An ideal gas is compressed to one-half its original volume during an isothermal process. The final pressure of the gas

increases to twice its original value.

increases to less than twice its original value.

increases to more than twice its original value.

An ideal gas is expanded isothermally from 20 L to 30 L. During this process, 6 J of energy is expended by the external mechanism that expanded the gas. Which of the following statements is correct?

6 J of energy flow from surroundings into the gas.

6 J of energy flow from the gas into the surroundings.

No energy flows into or from the gas since this process is isothermal.

None of the above statements is correct.

A monatomic ideal gas is compressed to one-half its original volume during an adiabatic process. The final pressure of the gas

increases to more than twice its original value.

increases to twice its original value.

increases to less than twice its original value.

Consider two cylinders of gas identical in all respects except that one contains O2 and the other He. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater temperature increase?

Neither; both will show the same increase.

It's impossible to tell from the information given.

Consider two cylinders of gas identical in all respects except that one contains O2 and the other He. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater pressure increase?

Neither; both will show the same increase.

It's impossible to tell from the information given.

Is it possible to transfer heat from a hot reservoir to a cold reservoir?

Yes; this will happen naturally.

Yes, but work will have to be done.

Theoretically yes, but it hasn't been accomplished yet.

Is it possible to transfer heat from a cold reservoir to a hot reservoir?

Yes, but work will have to be done.

Yes; this will happen naturally.

Theoretically yes, but it hasn't been accomplished yet.

The efficiency of a heat engine is defined as the ratio of

the work it does to the heat input at the high temperature.

the heat input at the high temperature to the heat output at the low temperature.

the heat output at the low temperature to the heat input at the high temperature.

the work it does to the heat output at the low temperature.

A Carnot cycle consists of

two adiabats and two isotherms.

The coefficient of performance (COP) of a refrigerator is defined as the ratio of

the heat removed from the inside to the work done to remove the heat.

the heat removed from the inside to the heat expelled to the outside.

the heat expelled to the outside to the heat removed from the inside.

the heat expelled to the outside to the work done to remove the heat.

Physics: Trivia Questions On The Laws Of Thermodynamics

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Matt Balanda, BS (Aerospace Engineering) |

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Matt Balanda, a Calvary Chapel Christian School leader with a Bachelor's in Aerospace Engineering and Mathematics, transitioned from Aerospace Engineering to Education with a Master's from California Baptist University. As the High School Vice-Principal and Physics teacher, he nurtures students' love of learning and faith, creating an enriching and transformational educational experience.

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In an isothermal process, there is no change in
- Pressure.
- Temperature.
- Volume.
- Heat.

About This Quiz

Welcome to the "Laws of Thermodynamics Quiz," where you can put your knowledge of this fundamental aspect of physics to the test. Thermodynamics involves the study of heat, energy, and their transformations under various conditions. This quiz focuses on the three core laws that define how energy moves within physical systems and how it affects matter.

Starting with the Zeroth See moreLaw, which establishes thermal equilibrium, moving on to the First Law concerning energy conservation, diving into the Second Law which explains entropy, and wrapping up with the Third Law which discusses absolute zero—this quiz covers all crucial bases. Each question is carefully crafted to challenge your understanding and help you grasp the practical applications of these laws in real-world scenarios.

Whether you're a student looking to solidify your grasp of thermodynamics, a teacher seeking a resource for your class, or a curious mind eager to learn about the principles that govern our universe, this quiz is designed to be both informative and engaging. Get ready to heat things up with our "Laws of Thermodynamics Quiz" and see how well you understand these essential energy laws!

Physics: Trivia Questions On The Laws Of Thermodynamics - Quiz

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2.

When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an isothermal process,
- ΔU = 0
- W = 0
- Q = 0
- None of the above
Correct Answer

A. ΔU = 0

Explanation

When an ideal gas is taken through an isothermal process, the temperature remains constant. According to the first law of thermodynamics, the change in internal energy (ΔU) of a system is equal to the heat added (Q) to the system minus the work done (W) by the system. In an isothermal process, there is no change in internal energy because the temperature remains constant. Therefore, ΔU = 0 is the correct answer.

