PU CET Lahore (Engineering & CS) Physics Heat & Thermodynamics — Set 3

PU CET Lahore (Engineering & CS) Physics Heat & Thermodynamics — Set 3

1. Question 1

   Q1. The coefficient of performance 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_hot / T_cold

   Answer: T_cold / (T_hot - T_cold)

   Explanation: COP = Q_cold / W = T_cold / (T_hot - T_cold), derived from the definition of COP and the Carnot refrigerator.

2. Question 2

   Q2. A gas expands isothermally. The work done is maximum when

   • A) The gas is ideal
   • B) The gas is real
   • C) The expansion is reversible
   • D) The expansion is irreversible

   Answer: The expansion is reversible

   Explanation: Maximum work is done in a reversible isothermal expansion, a fundamental principle in thermodynamics.

3. Question 3

   Q3. A Carnot engine operates between 300 K and 600 K. Its efficiency is

   • A) 50%
   • B) 60%
   • C) 40%
   • D) 30%

   Answer: 50%

   Explanation: Efficiency = 1 - (T_sink / T_source) = 1 - (300 / 600) = 0.5 or 50%.

4. Question 4

   Q4. The ratio of the specific heats (Cp/Cv) for a diatomic gas is

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

   Answer: 1.4

   Explanation: For a diatomic gas, Cp/Cv = 7/5 = 1.4, derived from the equipartition theorem.

5. Question 5

   Q5. The change in entropy when a hot body is brought into contact with a cold body is

   • A) Positive
   • B) Negative
   • C) Zero
   • D) Depends on the bodies

   Answer: Positive

   Explanation: When a hot body is brought into contact with a cold body, the total entropy increases, as the process is irreversible.

6. Question 6

   Q6. The temperature of a body is 300 K. Its temperature in Celsius is

   • A) 27°C
   • B) 273°C
   • C) 300°C
   • D) 573°C

   Answer: 27°C

   Explanation: T(°C) = T(K) - 273.15 = 300 - 273.15 = 26.85 ≈ 27°C.

7. Question 7

   Q7. A heat engine operates between 1000 K and 500 K. The maximum possible efficiency is

   • A) 50%
   • B) 40%
   • C) 60%
   • D) 30%

   Answer: 50%

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

8. Question 8

   Q8. The entropy of a system is a measure of its

   • A) Internal energy
   • B) Disorder or randomness
   • C) Temperature
   • D) Pressure

   Answer: Disorder or randomness

   Explanation: Entropy is a measure of the disorder or randomness of a system, a fundamental concept in thermodynamics.

9. Question 9

   Q9. A thermodynamic process is reversible if

   • A) It is isothermal
   • B) It is adiabatic
   • C) It is quasi-static and there is no dissipation
   • D) It is isobaric

   Answer: It is quasi-static and there is no dissipation

   Explanation: A reversible process is one that is quasi-static and has no dissipation, a key concept in thermodynamics.

10. Question 10

    Q10. A gas is compressed isothermally. The work done on the gas is 200 J. The change in internal energy is

    • A) 200 J
    • B) 0 J
    • C) -200 J
    • D) 100 J

    Answer: 0 J

    Explanation: In an isothermal process, the internal energy of an ideal gas remains constant, so ΔU = 0.

11. Question 11

    Q11. The root mean square speed of the molecules of a gas is proportional to

    • A) √T
    • B) T
    • C) 1 / √T
    • D) 1 / T

    Answer: √T

    Explanation: The rms speed is given by √(3RT/M), so it is proportional to √T.

12. Question 12

    Q12. A system undergoes a reversible adiabatic expansion. The entropy change is

    • A) Positive
    • B) Negative
    • C) Zero
    • D) Undefined

    Answer: Zero

    Explanation: In a reversible adiabatic process, ΔS = 0 because Q = 0 and the process is reversible.

13. Question 13

    Q13. A gas expands against a constant external pressure of 1 atm. If the volume change is 2 L, the work done by the gas is

    • A) -2 L atm
    • B) 2 L atm
    • C) -1 L atm
    • D) 1 L atm

    Answer: 2 L atm

    Explanation: W = P * ΔV = 1 atm * 2 L = 2 L atm.

14. Question 14

    Q14. The coefficient of performance of a refrigerator is 5. If the temperature inside the refrigerator is 250 K, the temperature outside is

    • A) 300 K
    • B) 320 K
    • C) 280 K
    • D) 310 K

    Answer: 300 K

    Explanation: COP = T_cold / (T_hot - T_cold), so 5 = 250 / (T_hot - 250), giving T_hot = 300 K.

15. Question 15

    Q15. The molar specific heat capacity at constant volume for a monatomic gas is

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

    Answer: 3/2 R

    Explanation: For a monatomic gas, Cv = (3/2)R.

16. Question 16

    Q16. A thermodynamic process is represented by a straight line on a P-V diagram. The process is

    • A) Isobaric
    • B) Isovolumetric
    • C) Isothermal
    • D) Adiabatic

    Answer: Isothermal

    Explanation: A straight line on a P-V diagram represents an isothermal process for an ideal gas.

17. Question 17

    Q17. The change in entropy of a system is given by ΔS = Q / T. This equation is valid for

    • A) Reversible processes
    • B) Irreversible processes
    • C) All thermodynamic processes
    • D) None of the above

    Answer: Reversible processes

    Explanation: ΔS = Q / T is valid for reversible processes.

18. Question 18

    Q18. A gas is heated at constant pressure. The fraction of heat energy used to do work is

    • A) R / Cp
    • B) Cp / R
    • C) R / Cv
    • D) Cv / R

    Answer: R / Cp

    Explanation: Fraction of heat used to do work = W / Q = (nRΔT) / (nCpΔT) = R / Cp.

19. Question 19

    Q19. The ratio of the specific heat capacities of a gas is 1.4. The gas is

    • A) Monatomic
    • B) Diatomic
    • C) Polyatomic
    • D) None of the above

    Answer: Diatomic

    Explanation: For a diatomic gas, Cp / Cv = 1.4 (or 7/5).

20. Question 20

    Q20. The work done in an adiabatic expansion is 100 J. The change in internal energy is

    • A) -100 J
    • B) 100 J
    • C) 0 J
    • D) 50 J

    Answer: -100 J

    Explanation: In an adiabatic process, Q = 0, so ΔU = -W = -100 J.