JSMU Karachi Entry Test Physics Thermodynamics — Set 2

JSMU Karachi Entry Test Physics Thermodynamics — Set 2

1. Question 1

   Q1. A heat engine operates between 227°C and 27°C. What is its maximum possible efficiency?

   • A) 20%
   • B) 40%
   • C) 50%
   • D) 60%

   Answer: 40%

   Explanation: Carnot efficiency = 1 - (T_cold / T_hot) = 1 - (300 / 500) = 0.4 or 40%. Other options are incorrect temperatures or calculations.

2. Question 2

   Q2. At constant volume, 200 J of heat is added to 2 moles of an ideal gas at 300 K. What is the change in internal energy?

   • A) 100 J
   • B) 200 J
   • C) 300 J
   • D) 400 J

   Answer: 200 J

   Explanation: ΔU = Q at constant volume. Q = 200 J, so ΔU = 200 J. Other options misinterpret the first law of thermodynamics.

3. Question 3

   Q3. A refrigerator's coefficient of performance is 5. How much heat is removed from the cold body if 1 kJ of work is done?

   • A) 2 kJ
   • B) 5 kJ
   • C) 10 kJ
   • D) 0.2 kJ

   Answer: 5 kJ

   Explanation: COP = Q_c / W, so Q_c = COP * W = 5 * 1 kJ = 5 kJ. Other options miscalculate COP or its implications.

4. Question 4

   Q4. For an adiabatic expansion of an ideal gas, what happens to the internal energy?

   • A) Increases
   • B) Decreases
   • C) Remains constant
   • D) Becomes zero

   Answer: Decreases

   Explanation: In adiabatic expansion, Q = 0, and W is done by the gas, so ΔU = -W, decreasing internal energy. Other options misinterpret adiabatic process.

5. Question 5

   Q5. The specific heat capacity of water is 4200 J/kg°C. How much heat is required to raise the temperature of 2 kg of water by 5°C?

   • A) 21000 J
   • B) 42000 J
   • C) 84000 J
   • D) 168000 J

   Answer: 42000 J

   Explanation: Q = mcΔT = 2 kg * 4200 J/kg°C * 5°C = 42000 J. Other options miscalculate or use wrong specific heat capacity.

6. Question 6

   Q6. In a cyclic process, the net change in internal energy is

   • A) Always positive
   • B) Always negative
   • C) Zero
   • D) Dependent on the path

   Answer: Zero

   Explanation: In a cyclic process, the system returns to its initial state, so ΔU = 0. Other options misunderstand the definition of a cyclic process.

7. Question 7

   Q7. The efficiency of a Carnot engine is 0.5. If the temperature of the hot reservoir is 500 K, what is the temperature of the cold reservoir?

   • A) 250 K
   • B) 300 K
   • C) 350 K
   • D) 400 K

   Answer: 250 K

   Explanation: η = 1 - (T_c / T_h), so T_c = T_h * (1 - η) = 500 K * (1 - 0.5) = 250 K. Other options miscalculate or misunderstand Carnot efficiency.

8. Question 8

   Q8. For a given isothermal expansion, the work done by an ideal gas is maximum when

   • A) Pressure is high
   • B) Pressure is low
   • C) Volume is small
   • D) Volume is large

   Answer: Pressure is low

   Explanation: Work done in isothermal expansion is given by W = nRT ln(V_f / V_i). For given V_i and V_f, work is maximum when pressure (and thus density) is low, allowing more expansion. Other options misunderstand the conditions for maximum work.

9. Question 9

   Q9. A system does 500 J of work and loses 200 J of heat. What is the change in its internal energy?

   • A) -700 J
   • B) -300 J
   • C) 300 J
   • D) 700 J

   Answer: -700 J

   Explanation: ΔU = Q - W = -200 J - 500 J = -700 J. Other options misapply the first law of thermodynamics.

10. Question 10

    Q10. The molar specific heat capacity at constant volume for an ideal monatomic gas is

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

    Answer: (3/2)R

    Explanation: For a monatomic ideal gas, C_v = (3/2)R. Other options are specific heats for different types of gases or processes.

11. Question 11

    Q11. During an adiabatic compression, the temperature of an ideal gas

    • A) Decreases
    • B) Remains constant
    • C) Increases
    • D) Becomes zero

    Answer: Increases

    Explanation: In adiabatic compression, work is done on the gas, increasing its internal energy and thus temperature. Other options misinterpret adiabatic compression.

12. Question 12

    Q12. The entropy change in a reversible adiabatic process 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. Other options misunderstand the definition of entropy change.

13. Question 13

    Q13. A heat engine operates between two temperatures, 500 K and 300 K. What is its Carnot efficiency?

    • A) 0.2
    • B) 0.4
    • C) 0.6
    • D) 0.8

    Answer: 0.4

    Explanation: η = 1 - (T_c / T_h) = 1 - (300 / 500) = 0.4. Other options miscalculate or misunderstand Carnot efficiency.

14. Question 14

    Q14. For an ideal gas, the internal energy depends on

    • A) Volume only
    • B) Pressure only
    • C) Temperature only
    • D) Both volume and temperature

    Answer: Temperature only

    Explanation: For an ideal gas, internal energy is a function of temperature only. Other options misunderstand the properties of an ideal gas.

15. Question 15

    Q15. In an isobaric process, the pressure remains

    • A) Constant
    • B) Increasing
    • C) Decreasing
    • D) Zero

    Answer: Constant

    Explanation: By definition, in an isobaric process, the pressure remains constant. Other options misinterpret the meaning of isobaric.

16. Question 16

    Q16. The work done in an isochoric process is

    • A) Maximum
    • B) Minimum
    • C) Zero
    • D) Undefined

    Answer: Zero

    Explanation: In an isochoric process, volume remains constant, so W = 0. Other options misunderstand the definition of an isochoric process.

17. Question 17

    Q17. For a reversible isothermal expansion of an ideal gas, the change in entropy is given by

    • A) nR ln(V_f / V_i)
    • B) nR ln(V_i / V_f)
    • C) nR ln(P_f / P_i)
    • D) nR ln(P_i / P_f)

    Answer: nR ln(V_f / V_i)

    Explanation: ΔS = Q / T = nRT ln(V_f / V_i) / T = nR ln(V_f / V_i). Other options misapply the formula for entropy change.

18. Question 18

    Q18. A Carnot refrigerator operates between 250 K and 300 K. What is its coefficient of performance?

    • A) 4
    • B) 5
    • C) 6
    • D) 7

    Answer: 5

    Explanation: COP = T_c / (T_h - T_c) = 250 / (300 - 250) = 5. Other options miscalculate or misunderstand the COP of a refrigerator.

19. Question 19

    Q19. A gas expands isothermally from 2L to 4L at 300K. What is the work done if there are 2 moles of gas?

    • A) 2300 J
    • B) 3456 J
    • C) -3456 J
    • D) 0 J

    Answer: -3456 J

    Explanation: Work done = nRT ln(Vf/Vi) = 2 * 8.314 * 300 * ln(4/2) = 3456 J, negative because it's expansion.

20. Question 20

    Q20. A heat engine operates between 500K and 300K. What is its maximum possible efficiency?

    • A) 20%
    • B) 40%
    • C) 60%
    • D) 80%

    Answer: 40%

    Explanation: Efficiency = 1 - (T_cold/T_hot) = 1 - (300/500) = 0.4 or 40%. Other options are incorrect temperatures or calculations.