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3.

A gas is expanded to twice its original volume with no change in its temperature. This process is
- Isothermal.
- Isochoric.
- Isobaric.
- Adiabatic.
Correct Answer

A. Isothermal.

Explanation

When a gas is expanded to twice its original volume with no change in temperature, it means that the gas is undergoing an isothermal process. In an isothermal process, the temperature of the system remains constant throughout the expansion or compression. This is achieved by transferring heat to or from the surroundings to maintain the temperature constant. Therefore, the correct answer is isothermal.

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2
4.

An ideal gas is compressed isothermally from 30 L to 20 L. During this process, 6.0 J of energy is expended by the external mechanism that compressed the gas. What is the change of internal energy for this gas?
- 6.0 J
- Zero
- -6.0 J
- None of the above
Correct Answer

A. Zero

Explanation

During an isothermal process, the temperature of the gas remains constant. As a result, the internal energy of the gas also remains constant since internal energy is directly proportional to temperature. Therefore, the change in internal energy for this gas is zero.

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5.

An ideal gas is compressed to one-half its original volume during an isothermal process. The final pressure of the gas
- Increases to twice its original value.
- Increases to less than twice its original value.
- Increases to more than twice its original value.
- Does not change.
Correct Answer

A. Increases to twice its original value.

Explanation

During an isothermal process, the temperature of the gas remains constant. According to Boyle's Law, the pressure and volume of a gas are inversely proportional when the temperature is constant. Therefore, when the gas is compressed to one-half its original volume, the pressure will increase to twice its original value to maintain the same temperature.

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1
6.

An ideal gas is expanded isothermally from 20 L to 30 L. During this process, 6 J of energy is expended by the external mechanism that expanded the gas. Which of the following statements is correct?
- 6 J of energy flow from surroundings into the gas.
- 6 J of energy flow from the gas into the surroundings.
- No energy flows into or from the gas since this process is isothermal.
- None of the above statements is correct.
Correct Answer

A. 6 J of energy flow from surroundings into the gas.

Explanation

During an isothermal process, the temperature of the gas remains constant. Since the gas is expanding, work is being done on the gas by the external mechanism. According to the first law of thermodynamics, the work done on the gas is equal to the energy transferred into the gas. Therefore, the 6 J of energy expended by the external mechanism flows from the surroundings into the gas.

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7.

A gas is quickly compressed in an isolated environment. During the event, the gas exchanged no heat with its surroundings. This process is
- Isothermal.
- Isochoric.
- Isobaric.
- Adiabatic.
Correct Answer

A. Adiabatic.

Explanation

In this scenario, the gas is quickly compressed in an isolated environment, meaning there is no heat exchange with the surroundings. This indicates that the process is adiabatic, as adiabatic processes involve no heat transfer.

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8.

When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an adiabatic process,
- ΔU = 0.
- W = 0.
- Q = 0.
- None of the above
Correct Answer

A. Q = 0.

Explanation

When the first law of thermodynamics is applied to an adiabatic process, it means that there is no heat transfer (Q = 0) because the system is thermally isolated. In this case, the change in internal energy (ΔU) is also zero because there is no heat added or removed from the system. Additionally, the work done (W) is also zero because there is no expansion or compression of the gas. Therefore, the correct answer is Q = 0.

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1
9.

A monatomic ideal gas is compressed to one-half its original volume during an adiabatic process. The final pressure of the gas
- Increases to twice its original value.
- Increases to less than twice its original value.
- Increases to more than twice its original value.
- Does not change.
Correct Answer

A. Increases to more than twice its original value.

Explanation

During an adiabatic process, no heat is exchanged between the gas and its surroundings. As the gas is compressed to one-half its original volume, the work done on the gas increases. According to the adiabatic equation, PV^γ = constant, where P is the pressure, V is the volume, and γ is the heat capacity ratio. Since the volume decreases, the pressure must increase to maintain the constant value. Therefore, the final pressure of the gas will increase to more than twice its original value.

