'b' corrects for molecular volume.
