Cycles, PV Loops, and Energy Accounting Quiz

Grade 11th

Reviewed by Ekaterina Yukhnovich

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Quizzes Created: 11119 | Total Attempts: 9,762,531

| Questions: 20 | Updated: Mar 6, 2026

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1. Which statements are correct about "work done by the system"?

Expansion tends to give W>0

Compression tends to give W<0

Work depends only on initial and final temperature

Work can be found from a PV diagram

Concept: work sign and path dependence. Expansion typically means ΔV>0, giving W>0, while compression gives ΔV<0 and W<0. Work is path-dependent, and on PV diagrams it is represented by the area under/within curves.

Explanation

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About This Quiz

Cycles, Pv Loops, And Energy Accounting Quiz - Quiz

This quiz features 20 questions covering cycles, PV loops, and energy accounting, essential topics in thermodynamics for students in Grade 11. Understanding these concepts helps you grasp how energy moves and transforms, which is crucial in many scientific and engineering fields. You'll explore key ideas like the first and second... see morelaws of thermodynamics and how to analyze different thermodynamic processes. As you tackle these questions, you'll build a stronger foundation for advanced studies and practical applications in science. Get ready to enhance your knowledge and confidence in thermodynamics!
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2. What first name or nickname would you like us to use?

You may optionally provide this to label your report, leaderboard, or certificate.

2. In the equation ΔU=Q-W, if Q and W are both positive, then ΔU must be:

Negative

Could be positive, negative, or zero depending on magnitudes

Always positive

Always zero

Concept: comparing magnitudes in ΔU=Q-W. If Q>W, ΔU is positive; if Q=W, ΔU is zero; if Q<W, ΔU is negative. The signs alone don’t determine ΔU—relative size matters.

Explanation

Concept: comparing magnitudes in ΔU=Q-W. If Q>W, ΔU is positive; if Q=W, ΔU is zero; if Q<W, ΔU is negative. The signs alone don’t determine ΔU—relative size matters.

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3. Which are "red flags" that your first law setup might be wrong?

Final temperature outside realistic limits with small energies

A cycle giving ΔU≠0

A rigid container calculation giving nonzero work

Units mismatch (J vs kJ vs Pa·m³)

Concept: sanity checks in thermodynamics. Cycles must give ΔU=0 and rigid containers must give W=0, so violating those is a clear setup error. Unrealistic outputs and unit mismatches often indicate sign mistakes or inconsistent conversions.

Explanation

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4. A system has Q=-80 J and ΔU=+40 J. Work done by the system is:

+120 J

+40 J

-120 J

-40 J

Concept: solving for W from ΔU and Q. Use ΔU=Q-W: 40=-80-W. Adding 80 gives 120=-W, so W=-120 J, meaning work was done on the system.

Explanation

Concept: solving for W from ΔU and Q. Use ΔU=Q-W: 40=-80-W. Adding 80 gives 120=-W, so W=-120 J, meaning work was done on the system.

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5. Internal energy can increase even when the system releases heat, if enough work is done on it.

True

False

Concept: competing effects of Q and W. Releasing heat makes Q negative, which tends to lower U. But if the surroundings do enough work on the system (W very negative), then ΔU=Q-W can still be positive.

Explanation

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6. An insulated gas expands and does 250 J of work. ΔU is:

+250 J

-250 J

+500 J

Concept: adiabatic expansion. Insulated means Q=0, and expansion with work done by the gas means W>0. Then ΔU=0-250=-250 J, so internal energy decreases.

Explanation

Concept: adiabatic expansion. Insulated means Q=0, and expansion with work done by the gas means W>0. Then ΔU=0-250=-250 J, so internal energy decreases.

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7. If the system is insulated, then Q=___.

Concept: insulation implies adiabatic. An insulated boundary prevents heat transfer between system and surroundings. Therefore Q=0 by definition of an adiabatic condition.

Explanation

Concept: insulation implies adiabatic. An insulated boundary prevents heat transfer between system and surroundings. Therefore Q=0 by definition of an adiabatic condition.

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8. A process has ΔU=+600 J and the system does W=+900 J. Heat Q is:

300 J

1,500 J

-300 J

-1,500 J

Concept: solving for Q. Rearrange ΔU=Q-W to Q=ΔU+W. Q=600+900=1500 J, meaning heat input must cover both the internal energy increase and the work output.

Explanation

Concept: solving for Q. Rearrange ΔU=Q-W to Q=ΔU+W. Q=600+900=1500 J, meaning heat input must cover both the internal energy increase and the work output.

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9. If W=0 for every step in a cycle, then net heat must also be zero.

True

False

Concept: cycle with ΔU=0. Over a cycle, ΔU=0. If net W=0 as well, then the first law gives Q_net=0.