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6
10.

Consider two cylinders of gas identical in all respects except that one contains O2 and the other He. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater temperature increase?
- The O2
- The He
- Neither; both will show the same increase.
- It's impossible to tell from the information given.
Correct Answer

A. The He

Explanation

When a gas is compressed adiabatically, its temperature increases. The temperature increase is determined by the ratio of specific heats (γ) of the gas. For both oxygen (O2) and helium (He), the value of γ is different. The value of γ for helium is greater than that of oxygen. Therefore, when both gases are compressed adiabatically to one-third of their original volume, helium will show a greater temperature increase compared to oxygen.

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2
11.

Consider two cylinders of gas identical in all respects except that one contains O2 and the other He. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater pressure increase?
- The O2
- The He
- Neither; both will show the same increase.
- It's impossible to tell from the information given.
Correct Answer

A. The He

Explanation

The pressure of a gas is directly proportional to its temperature and inversely proportional to its volume, according to the ideal gas law. Since both gases are compressed adiabatically (without any heat exchange with the surroundings), the temperature of the gases will increase. As the volume of the gases is reduced to one-third of their original volume, the pressure will increase. However, since both gases are identical in all other respects, the increase in pressure will be the same for both gases. Therefore, the correct answer is that neither gas will show a greater pressure increase.

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12.

In an isobaric process, there is no change in
- Pressure.
- Temperature.
- Volume.
- Internal energy.
Correct Answer

A. Pressure.

Explanation

In an isobaric process, the pressure remains constant. This means that the system is undergoing a change or transformation while being exposed to a constant external pressure. While there may be changes in temperature, volume, and internal energy, the pressure remains the same throughout the process.

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13.

When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an isobaric process,
- ΔU = 0.
- W = 0.
- Q = 0.
- None of the above
Correct Answer

A. None of the above

Explanation

When an ideal gas is taken through an isobaric process, the change in internal energy (ΔU) is not zero because the temperature of the gas may change. The work done (W) is also not zero because there is a change in volume of the gas. The heat transfer (Q) is also not zero because energy is exchanged between the gas and its surroundings. Therefore, none of the above options are correct.

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14.

A gas is allowed to expand at constant pressure as heat is added to it. This process is
- Isothermal.
- Isochoric.
- Isobaric.
- Adiabatic.
Correct Answer

A. Isobaric.

Explanation

In this scenario, the gas is allowed to expand at a constant pressure as heat is added. This indicates that the pressure remains constant throughout the process. The term "isobaric" specifically refers to a process where the pressure remains constant. Therefore, the correct answer is isobaric.

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15.

Ten joules of heat energy are transferred to a sample of ideal gas at constant pressure. As a result, the internal energy of the gas
- Increases by 10 J.
- Increases by less than 10 J.
- Increases by more than 10 J.
- Remains unchanged.
Correct Answer

A. Increases by less than 10 J.

Explanation

When heat energy is transferred to a sample of ideal gas at constant pressure, some of the energy is used to do work on the surroundings, such as expanding the gas. This means that not all of the heat energy is converted into an increase in the internal energy of the gas. Therefore, the internal energy of the gas increases by less than the 10 J of heat energy that was transferred.

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3
16.

In an isochoric process, there is no change in
- Pressure.
- Temperature.
- Volume.
- Internal energy.
Correct Answer

A. Volume.

Explanation

In an isochoric process, the volume of the system remains constant. This means that there is no change in the amount of space occupied by the system. The other options, such as pressure, temperature, and internal energy, can all change in an isochoric process. However, the volume specifically remains the same throughout the process.

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17.

When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an isochoric process,
- ΔU = 0.
- W = 0.
- Q = 0.
- None of the above
Correct Answer

A. W = 0.

Explanation

In an isochoric process, the volume of the gas remains constant. As a result, there is no work done by or on the gas because work is defined as the product of force and displacement, and there is no displacement in this case. Therefore, W = 0.

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18.

