# Unit 6 Concepts Quiz

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• 1.

### In an isothermal process, there is no change in

• A.

Pressure

• B.

Temperature

• C.

Volume

• D.

Heat

B. Temperature
Explanation
In an isothermal process, the temperature remains constant. This means that there is no change in the average kinetic energy of the particles in the system. The pressure, volume, and heat can still change during an isothermal process, but the temperature will stay the same.

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• 2.

### A gas is expanded to twice its original volume with no change in its temperature.  This process is ...

• A.

Isothermal

• B.

Isobaric

• C.

• D.

Isovolumetric

A. Isothermal
Explanation
When a gas is expanded to twice its original volume with no change in temperature, it means that the process is occurring at a constant temperature. In an isothermal process, the temperature remains constant throughout the expansion or compression. This is because the gas is in thermal equilibrium with its surroundings and any heat absorbed or released during the expansion is balanced by the surroundings. Therefore, the correct answer is isothermal.

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• 3.

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

• A.

Increases to twice its original value

• B.

Increases to less than twice its originial value

• C.

Increases to more than twice its original value

• D.

Does not change

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. As the gas is compressed to one-half its original volume, the pressure must increase to maintain the same temperature. Therefore, the final pressure of the gas increases to twice its original value.

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• 4.

### When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an adiabatic process,

• A.

ΔU = 0

• B.

W = 0

• C.

Q = 0

• D.

None of the above

C. Q = 0
Explanation
When the first law of thermodynamics is applied to an adiabatic process of an ideal gas, it means that no heat is transferred to or from the system (Q = 0). Since there is no heat transfer, the change in internal energy (ΔU) and work done (W) must also be zero. Therefore, the correct answer is Q = 0, indicating that no heat is exchanged during the adiabatic process.

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• 5.

### When the first law of thermodynamics, Q = ΔU + W, is applied to an ideal gas that is taken through an isovolumetric process

• A.

ΔU = 0

• B.

W = 0

• C.

Q = 0

• D.

None of the above

B. W = 0
Explanation
In an isovolumetric process, the volume of the gas remains constant. Since work (W) is defined as the product of force and displacement, and there is no change in volume, no work is done by or on the gas. Therefore, W = 0.

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• 6.

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

• A.

No

• B.

Yes; this will happen naturally

• C.

Yes, but work will have to be done.

• D.

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

C. 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 will have to be done. This is because heat naturally flows from a higher temperature to a lower temperature, so in order to transfer heat in the opposite direction, external energy or work needs to be applied to the system. This can be achieved 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 hotter reservoir.

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• 7.

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

• A.

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

• B.

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

• C.

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

• D.

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

C. 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 is determined by how much useful work is produced by the engine compared to the amount of heat energy that is input into the engine at the high temperature. The higher the ratio, the more efficient the engine is at converting heat energy into useful work.

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• 8.

### A Carnot cycle consists of

• A.

• B.

Two isobars and two isotherms

• C.

Two isotherms and two isomets

• D.

D. Two adiabats and two isotherms
Explanation
A Carnot cycle consists of two adiabats and two isotherms. This is because a Carnot cycle is an idealized thermodynamic cycle that represents the most efficient heat engine possible. The adiabats represent the processes in which no heat is exchanged with the surroundings, and the isotherms represent the processes in which the temperature remains constant. By combining these processes, the Carnot cycle maximizes the conversion of heat into work.

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• 9.

### During an isothermal process, 5.0 J of heat is removed from an ideal gas.  What is the change in internal energy?

• A.

Zero

• B.

2.5 J

• C.

5.0 J

• D.

10 J

A. Zero
Explanation
if its isothermal there's no change in temperature. if there's no change in temperature there's no change in internal energy.

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• 10.

### 200 J of work is done in compressing a gas adiabatically.  What is the change in internal energy of the gas?

• A.

Zero

• B.

100

• C.

200

• D.

There is not enough information

C. 200
Explanation
In an adiabatic process, no heat is exchanged between the system and its surroundings. Therefore, the change in internal energy of the gas is equal to the work done on the gas, which in this case is 200 J.

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• 11.

### 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?

• A.

Zero

• B.

52.5 J

• C.

97.5 J

• D.

150 J

B. 52.5 J
Explanation
The efficiency of a heat engine is given by the formula: efficiency = (work output/heat input) x 100%. In this case, the efficiency is 35.0%. We are given the heat input per cycle, which is 150 J. To find the work output, we can rearrange the formula to solve for work output: work output = (efficiency/100%) x heat input. Plugging in the values, we get work output = (35.0/100) x 150 J = 52.5 J. Therefore, the heat engine performs 52.5 J of work in each cycle.

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• Current Version
• Mar 21, 2023
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• Dec 20, 2010
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