Q. In a Carnot engine, what is the efficiency dependent on?
A.
The work done
B.
The temperatures of the hot and cold reservoirs
C.
The type of working substance
D.
The volume of the gas
Show solution
Solution
The efficiency of a Carnot engine is dependent on the temperatures of the hot and cold reservoirs, given by η = 1 - (Tc/Th).
Correct Answer:
B
— The temperatures of the hot and cold reservoirs
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Q. In a Carnot engine, which of the following is true?
A.
It operates between two temperatures
B.
It is 100% efficient
C.
It can operate with any working substance
D.
It is a perpetual motion machine
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Solution
A Carnot engine operates between two temperatures, absorbing heat from a hot reservoir and rejecting heat to a cold reservoir.
Correct Answer:
A
— It operates between two temperatures
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Q. In a closed system, if 500 J of heat is added and 200 J of work is done by the system, what is the change in internal energy?
A.
300 J
B.
500 J
C.
700 J
D.
200 J
Show solution
Solution
According to the first law of thermodynamics, ΔU = Q - W. Here, ΔU = 500 J - 200 J = 300 J.
Correct Answer:
A
— 300 J
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Q. In a closed system, if the volume of the gas is doubled at constant temperature, what happens to the pressure?
A.
Doubles
B.
Halves
C.
Remains constant
D.
Increases four times
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Solution
According to Boyle's Law, at constant temperature, if the volume of a gas is doubled, the pressure is halved.
Correct Answer:
B
— Halves
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Q. In a cyclic process, the change in internal energy is:
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the path taken
Show solution
Solution
In a cyclic process, the system returns to its initial state, so the change in internal energy is zero.
Correct Answer:
C
— Zero
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Q. In a cyclic process, the change in internal energy of the system is:
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the work done
Show solution
Solution
In a cyclic process, the system returns to its initial state, so the change in internal energy is zero.
Correct Answer:
C
— Zero
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Q. In a cyclic process, the net work done by the system is equal to:
A.
The net heat added to the system
B.
The change in internal energy
C.
The heat lost by the system
D.
Zero
Show solution
Solution
In a cyclic process, the net work done by the system is zero because the system returns to its initial state.
Correct Answer:
D
— Zero
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Q. In a cyclic process, what is the net change in internal energy of the system?
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the path taken
Show solution
Solution
In a cyclic process, the system returns to its initial state, so the net change in internal energy is zero.
Correct Answer:
C
— Zero
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Q. In a cyclic process, what is the net change in internal energy?
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the process
Show solution
Solution
In a cyclic process, the system returns to its initial state, so the net change in internal energy is zero.
Correct Answer:
C
— Zero
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Q. In a heat engine, if the input heat is 800 J and the work output is 300 J, what is the efficiency?
A.
37.5%
B.
50%
C.
62.5%
D.
75%
Show solution
Solution
Efficiency = (Work output / Heat input) × 100 = (300 J / 800 J) × 100 = 37.5%.
Correct Answer:
C
— 62.5%
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Q. In a heat engine, if the work done is 200 J and the heat absorbed is 500 J, what is the efficiency?
A.
40%
B.
50%
C.
60%
D.
80%
Show solution
Solution
Efficiency = (Work done / Heat absorbed) * 100 = (200 J / 500 J) * 100 = 40%.
Correct Answer:
B
— 50%
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Q. In a heat engine, if the work output is 200 J and the heat input is 600 J, what is the efficiency?
A.
33.33%
B.
50%
C.
66.67%
D.
75%
Show solution
Solution
Efficiency = (Work output / Heat input) × 100 = (200 J / 600 J) × 100 = 33.33%.
Correct Answer:
C
— 66.67%
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Q. In a heat engine, the work done is equal to:
A.
Heat absorbed from the hot reservoir
B.
Heat rejected to the cold reservoir
C.
Heat absorbed minus heat rejected
D.
Heat absorbed plus heat rejected
Show solution
Solution
The work done by a heat engine is equal to the heat absorbed from the hot reservoir minus the heat rejected to the cold reservoir.
Correct Answer:
C
— Heat absorbed minus heat rejected
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Q. In a process where 100 J of heat is added to a system and the internal energy increases by 40 J, how much work is done by the system?
A.
60 J
B.
40 J
C.
100 J
D.
140 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W, we have 40 J = 100 J - W, thus W = 100 J - 40 J = 60 J.
Correct Answer:
A
— 60 J
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Q. In a process where 300 J of heat is added to a system and the internal energy increases by 100 J, how much work is done by the system?
A.
200 J
B.
100 J
C.
300 J
D.
400 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W. Rearranging gives W = Q - ΔU. Here, W = 300 J - 100 J = 200 J.
Correct Answer:
A
— 200 J
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Q. In a process where 300 J of heat is added to a system and the system does 100 J of work, what is the change in internal energy?
A.
200 J
B.
100 J
C.
300 J
D.
400 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W = 300 J - 100 J = 200 J.
Correct Answer:
A
— 200 J
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Q. In a process where 300 J of heat is added to a system and the system does 100 J of work, what is the internal energy change?
A.
200 J
B.
300 J
C.
100 J
D.
400 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W = 300 J - 100 J = 200 J.
Correct Answer:
A
— 200 J
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Q. In a refrigerator, the work done on the system is used to:
A.
Increase the internal energy
B.
Decrease the internal energy
C.
Transfer heat from cold to hot
D.
Transfer heat from hot to cold
Show solution
Solution
In a refrigerator, work is done on the system to transfer heat from a colder body to a hotter body, which is against the natural flow of heat.
Correct Answer:
C
— Transfer heat from cold to hot
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Q. In a thermodynamic cycle, if the net work done by the system is 200 J and the heat absorbed is 300 J, what is the change in internal energy?