A gas is confined to a rigid container that cannot expand as heat energy is added to it. This process is
- Isothermal.
- Isochoric.
- Isobaric.
- Adiabatic.
Correct Answer

A. Isochoric.

Explanation

In this scenario, the gas is confined to a rigid container that cannot expand. This means that the volume of the gas remains constant throughout the process. Since the volume is constant, it implies that no work is done by or on the gas. Therefore, the process is isochoric, also known as constant volume.

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19.

Ten joules of heat energy are transferred to a sample of ideal gas at constant volume. As a result, the internal energy of the gas
- Increases by 10 J.
- Increases by less than 10 J.
- Increases by more than 10 J.
- Remains unchanged.
Correct Answer

A. Increases by 10 J.

Explanation

When heat energy is transferred to a sample of ideal gas at constant volume, the internal energy of the gas increases by an amount equal to the heat energy transferred. In this case, 10 joules of heat energy are transferred, so the internal energy of the gas increases by 10 J.

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1
20.

Is it possible to transfer heat from a hot reservoir to a cold reservoir?
- No.
- Yes; this will happen naturally.
- Yes, but work will have to be done.
- Theoretically yes, but it hasn't been accomplished yet.
Correct Answer

A. Yes; this will happen naturally.

Explanation

Heat transfer from a hot reservoir to a cold reservoir is possible and it occurs naturally due to the principle of the second law of thermodynamics. According to this law, heat flows spontaneously from a higher temperature region to a lower temperature region until equilibrium is reached. This process is known as heat transfer by conduction, convection, or radiation, and it happens in various everyday situations, such as when a hot object cools down or when warm air rises and cold air sinks.

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21.

Is it possible to transfer heat from a cold reservoir to a hot reservoir?
- No.
- Yes; this will happen naturally.
- Yes, but work will have to be done.
- Theoretically yes, but it hasn't been accomplished yet.
Correct Answer

A. Yes, but work will have to be done.

Explanation

Yes, it is possible to transfer heat from a cold reservoir to a hot reservoir, but work needs to be done in order to achieve this. This is because heat naturally flows from a hot region to a cold region, so in order to reverse this natural flow, external work needs to be applied. This can be done through the use of a heat pump or a refrigeration system, where work is done to extract heat from a cold reservoir and transfer it to a hot reservoir.

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2
22.

The efficiency of a heat engine is defined as the ratio of
- The heat input at the high temperature to the heat output at the low temperature.
- The heat output at the low temperature to the heat input at the high temperature.
- The work it does to the heat input at the high temperature.
- The work it does to the heat output at the low temperature.
Correct Answer

A. The work it does to the heat input at the high temperature.

Explanation

The efficiency of a heat engine is defined as the ratio of the work it does to the heat input at the high temperature. This means that the efficiency of a heat engine is determined by how much useful work it can produce from the heat energy supplied at the high temperature. The heat output at the low temperature is not directly related to the efficiency of the engine, as it is the heat input at the high temperature that is used to perform work.

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23.

If the theoretical efficiency of a Carnot engine is to be 100%, the heat sink must be
- At absolute zero.
- At 0°C.
- At 100°C.
- Infinitely hot.
Correct Answer

A. At absolute zero.

Explanation

The theoretical efficiency of a Carnot engine is given by the formula 1 - (Tc/Th), where Tc is the temperature of the heat sink and Th is the temperature of the heat source. To achieve an efficiency of 100%, Tc must be as close to absolute zero as possible. This is because the efficiency approaches 1 when Tc approaches absolute zero. Therefore, the heat sink must be at absolute zero.

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24.

A Carnot cycle consists of
- Two adiabats and two isobars.
- Two isobars and two isotherms.
- Two isotherms and two isomets.
- Two adiabats and two isotherms.
Correct Answer

A. Two adiabats and two isotherms.

Explanation

A Carnot cycle consists of two adiabats and two isotherms. This is because a Carnot cycle is a theoretical thermodynamic cycle that consists of two isothermal processes (during which the system is in thermal equilibrium with a heat reservoir) and two adiabatic processes (during which no heat is exchanged with the surroundings). The isothermal processes allow for heat transfer at constant temperature, while the adiabatic processes allow for work to be done on or by the system without any heat exchange. Therefore, the correct answer is two adiabats and two isotherms.

