Q. During an isochoric process, the volume of the gas:
A.
Increases
B.
Decreases
C.
Remains constant
D.
Varies with temperature
Show solution
Solution
In an isochoric process, the volume of the gas remains constant.
Correct Answer:
C
— Remains constant
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Q. During an isothermal expansion of an ideal gas, what happens to the internal energy?
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 internal energy remains constant because the temperature does not change.
Correct Answer:
C
— Remains constant
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Q. If 100 J of heat is added to a system and 40 J of work is done by the system, what is the change in internal energy?
A.
60 J
B.
40 J
C.
100 J
D.
140 J
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Solution
According to the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 100 J - 40 J = 60 J.
Correct Answer:
A
— 60 J
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Q. If 500 J of heat is added to a system 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. If a gas expands against a constant external pressure, the work done by the gas is given by:
A.
W = P_ext * ΔV
B.
W = ΔU + Q
C.
W = Q - ΔU
D.
W = P_ext / ΔV
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Solution
The work done by the gas during expansion against a constant external pressure is given by W = P_ext * ΔV.
Correct Answer:
A
— W = P_ext * ΔV
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Q. If a gas expands and does 150 J of work while absorbing 100 J of heat, what is the change in internal energy?
A.
-50 J
B.
50 J
C.
250 J
D.
100 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W = 100 J - 150 J = -50 J.
Correct Answer:
A
— -50 J
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Q. If a gas expands and does 50 J of work while absorbing 30 J of heat, what is the change in internal energy?
A.
-20 J
B.
20 J
C.
80 J
D.
30 J
Show solution
Solution
Using the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 30 J - 50 J = -20 J.
Correct Answer:
B
— 20 J
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Q. If a system absorbs 100 J of heat and does 40 J of work, what is the change in internal energy?
A.
60 J
B.
40 J
C.
100 J
D.
140 J
Show solution
Solution
Using the first law of thermodynamics, ΔU = Q - W = 100 J - 40 J = 60 J.
Correct Answer:
A
— 60 J
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Q. If a system absorbs 200 J of heat and does 50 J of work, what is the change in internal energy?
A.
150 J
B.
250 J
C.
200 J
D.
50 J
Show solution
Solution
Using the First Law of Thermodynamics, ΔU = Q - W = 200 J - 50 J = 150 J.
Correct Answer:
A
— 150 J
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Q. If a system does 150 J of work on the surroundings and absorbs 100 J of heat, what is the change in internal energy?
A.
-50 J
B.
50 J
C.
100 J
D.
250 J
Show solution
Solution
Using the first law, ΔU = Q - W = 100 J - 150 J = -50 J.
Correct Answer:
A
— -50 J
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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, 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 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 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:
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 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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Q. In an isothermal process, the change in internal energy of an ideal gas is:
A.
Positive
B.
Negative
C.
Zero
D.
Depends on the amount of gas
Show solution
Solution
In an isothermal process for an ideal gas, the temperature remains constant, hence the change in internal energy is zero.
Correct Answer:
C
— Zero
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Q. In an isothermal process, the internal energy of an ideal gas:
A.
Increases
B.
Decreases
C.
Remains constant
D.
Depends on the amount of gas
Show solution
Solution
For an ideal gas, the internal energy is a function of temperature. In an isothermal process, the temperature remains constant, hence the internal energy remains constant.
Correct Answer:
C
— Remains constant
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Q. What does the first law of thermodynamics state?
A.
Energy can be created and destroyed.
B.
The total energy of an isolated system is constant.
C.
Heat cannot be converted into work.
D.
The internal energy of a system is independent of its state.
Show solution
Solution
The first law of thermodynamics states that the total energy of an isolated system is constant, meaning energy can neither be created nor destroyed.
Correct Answer:
B
— The total energy of an isolated system is constant.
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Showing 1 to 30 of 46 (2 Pages)
First Law of Thermodynamics MCQ & Objective Questions
The First Law of Thermodynamics is a fundamental principle that every student must grasp for a solid foundation in physics. Understanding this law is crucial for scoring well in exams, as it frequently appears in various formats, including MCQs and objective questions. By practicing these questions, students can enhance their exam preparation and boost their confidence in tackling important questions related to energy conservation and thermodynamic processes.
What You Will Practise Here
Definition and significance of the First Law of Thermodynamics
Key concepts: internal energy, heat, and work
Mathematical formulation and related equations
Applications of the First Law in real-world scenarios
Diagrams illustrating thermodynamic processes
Examples of energy transfer in closed and open systems
Common misconceptions and clarifications
Exam Relevance
The First Law of Thermodynamics is a vital topic in the curriculum for CBSE, State Boards, NEET, and JEE. Students can expect questions that test their understanding of energy conservation, as well as numerical problems involving calculations of work done and heat transfer. Common question patterns include direct applications of the law, theoretical explanations, and problem-solving scenarios that require a clear grasp of the concepts.
Common Mistakes Students Make
Confusing heat transfer with work done
Misunderstanding the concept of internal energy
Overlooking the signs of heat and work in calculations
Neglecting to apply the law to real-life examples
Failing to differentiate between closed and open systems
FAQs
Question: What is the First Law of Thermodynamics?Answer: The First Law of Thermodynamics states that energy cannot be created or destroyed, only transformed from one form to another.
Question: How is the First Law applied in practical scenarios?Answer: It is applied in various processes, such as engines and refrigerators, where energy conversion takes place.
Question: Why is it important to practice MCQs on this topic?Answer: Practicing MCQs helps reinforce concepts, improve problem-solving skills, and prepare effectively for exams.
Now is the time to solidify your understanding of the First Law of Thermodynamics. Dive into our practice MCQs and test your knowledge to excel in your upcoming exams!
