Which of the following statements is correct regarding the first law of thermodynamics?
Answer: C
The first law states that dU = Q - W, where heat (Q) and work (W) are energy transfer mechanisms, not state properties. Internal energy (U) is a state property.
Q.42Medium
During an adiabatic expansion of an ideal gas, the temperature of the gas decreases. This is because:
Answer: B
In adiabatic process, Q = 0. From first law: dU = -W. During expansion, W > 0, so dU < 0, meaning internal energy and temperature decrease.
Q.43Easy
What is the SI unit of entropy?
Answer: A
Entropy is defined as dS = dQ_rev/T. Units: Joules/Kelvin = J/K. This is the fundamental SI unit for entropy.
Q.44Hard
For a closed system undergoing a reversible isothermal process, the change in Gibbs free energy (ΔG) is:
Answer: C
For isothermal process: ΔG = ΔH - TΔS = W_useful (non-PV work). This represents maximum useful work available.
Q.45Hard
A Rankine cycle (ideal steam cycle) is used in thermal power plants. Which process has the highest irreversibility?
Answer: B
Heat transfer across finite temperature differences in the boiler creates maximum entropy generation and irreversibility among the four processes.
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Q.46Easy
If the specific heat at constant volume (Cv) for a diatomic ideal gas is 5R/2, what is its specific heat at constant pressure (Cp)?
Answer: A
Using Mayer's relation: Cp - Cv = R. Therefore, Cp = 5R/2 + R = 7R/2
Q.47Medium
During throttling of a real gas through an expansion valve, which thermodynamic property remains constant?
Answer: C
Throttling is an isenthalpic process (constant enthalpy). Temperature may change for real gases due to Joule-Thomson effect.
Q.48Easy
A system undergoes a cyclic process. The net work done by the system is 150 kJ and the net heat absorbed is 150 kJ. Which statement is correct?
Answer: B
For any cycle: ΔU_cycle = 0. First law: Q_net = W_net confirms energy conservation. 150 = 150 ✓
Q.49Medium
In a turbocharger application, compressed air enters a turbine. If the isentropic efficiency of the turbine is 0.85, what does this indicate?
Answer: B
Isentropic efficiency = Actual work output / Isentropic work output = 0.85. It compares real process with ideal isentropic process.
Q.50Medium
For an ideal gas in a constant pressure (isobaric) process from state 1 to state 2, the change in specific entropy is:
Answer: A
For isobaric process: dS = dQ/T = nCp dT/T. Integrating: ΔS = nCp ln(T2/T1) or specific entropy Δs = Cp ln(T2/T1)
Q.51Easy
Which of the following processes has zero work done by/on the gas?
Answer: B
Work W = ∫P dV. In isochoric process, dV = 0, therefore W = 0 regardless of pressure or temperature changes.
Q.52Medium
A heat pump operating between 280 K and 350 K has a coefficient of performance (COP) of 5. How much work input is required to transfer 1000 J of heat to the hot reservoir?
Answer: A
COP = Q_h / W = 5. Therefore, W = Q_h / COP = 51000 = 200 J
Q.53Hard
For a real gas undergoing Joule-Thomson expansion through a throttle valve, the Joule-Thomson coefficient (μ_JT) is negative. This means:
Answer: A
μ_JT = (∂T/∂P)_H. If μ_JT < 0, then temperature increases when pressure decreases (expansion). Negative μ_JT occurs above inversion temperature.
Q.54Medium
Steam at 80 kPa with enthalpy 2660 kJ/kg and entropy 7.31 kJ/kg·K. What is the thermodynamic state of this steam?
Answer: D
At 80 kPa, saturated vapor properties are h_g ≈ 2660.3 kJ/kg and s_g ≈ 7.31 kJ/kg·K. Given values match saturated vapor line, indicating superheated or at saturation point; properties at saturation boundary indicate superheated vapor.
Q.55Easy
Which statement correctly describes the second law of thermodynamics?
Answer: B
The Clausius statement of second law: For isolated systems, dS_universe ≥ 0. dS = 0 for reversible, dS > 0 for irreversible processes.
Q.56Hard
In a diesel cycle, the expansion process (power stroke) is adiabatic. If the pressure and temperature at the end of compression are 40 bar and 850 K respectively, and the expansion ratio is 8, what is approximately the temperature at the end of expansion? (Take γ = 1.4)
Answer: A
For adiabatic process: T2/T1 = (P2/P1)^((γ-1)/γ) or T2 = T1 × r^(-(γ-1)) = 850 × 8^(-0.14.4) ≈ 850 × 0.485 ≈ 412 K
Q.57Hard
A compressor compresses air from 1 bar and 25°C to 8 bar. If the isentropic efficiency is 0.80 and the process is adiabatic, what is the actual temperature of air after compression? (γ = 1.4, R = 287 J/kg·K)