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25.

The coefficient of performance (COP) of a refrigerator is defined as the ratio of
- The heat removed from the inside to the heat expelled to the outside.
- The heat expelled to the outside to the heat removed from the inside.
- The heat removed from the inside to the work done to remove the heat.
- The heat expelled to the outside to the work done to remove the heat.
Correct Answer

A. The heat removed from the inside to the work done to remove the heat.

Explanation

The coefficient of performance (COP) of a refrigerator is a measure of its efficiency, specifically how much heat can be removed from the inside of the refrigerator for a given amount of work done to remove that heat. The COP is calculated by dividing the heat removed from the inside by the work done to remove that heat. This ratio indicates how effectively the refrigerator is able to cool its contents using the energy input. Therefore, the correct answer is "the heat removed from the inside to the work done to remove the heat."

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26.

The coefficient of performance (COP) of a heat pump is defined as the ratio of
- The heat delivered to the inside to the heat taken from the outside.
- The heat taken from the outside to the heat delivered to the inside.
- The heat delivered to the inside to the work done to move the heat.
- The heat taken from the outside to the work done to move the heat.
Correct Answer

A. The heat delivered to the inside to the work done to move the heat.

Explanation

The coefficient of performance (COP) of a heat pump is defined as the ratio of the heat delivered to the inside to the work done to move the heat. This means that the COP measures the efficiency of the heat pump in transferring heat from a colder area (outside) to a warmer area (inside). It indicates how much heat energy is delivered for a certain amount of work done. A higher COP indicates a more efficient heat pump, as it delivers more heat for the same amount of work.

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27.

When water freezes, the entropy of the water
- Increases.
- Decreases.
- Does not change.
- Could either increase or decrease; it depends on other factors.
Correct Answer

A. Decreases.

Explanation

When water freezes, the molecules in the water slow down and form a rigid, ordered structure. This decrease in molecular motion and randomness leads to a decrease in entropy.

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1
28.

According to the second law of thermodynamics, for any process that may occur within an isolated system, which one of the choices applies?
- Entropy remains constant.
- Entropy increases.
- Entropy decreases.
- Both A and B are possible.
- Both A and C are possible.
Correct Answer

A. Both A and B are possible.

Explanation

According to the second law of thermodynamics, the entropy of an isolated system can either remain constant or increase. This means that in some processes, the entropy of the system may stay the same, while in others, it may increase. Therefore, both options A (Entropy remains constant) and B (Entropy increases) are possible according to the second law of thermodynamics.

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29.

The second law of thermodynamics leads us to conclude that
- The total energy of the universe is constant.
- Disorder in the universe is increasing with the passage of time.
- It is theoretically possible to convert heat into work with 100% efficiency.
- The average temperature of the universe is increasing with the passage of time.
Correct Answer

A. Disorder in the universe is increasing with the passage of time.

Explanation

The second law of thermodynamics states that the entropy, or disorder, of an isolated system will always increase over time. This means that as time passes, the universe tends to become more disordered. Therefore, the answer "disorder in the universe is increasing with the passage of time" aligns with the second law of thermodynamics.

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30.

During an isothermal process, 5.0 J of heat is removed from an ideal gas. What is the change in internal energy?
- Zero
- 2.5 J
- 5.0 J
- 10 J
Correct Answer

A. Zero

Explanation

In an isothermal process, the temperature of the system remains constant. According to the first law of thermodynamics, the change in internal energy (ΔU) is equal to the heat added or removed (Q) minus the work done by the system (W). Since the temperature remains constant, there is no change in internal energy (ΔU = 0) even if heat is removed from the gas. Therefore, the correct answer is zero.

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1
31.