Explanation

Concept: cycle with ΔU=0. Over a cycle, ΔU=0. If net W=0 as well, then the first law gives Q_net=0.

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10. A rigid tank is taken through a cycle of heating and cooling (volume constant the whole time). Net work over the cycle is:

Positive

Negative

Zero

Cannot be determined

Concept: no PV work at constant volume. If volume never changes, dV=0 throughout the cycle. Therefore W=∫PdV=0 for every step and the net work is zero.

Explanation

Concept: no PV work at constant volume. If volume never changes, dV=0 throughout the cycle. Therefore W=∫PdV=0 for every step and the net work is zero.

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11. For any complete cycle, the change in internal energy is:

Positive

Negative

Zero

Depends on the path

Concept: state function over a cycle. Internal energy U depends only on the state, so returning to the starting state makes ΔU=0. This is true regardless of what happens in between.

Explanation

Concept: state function over a cycle. Internal energy U depends only on the state, so returning to the starting state makes ΔU=0. This is true regardless of what happens in between.

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12. A system releases Q=-400 J and has W=-100 J. ΔU is:

+500 J

+300 J

-300 J

-500 J

Concept: substituting signs correctly. ΔU=Q-W=-400-(-100)=-400+100=-300 J. Even though compression (negative W) adds energy, the heat loss is larger, so ΔU is negative.

Explanation

Concept: substituting signs correctly. ΔU=Q-W=-400-(-100)=-400+100=-300 J. Even though compression (negative W) adds energy, the heat loss is larger, so ΔU is negative.

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13. If a system does more work than the heat it absorbs, its internal energy must decrease.

True

False

Concept: comparing Q and W in ΔU=Q-W. If W>Q, then Q-W is negative, so ΔU<0. That means internal energy decreases because more energy left as work than entered as heat.

Explanation

Concept: comparing Q and W in ΔU=Q-W. If W>Q, then Q-W is negative, so ΔU<0. That means internal energy decreases because more energy left as work than entered as heat.

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14. A system absorbs Q=+1,500 J and its internal energy decreases by 200 J. Work done by the system is:

1,300 J

1,700 J

-1,700 J

-1,300 J

Concept: using W=Q-ΔU. Internal energy decreases means ΔU=-200 J. W=Q-ΔU=1500-(-200)=1700 J, so the system did 1700 J of work.

Explanation

Concept: using W=Q-ΔU. Internal energy decreases means ΔU=-200 J. W=Q-ΔU=1500-(-200)=1700 J, so the system did 1700 J of work.

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15. In any process, rearranging ΔU=Q-W gives W=-____.

Concept: algebraic rearrangement of the first law. Starting from ΔU=Q-W, add W to both sides and subtract ΔU from both sides. This gives W=Q-ΔU, useful when solving for work.

Explanation

Concept: algebraic rearrangement of the first law. Starting from ΔU=Q-W, add W to both sides and subtract ΔU from both sides. This gives W=Q-ΔU, useful when solving for work.

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16. Over a cycle, if the system releases 200 J of heat overall, then the net work done by the system is:

+200 J

-200 J

0 J

+400 J

Concept: cycle energy balance. Releasing heat overall means Q_net=-200 J. Since Q_net=W_net for a cycle, W_net must also be -200 J.

Explanation

Concept: cycle energy balance. Releasing heat overall means Q_net=-200 J. Since Q_net=W_net for a cycle, W_net must also be -200 J.

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17. If a PV cycle is clockwise, the net work done by the gas is typically positive.

True

False

Concept: direction of PV loop. A clockwise loop indicates expansion occurs at higher pressure than compression. That yields positive net area and therefore positive net work done by the gas.

Explanation

Concept: direction of PV loop. A clockwise loop indicates expansion occurs at higher pressure than compression. That yields positive net area and therefore positive net work done by the gas.

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18. A cycle has net work done by the system of +700 J. Then the net heat transfer to the system is:

-700 J

0 J

+700 J

+1,400 J

Concept: Q_net equals W_net in a cycle. For a cycle, ΔU=0, so Q_net=W_net. If net work done by the system is +700 J, the system must have net heat input of +700 J.

Explanation

Concept: Q_net equals W_net in a cycle. For a cycle, ΔU=0, so Q_net=W_net. If net work done by the system is +700 J, the system must have net heat input of +700 J.

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19. On a PV diagram, the net work done over one cycle is represented by:

The highest pressure reached

The average volume

The area enclosed by the loop (with sign)

The slope of the curve

Concept: PV loop area equals net work. Net work over a cycle is ∫PdV, which is the signed area inside the loop. Clockwise vs counterclockwise direction sets whether the work is positive or negative.

Explanation

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