Thermodynamics
Q. In a Carnot engine, what is the efficiency dependent on?
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A.
The work done
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B.
The temperatures of the hot and cold reservoirs
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C.
The type of working substance
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D.
The volume of the gas
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?
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A.
It operates between two temperatures
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B.
It is 100% efficient
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C.
It can operate with any working substance
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D.
It is a perpetual motion machine
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?
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A.
300 J
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B.
500 J
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C.
700 J
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D.
200 J
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?
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A.
Doubles
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B.
Halves
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C.
Remains constant
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D.
Increases four times
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:
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A.
Positive
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B.
Negative
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C.
Zero
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D.
Depends on the path taken
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:
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A.
Positive
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B.
Negative
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C.
Zero
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D.
Depends on the work done
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:
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A.
The net heat added to the system
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B.
The change in internal energy
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C.
The heat lost by the system
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D.
Zero
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?
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A.
Positive
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B.
Negative
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C.
Zero
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D.
Depends on the path taken
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?
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A.
Positive
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B.
Negative
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C.
Zero
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D.
Depends on the process
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?
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A.
37.5%
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B.
50%
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C.
62.5%
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D.
75%
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?
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A.
40%
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B.
50%
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C.
60%
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D.
80%
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?
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A.
33.33%
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B.
50%
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C.
66.67%
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D.
75%
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:
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A.
Heat absorbed from the hot reservoir
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B.
Heat rejected to the cold reservoir
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C.
Heat absorbed minus heat rejected
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D.
Heat absorbed plus heat rejected
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?
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A.
60 J
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B.
40 J
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C.
100 J
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D.
140 J
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?
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A.
200 J
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B.
100 J
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C.
300 J
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D.
400 J
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?
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A.
200 J
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B.
100 J
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C.
300 J
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D.
400 J
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?
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A.
200 J
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B.
300 J
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C.
100 J
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D.
400 J
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:
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A.
Increase the internal energy
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B.
Decrease the internal energy
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C.
Transfer heat from cold to hot
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D.
Transfer heat from hot to cold
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?
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A.
100 J
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B.
200 J
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C.
300 J
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D.
500 J
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:
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A.
Net heat added to the system
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B.
Net heat removed from the system
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C.
Change in internal energy
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D.
Change in entropy
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?
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A.
Net heat added to the system
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B.
Net heat rejected by the system
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C.
Change in internal energy
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D.
Change in enthalpy
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:
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A.
Net heat added to the system
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B.
Net change in internal energy
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C.
Net heat removed from the system
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D.
None of the above
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?
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A.
Heat is added to the system
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B.
Work is done by the system
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C.
Both heat is added and work is done by the system
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D.
Work is done on the system
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?
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A.
Conduction
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B.
Convection
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C.
Radiation
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D.
All of the above
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:
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A.
Increases
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B.
Decreases
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C.
Remains constant
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D.
Varies with temperature
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?
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A.
It decreases
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B.
It remains constant
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C.
It increases
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D.
It depends on the gas
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?
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A.
It decreases.
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B.
It remains constant.
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C.
It increases.
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D.
It depends on the amount of heat added.
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?
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A.
The internal energy remains constant.
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B.
The temperature increases.
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C.
The pressure decreases.
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D.
The volume remains constant.
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?
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A.
Increases
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B.
Decreases
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C.
Remains constant
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D.
Depends on the amount of gas
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
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C.
Zero
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D.
Depends on the system
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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