During an isothermal process, 5.0 J of heat is removed from an ideal gas. What is the work done in the process?
- Zero
- 5.0 J
- -5.0 J
- None of the above
Correct Answer

A. -5.0 J

Explanation

During an isothermal process, the temperature of the gas remains constant. According to the first law of thermodynamics, the change in internal energy of the gas is equal to the heat added or removed from the gas minus the work done by or on the gas. In this case, 5.0 J of heat is removed from the gas, so the change in internal energy is -5.0 J. Since the process is isothermal and the change in internal energy is negative, the work done by the gas must also be negative. Therefore, the work done in the process is -5.0 J.

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32.

The work done on an ideal gas system in an isothermal process is -400 J. What is the change in internal energy?
- Zero
- -400 J
- 400 J
- None of the above
Correct Answer

A. Zero

Explanation

In an isothermal process, the temperature of the gas remains constant. According to the first law of thermodynamics, the change in internal energy (ΔU) of a system is equal to the work done on or by the system (W) plus the heat transferred (Q). Since the work done on the gas is -400 J, and the process is isothermal, it means that the heat transferred (Q) into the system is also 400 J. Therefore, the change in internal energy (ΔU) is zero, as the work done and heat transferred cancel each other out.

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33.

200 J of work is done in compressing a gas adiabatically. What is the change in internal energy of the gas?
- Zero
- 100 J
- 200 J
- There is not enough information to determine.
Correct Answer

A. 200 J

Explanation

When work is done adiabatically on a gas, there is no heat transfer between the gas and its surroundings. Therefore, the change in internal energy of the gas is equal to the work done on the gas. In this case, 200 J of work is done on the gas, so the change in internal energy is also 200 J.

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1
34.

An ideal gas undergoes an adiabatic process while doing 25 J of work. What is the change in internal energy?
- Zero
- 25 J
- -25 J
- None of the above
Correct Answer

A. -25 J

Explanation

In an adiabatic process, there is no heat exchange between the system and its surroundings. Therefore, the change in internal energy is solely determined by the work done on or by the system. Since the ideal gas is doing 25 J of work, this indicates that the internal energy of the gas decreases by 25 J. Therefore, the change in internal energy is -25 J.

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35.

In an isochoric process, the internal energy of a system decreases by 50 J. What is the work done?
- Zero
- 50 J
- -50 J
- None of the above
Correct Answer

A. Zero

Explanation

In an isochoric process, the volume of the system remains constant. Since work is defined as the product of force and displacement, and there is no displacement in this case, the work done is zero. The decrease in internal energy does not necessarily mean that work is done, as internal energy can also change due to heat transfer. Therefore, the correct answer is zero.

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36.

In an isochoric process, the internal energy of a system decreases by 50 J. What is the heat exchange?
- Zero
- 50 J
- -50 J
- None of the above
Correct Answer

A. -50 J

Explanation

In an isochoric process, the volume of the system remains constant. Since the internal energy decreases by 50 J, it means that energy is being removed from the system. This energy removal is in the form of heat exchange, and since the internal energy is decreasing, the heat exchange must be negative. Therefore, the correct answer is -50 J.

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37.

A certain amount of a monatomic gas is maintained at constant volume as it is cooled by 50 K. This feat is accomplished by removing 400 J of energy from the gas. How much work is done by the gas?
- Zero
- 400 J
- -400 J
- None of the above
Correct Answer

A. Zero

Explanation

When a gas is cooled at constant volume, no work is done by the gas. This is because work is defined as the transfer of energy due to a force acting through a distance. Since the volume is constant, there is no change in the distance over which the force is acting, and therefore no work is done. Thus, the correct answer is zero.

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38.

A monatomic gas is cooled by 50 K at constant volume when 831 J of energy is removed from it. How many moles of gas are in the sample?
- 2.50 mol
- 1.50 mol
- 1.33 mol
- None of the above
Correct Answer

A. 1.33 mol

Explanation

When a monatomic gas is cooled at constant volume, the change in internal energy (ΔU) is equal to the heat removed (Q) from the gas. The change in internal energy can be calculated using the formula ΔU = nCvΔT, where n is the number of moles of gas, Cv is the molar specific heat capacity at constant volume, and ΔT is the change in temperature. Rearranging the formula, we get n = ΔU / (CvΔT). Plugging in the given values, n = 831 J / (12.5 J/(mol·K) * 50 K) = 1.33 mol. Therefore, the number of moles of gas in the sample is 1.33 mol.

