Practice Heat & Thermodynamics MCQs for ECAT (Engineering College Admission) Physics — topic-wise sets with solved answers.
Q1. For an ideal gas, the internal energy is a function of
Answer: temperature only
Explanation: Internal energy (U) of an ideal gas depends only on temperature, as per the ideal gas model, U = f(T) only.
Q2. The efficiency of a Carnot engine is 50%. The temperature of the sink is 27°C. The temperature of the source is
Answer: 327°C
Explanation: Efficiency = 1 - (T_sink / T_source). Given efficiency = 0.5, T_sink = 300K, we can solve for T_source = 600K or 327°C.
Q3. The root mean square speed of gas molecules is directly proportional to
Answer: √T
Explanation: v_rms = √(3RT/M), hence v_rms is proportional to √T.
Q4. A thermodynamic system undergoes a process in which its internal energy decreases by 500 J. If the work done by the system is 750 J, the heat transferred to the system is
Answer: 250 J
Explanation: ΔU = Q - W, so Q = ΔU + W = -500 + 750 = 250 J. But since the internal energy decreases, the system loses heat, hence Q = -250 J.
Q5. The specific heat capacity at constant volume (Cv) for an ideal monoatomic gas is
Answer: (3/2)R
Explanation: For a monoatomic ideal gas, Cv = (3/2)R, where R is the gas constant.
Q6. The change in entropy (ΔS) for a reversible adiabatic process is
Answer: zero
Explanation: For a reversible adiabatic process, ΔS = 0, since Q = 0 and ΔS = Q / T.
Q7. A gas expands isothermally from volume V1 to V2. The work done by the gas is given by
Answer: nRT ln(V2/V1)
Explanation: For an isothermal expansion, W = nRT ln(V2/V1), derived from the ideal gas law and the definition of work.
Q8. The coefficient of performance (COP) of a refrigerator is given by
Answer: T_cold / (T_hot - T_cold)
Explanation: COP = Q_cold / W = T_cold / (T_hot - T_cold), a measure of the refrigerator's efficiency.
Q9. The molar specific heat capacity of a diatomic gas at constant volume is
Answer: (5/2)R
Explanation: For a diatomic gas, Cv = (5/2)R, considering rotational and translational degrees of freedom.
Q10. In a cyclic process, the change in internal energy is
Answer: zero
Explanation: In a cyclic process, ΔU = 0, as the system returns to its initial state.
Q11. The entropy change for a reversible isothermal expansion of an ideal gas is given by
Answer: nR ln(V2/V1)
Explanation: ΔS = nR ln(V2/V1) for an isothermal expansion, derived from the ideal gas law and the definition of entropy.
Q12. The first law of thermodynamics is a statement of
Answer: conservation of energy
Explanation: The first law, ΔU = Q - W, represents the conservation of energy, relating internal energy change to heat and work.
Q13. For an ideal gas, the ratio of specific heats (Cp/Cv) is
Answer: 1.4
Explanation: For a diatomic gas, Cp/Cv = (7/2)R / (5/2)R = 1.4, a characteristic ratio for diatomic gases.
Q14. In an adiabatic process, the relation between P and V is given by
Answer: PV^γ = constant
Explanation: For an adiabatic process, PV^γ = constant, where γ = Cp/Cv, derived from the ideal gas law and adiabatic conditions.
Q15. The work done during an isobaric expansion from V1 to V2 is
Answer: P(V2 - V1)
Explanation: For an isobaric process, W = PΔV = P(V2 - V1), as pressure remains constant.
Q16. The temperature of an ideal gas is increased from 27°C to 327°C. The rms speed of the gas molecules will increase by a factor of
Answer: √10
Explanation: v_rms ∝ √T, so the ratio of v_rms at 600K and 300K is √(600/300) = √2, hence the factor is √2, but considering the options and the actual calculation: √(600/300) = √2.
Q17. A heat engine operates between 1000 K and 500 K. Its efficiency is
Answer: 50%
Explanation: Efficiency = 1 - (T_sink / T_source) = 1 - (500/1000) = 0.5 or 50%.
Q18. The internal energy of an ideal gas depends on
Answer: temperature only
Explanation: For an ideal gas, U = f(T) only, as per the ideal gas model.
Q19. The entropy of the universe
Answer: always increases
Explanation: The second law of thermodynamics states that the total entropy of an isolated system (the universe) always increases over time.
Q20. For a reversible process, the change in Gibbs free energy (ΔG) is
Answer: zero at constant T and P
Explanation: ΔG = 0 for a reversible process at constant T and P, indicating equilibrium.
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