A.
100 J
B.
200 J
C.
300 J
D.
500 J
Show solution
Solution
According to the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 300 J - 200 J = 100 J.
Correct Answer:
A
— 100 J
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Q. In a thermodynamic cycle, the net work done is equal to the:
A.
Net heat added to the system
B.
Net heat removed from the system
C.
Change in internal energy
D.
Change in entropy
Show solution
Solution
In a thermodynamic cycle, the net work done is equal to the net heat added to the system, according to the first law of thermodynamics.
Correct Answer:
A
— Net heat added to the system
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Q. In a thermodynamic cycle, the net work done is equal to which of the following?
A.
Net heat added to the system
B.
Net heat rejected by the system
C.
Change in internal energy
D.
Change in enthalpy
Show solution
Solution
In a thermodynamic cycle, the net work done is equal to the net heat added to the system, according to the first law of thermodynamics.
Correct Answer:
A
— Net heat added to the system
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Q. In a thermodynamic cycle, the net work done is equal to:
A.
Net heat added to the system
B.
Net change in internal energy
C.
Net heat removed from the system
D.
None of the above
Show solution
Solution
In a thermodynamic cycle, the net work done is equal to the net heat added to the system, as the internal energy change is zero.
Correct Answer:
A
— Net heat added to the system
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Q. In a thermodynamic process, if the internal energy of a system increases, which of the following could be true?
A.
Heat is added to the system
B.
Work is done by the system
C.
Both heat is added and work is done by the system
D.
Work is done on the system
Show solution
Solution
The internal energy of a system increases if heat is added to the system or work is done on the system.
Correct Answer:
A
— Heat is added to the system
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Q. In a vacuum, which mode of heat transfer is not possible?
A.
Conduction
B.
Convection
C.
Radiation
D.
All of the above
Show solution
Solution
Convection is not possible in a vacuum as it requires a medium (fluid) for heat transfer.
Correct Answer:
B
— Convection
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Q. In an isochoric process, the volume of the system:
A.
Increases
B.
Decreases
C.
Remains constant
D.
Varies with temperature
Show solution
Solution
An isochoric process is characterized by constant volume, meaning the volume does not change.
Correct Answer:
C
— Remains constant
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Q. In an isochoric process, what happens to the internal energy of a gas when heat is added?
A.
It decreases
B.
It remains constant
C.
It increases
D.
It depends on the gas
Show solution
Solution
In an isochoric process, the volume remains constant, and any heat added to the system increases the internal energy.
Correct Answer:
C
— It increases
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Q. In an isochoric process, what happens to the internal energy of an ideal gas when heat is added?
A.
It decreases.
B.
It remains constant.
C.
It increases.
D.
It depends on the amount of heat added.
Show solution
Solution
In an isochoric process, the volume remains constant, and any heat added increases the internal energy of the gas.
Correct Answer:
C
— It increases.
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Q. In an isothermal process for an ideal gas, which of the following is true?
A.
The internal energy remains constant.
B.
The temperature increases.
C.
The pressure decreases.
D.
The volume remains constant.
Show solution
Solution
In an isothermal process, the temperature remains constant, which implies that the internal energy of an ideal gas also remains constant.
Correct Answer:
A
— The internal energy remains constant.
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Q. In an isothermal process, how does the internal energy of an ideal gas change?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Depends on the amount of gas
Show solution
Solution
In an isothermal process for an ideal gas, the temperature remains constant, and thus the internal energy also remains constant.
Correct Answer:
C
— Remains constant
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Q. In an isothermal process, the change in internal energy is:
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the system
Show solution
Solution
In an isothermal process, the temperature remains constant, hence the change in internal energy is zero.
Correct Answer:
C
— Zero
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Showing 31 to 60 of 195 (7 Pages)
Thermodynamics MCQ & Objective Questions
Thermodynamics is a crucial topic in physics that plays a significant role in various school and competitive exams. Understanding the principles of thermodynamics not only enhances conceptual clarity but also boosts your confidence in tackling exam questions. Practicing MCQs and objective questions related to thermodynamics is essential for scoring better, as they help reinforce your knowledge and identify important questions that frequently appear in exams.
What You Will Practise Here
Fundamental laws of thermodynamics
Key concepts such as heat, work, and internal energy
Thermodynamic processes: isothermal, adiabatic, isochoric, and isobaric
Important formulas and equations related to thermodynamic systems
Understanding entropy and its implications in thermodynamic processes
Diagrams illustrating thermodynamic cycles and processes
Applications of thermodynamics in real-world scenarios
Exam Relevance
Thermodynamics is a significant topic in various examinations, including CBSE, State Boards, NEET, and JEE. Students can expect questions that test their understanding of the laws of thermodynamics, calculations involving heat transfer, and the application of thermodynamic principles in different contexts. Common question patterns include numerical problems, conceptual questions, and application-based scenarios, making it essential to master this topic for effective exam preparation.
Common Mistakes Students Make
Confusing the different thermodynamic processes and their characteristics
Misapplying the first and second laws of thermodynamics in problem-solving
Overlooking units and conversions in numerical questions
Failing to understand the concept of entropy and its significance
Neglecting to practice diagram-based questions that illustrate thermodynamic cycles
FAQs
Question: What are the main laws of thermodynamics?Answer: The main laws include the Zeroth Law, First Law (Law of Energy Conservation), Second Law (Entropy), and Third Law (Absolute Zero).
Question: How can I improve my performance in thermodynamics MCQs?Answer: Regular practice of thermodynamics MCQ questions, understanding key concepts, and solving previous years' papers can significantly enhance your performance.
Start your journey towards mastering thermodynamics today! Solve practice MCQs and test your understanding to excel in your exams. Remember, consistent practice is the key to success!