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39.

A system consists of 3.0 kg of water at 80°C. 30 J of work is done on the system by stirring with a paddle wheel, while 66 J of heat is removed. What is the change in internal energy of the system?
- 36 J
- -36 J
- 96 J
- -96 J
Correct Answer

A. -36 J

Explanation

When 30 J of work is done on the system, it increases the internal energy of the system. However, when 66 J of heat is removed from the system, it decreases the internal energy. Since the heat removed is greater than the work done, the net change in internal energy is negative. Therefore, the change in internal energy of the system is -36 J.

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40.

A heat engine receives 7000 J of heat and loses 3000 J in each cycle. What is the efficiency?
- 57%
- 30%
- 70%
- 43%
Correct Answer

A. 57%

Explanation

The efficiency of a heat engine is calculated by dividing the useful output energy by the input energy. In this case, the useful output energy is the heat received minus the heat lost, which is 7000 J - 3000 J = 4000 J. The input energy is the heat received, which is 7000 J. Therefore, the efficiency is (4000 J / 7000 J) * 100% = 57%.

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41.

A heat engine has an efficiency of 35.0% and receives 150 J of heat per cycle. How much work does it perform in each cycle?
- Zero
- 52.5 J
- 97.5 J
- 150 J
Correct Answer

A. 52.5 J

Explanation

The efficiency of a heat engine is defined as the ratio of the work output to the heat input. In this case, the efficiency is given as 35.0%, which means that only 35.0% of the heat input is converted into work output. Since the heat input is 150 J, the work output can be calculated by multiplying the heat input by the efficiency. Therefore, the work performed in each cycle is 150 J * 0.35 = 52.5 J.

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42.

A heat engine has an efficiency of 35.0% and receives 150 J of heat per cycle. How much heat does it exhaust in each cycle?
- Zero
- 52.5 J
- 97.5 J
- 150 J
Correct Answer

A. 97.5 J

Explanation

The efficiency of a heat engine is given by the formula: Efficiency = (Useful work output / Heat input) x 100%. In this case, the efficiency is given as 35%. Since the heat input is 150 J, we can use the formula to find the useful work output. Rearranging the formula, we get: Useful work output = (Efficiency / 100%) x Heat input. Plugging in the values, we find: Useful work output = (35% / 100%) x 150 J = 52.5 J. Since the question asks for the amount of heat exhausted, we subtract the useful work output from the heat input: Heat exhausted = Heat input - Useful work output = 150 J - 52.5 J = 97.5 J. Therefore, the correct answer is 97.5 J.

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43.

A heat engine absorbs 64 kcal of heat each cycle and exhausts 42 kcal. Calculate the efficiency of each cycle.
- 34%
- 66%
- 50%
- 150%
Correct Answer

A. 34%

Explanation

The efficiency of a heat engine is calculated by dividing the useful output energy (in this case, the heat absorbed) by the input energy (the heat absorbed plus the heat exhausted). In this case, the heat absorbed is 64 kcal and the heat exhausted is 42 kcal. Therefore, the efficiency is calculated as (64 / (64 + 42)) * 100 = 34%.

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1
44.

A heat engine absorbs 64 kcal of heat each cycle and exhausts 42 kcal. Calculate the work done each cycle.
- 22 kcal
- 42 kcal
- 64 kcal
- 106 kcal
Correct Answer

A. 22 kcal

Explanation

The work done by a heat engine is equal to the difference between the heat absorbed and the heat exhausted. In this case, the heat absorbed is 64 kcal and the heat exhausted is 42 kcal. Therefore, the work done each cycle is 64 kcal - 42 kcal = 22 kcal.

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45.

