Air University Entry Test Physics: Heat & Thermodynamics MCQs

Air University Entry Test Physics: Heat & Thermodynamics MCQs — sample questions

1. Question 1

   Q1. For an ideal gas, the internal energy is a function of

   • A) temperature only
   • B) pressure only
   • C) volume only
   • D) both temperature and pressure

   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.

2. Question 2

   Q2. The efficiency of a Carnot engine is 50%. The temperature of the sink is 27°C. The temperature of the source is

   • A) 327°C
   • B) 127°C
   • C) 227°C
   • D) 300°C

   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.

3. Question 3

   Q3. The root mean square speed of gas molecules is directly proportional to

   • A) √T
   • B) √P
   • C) T
   • D) P

   Answer: √T

   Explanation: v_rms = √(3RT/M), hence v_rms is proportional to √T.

4. Question 4

   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

   • A) -1250 J
   • B) 250 J
   • C) -250 J
   • D) 1250 J

   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.

5. Question 5

   Q5. The specific heat capacity at constant volume (Cv) for an ideal monoatomic gas is

   • A) (3/2)R
   • B) (5/2)R
   • C) R
   • D) (7/2)R

   Answer: (3/2)R

   Explanation: For a monoatomic ideal gas, Cv = (3/2)R, where R is the gas constant.

6. Question 6

   Q6. The change in entropy (ΔS) for a reversible adiabatic process is

   • A) positive
   • B) negative
   • C) zero
   • D) dependent on the process

   Answer: zero

   Explanation: For a reversible adiabatic process, ΔS = 0, since Q = 0 and ΔS = Q / T.

7. Question 7

   Q7. A gas expands isothermally from volume V1 to V2. The work done by the gas is given by

   • A) nRT ln(V2/V1)
   • B) nRT ln(V1/V2)
   • C) nRT(V2 - V1)
   • D) P(V2 - V1)

   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.

8. Question 8

   Q8. The coefficient of performance (COP) of a refrigerator is given by

   • A) T_cold / (T_hot - T_cold)
   • B) T_hot / (T_hot - T_cold)
   • C) (T_hot - T_cold) / T_cold
   • D) T_cold / T_hot

   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.

9. Question 9

   Q9. The molar specific heat capacity of a diatomic gas at constant volume is

   • A) (5/2)R
   • B) (3/2)R
   • C) (7/2)R
   • D) R

   Answer: (5/2)R

   Explanation: For a diatomic gas, Cv = (5/2)R, considering rotational and translational degrees of freedom.

10. Question 10

    Q10. In a cyclic process, the change in internal energy is

    • A) positive
    • B) negative
    • C) zero
    • D) dependent on the path

    Answer: zero

    Explanation: In a cyclic process, ΔU = 0, as the system returns to its initial state.

11. Question 11

    Q11. The entropy change for a reversible isothermal expansion of an ideal gas is given by

    • A) nR ln(V2/V1)
    • B) nR ln(V1/V2)
    • C) nR(V2 - V1)
    • D) nCv ln(T2/T1)

    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.

12. Question 12

    Q12. The first law of thermodynamics is a statement of

    • A) conservation of energy
    • B) conservation of momentum
    • C) conservation of angular momentum
    • D) conservation of charge

    Answer: conservation of energy

    Explanation: The first law, ΔU = Q - W, represents the conservation of energy, relating internal energy change to heat and work.

13. Question 13

    Q13. For an ideal gas, the ratio of specific heats (Cp/Cv) is

    • A) 1.67
    • B) 1.4
    • C) 1.33
    • D) 1.29

    Answer: 1.4

    Explanation: For a diatomic gas, Cp/Cv = (7/2)R / (5/2)R = 1.4, a characteristic ratio for diatomic gases.

14. Question 14

    Q14. In an adiabatic process, the relation between P and V is given by

    • A) PV^γ = constant
    • B) PV = constant
    • C) P/V = constant
    • D) P + V = constant

    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.

15. Question 15

    Q15. The work done during an isobaric expansion from V1 to V2 is

    • A) P(V2 - V1)
    • B) P(V1 - V2)
    • C) nRT ln(V2/V1)
    • D) 0

    Answer: P(V2 - V1)

    Explanation: For an isobaric process, W = PΔV = P(V2 - V1), as pressure remains constant.

16. Question 16

    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

    • A) √2
    • B) 2
    • C) √10
    • D) 4

    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.

17. Question 17

    Q17. A heat engine operates between 1000 K and 500 K. Its efficiency is

    • A) 50%
    • B) 25%
    • C) 75%
    • D) 100%

    Answer: 50%

    Explanation: Efficiency = 1 - (T_sink / T_source) = 1 - (500/1000) = 0.5 or 50%.

18. Question 18

    Q18. The internal energy of an ideal gas depends on

    • A) temperature, pressure, and volume
    • B) temperature only
    • C) pressure and volume
    • D) pressure only

    Answer: temperature only

    Explanation: For an ideal gas, U = f(T) only, as per the ideal gas model.

19. Question 19

    Q19. The entropy of the universe

    • A) always increases
    • B) always decreases
    • C) remains constant
    • D) can increase or decrease

    Answer: always increases

    Explanation: The second law of thermodynamics states that the total entropy of an isolated system (the universe) always increases over time.

20. Question 20

    Q20. For a reversible process, the change in Gibbs free energy (ΔG) is

    • A) zero at constant T and P
    • B) positive at constant T and P
    • C) negative at constant T and P
    • D) independent of T and P

    Answer: zero at constant T and P

    Explanation: ΔG = 0 for a reversible process at constant T and P, indicating equilibrium.