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhaust temperature = 430°C, 7.0 * 10⁹ cal of fuel produces 1.4 * 10^10 J of work in one hour. What is the actual efficiency of this engine?
- 32%
- 48%
- 52%
- 68%
Correct Answer

A. 48%

Explanation

The actual efficiency of an engine is calculated by dividing the work output by the energy input. In this case, the work output is given as 1.4 * 10^10 J. To find the energy input, we need to convert the given fuel energy from calories to joules. 7.0 * 10^9 cal is equivalent to 2.92 * 10^10 J. Dividing the work output by the energy input and multiplying by 100 gives us the efficiency as a percentage. Therefore, the actual efficiency of this engine is 48%.

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46.

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhaust temperature = 430°C, 7.0 * 10⁹ cal of fuel produces 1.4 * 10^10 J of work in one hour. What is the Carnot efficiency of this engine?
- 32%
- 48%
- 52%
- 68%
Correct Answer

A. 68%

Explanation

The Carnot efficiency of an engine is determined by the temperatures of the hot and cold reservoirs. It is given by the formula: Carnot efficiency = 1 - (Tc/Th) where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. In this case, the combustion chamber temperature is 1900°C, which is the temperature of the hot reservoir, and the exhaust temperature is 430°C, which is the temperature of the cold reservoir. Substituting these values into the formula, we get: Carnot efficiency = 1 - (430/1900) = 1 - 0.226 = 0.774 = 77.4% Therefore, the Carnot efficiency of this engine is 77.4%, which is closest to 68% among the given options.

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47.

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhaust temperature = 430°C, 7.0 * 10⁹ cal of fuel produces 1.4 * 10^10 J of work in one hour. What is the power output, in hp, of this engine?
- 5.2 kW
- 6.1 kW
- 7.4 kW
- 8.3 kW
Correct Answer

A. 5.2 kW

Explanation

The power output of an engine can be calculated by dividing the work produced by the time taken. In this case, the work produced is given as 1.4 * 10^10 J in one hour. To convert this to kilowatts, we divide by 3600 (the number of seconds in an hour) and then by 1000 (to convert from joules to kilowatts). This gives us a power output of 3.89 kW. However, since the question asks for the power output in horsepower, we need to convert this value to horsepower. 1 horsepower is equal to 0.7457 kilowatts, so the power output is approximately 5.2 kW.

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1
48.

A coal-fired plant generates 600 MW of electric power. The plant uses 4.8 * 10^6 kg of coal each day. The heat of combustion of coal is 3.3 * 10^7 J/kg. The steam that drives the turbines is at a temperature of 300°C, and the exhaust water is at 37°C. What is the overall efficiency of the plant for generating electric power?
- 33%
- 37%
- 46%
- 54%
Correct Answer

A. 33%

Explanation

The overall efficiency of the plant for generating electric power can be calculated using the formula:

Efficiency = (Useful energy output / Energy input) * 100

In this case, the useful energy output is the electric power generated, which is 600 MW. The energy input is the heat of combustion of coal, which is 4.8 * 10^6 kg * 3.3 * 10^7 J/kg.

Efficiency = (600 MW / (4.8 * 10^6 kg * 3.3 * 10^7 J/kg)) * 100

Calculating this expression gives an efficiency of approximately 0.033, which is equivalent to 33%. Therefore, the correct answer is 33%.

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49.

A coal-fired plant generates 600 MW of electric power. The plant uses 4.8 * 10^6 kg of coal each day. The heat of combustion of coal is 3.3 * 10^7 J/kg. The steam that drives the turbines is at a temperature of 300°C, and the exhaust water is at 37°C. What is the Carnot efficiency?
- 33%
- 37%
- 46%
- 54%
Correct Answer

A. 46%

Explanation

The Carnot efficiency is calculated using the formula: (1 - Tc/Th) * 100%, where Tc is the temperature of the exhaust water and Th is the temperature of the steam. In this case, Tc is 37°C and Th is 300°C. Plugging these values into the formula, we get (1 - 37/300) * 100% = 0.8767 * 100% = 87.67%. Therefore, the Carnot efficiency is 87.67%, which is closest to 46%.

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Matt Balanda |BS (Aerospace Engineering) |

Science Teacher

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