Thermodynamic Processes Practice Test: Test Your Physics Skills

Grade 11th

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

| Attempts: 28 | Questions: 20 | Updated: Mar 17, 2026

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1. Which process has w=0 for a gas in a rigid container?

Isothermal

Isobaric

Isochoric

Adiabatic

In a rigid container, volume does not change, so Δv=0. Since pv work is w=∫p dv, no volume change means w=0.

Explanation

In a rigid container, volume does not change, so Δv=0. Since pv work is w=∫p dv, no volume change means w=0.

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About This Quiz

Thermodynamic Processes Practice Test: Test Your Physics Skills - Quiz

This quiz features 20 questions about thermodynamic processes, designed for students in Grade 11. You will explore concepts like the laws of thermodynamics, heat transfer, and work done in systems, all of which are key to understanding how energy moves and changes form. Mastering these topics is important for success... see morein physics and can help you in future science courses. Take your time, think critically, and see how well you can apply these principles to solve problems effectively.
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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. If a gas is heated in a rigid container, all the added heat goes into increasing internal energy.

True

False

In a rigid container, the gas cannot do pv work because volume is constant. Therefore Δu=q, meaning all the heat added increases internal energy.

Explanation

In a rigid container, the gas cannot do pv work because volume is constant. Therefore Δu=q, meaning all the heat added increases internal energy.

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3. A system has Δu=-250 J and does w=+400 J. Heat q is:

250 J

150 J

−650 J

−150 J

Rearranging Δu=q−w gives q=Δu+w. Substituting gives q=-250+400=150 J, meaning net heat enters even though internal energy drops because work done is larger.

Explanation

Rearranging Δu=q−w gives q=Δu+w. Substituting gives q=-250+400=150 J, meaning net heat enters even though internal energy drops because work done is larger.

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4. Work and heat are 'methods' of energy transfer, not energy stored in the system.

True

False

Heat and work describe energy crossing the system boundary due to temperature difference or mechanical interaction. Internal energy u is the stored energy in the system and is a state function.

Explanation

Heat and work describe energy crossing the system boundary due to temperature difference or mechanical interaction. Internal energy u is the stored energy in the system and is a state function.

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5. In an isothermal process for an ideal gas, the temperature is constant so Δu is approximately ___.

For an ideal gas, internal energy is primarily a function of temperature. If temperature stays constant, the change in internal energy is approximately zero.

Explanation

For an ideal gas, internal energy is primarily a function of temperature. If temperature stays constant, the change in internal energy is approximately zero.

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6. In any process, if q=w, then Δu=0.

True

False

The first law states Δu=q−w. If q equals w, the difference is zero, so internal energy does not change.

Explanation

The first law states Δu=q−w. If q equals w, the difference is zero, so internal energy does not change.

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7. For an ideal gas isothermal compression, q is negative and w(by the gas) is negative.

True

False

Compression means the gas has Δv<0, so work done by the gas is negative (w<0). To keep temperature constant for an ideal gas (Δu≈0), heat must leave the gas, making q<0.

Explanation

Compression means the gas has Δv<0, so work done by the gas is negative (w<0). To keep temperature constant for an ideal gas (Δu≈0), heat must leave the gas, making q<0.

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8. Which process has q=0?

Isothermal

Isobaric

Isochoric

Adiabatic

An adiabatic process is defined by no heat transfer between the system and surroundings, so q=0. Energy can still change due to work, which then must come from (or go into) internal energy.

Explanation

An adiabatic process is defined by no heat transfer between the system and surroundings, so q=0. Energy can still change due to work, which then must come from (or go into) internal energy.

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9. A gas is compressed in an insulated cylinder. Which is most likely true?

Δu<0

Δu=0

Δu>0

Q>0

Insulated means q=0, and compression means work done by the system is negative (w<0). Then Δu=q−w=0−(negative)>0, so internal energy increases.

Explanation

Insulated means q=0, and compression means work done by the system is negative (w<0). Then Δu=q−w=0−(negative)>0, so internal energy increases.

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10. If a system cools while doing work, then Δu is likely ___ (positive/negative).

Cooling suggests heat is leaving the system (q<0) and doing work means w>0. Both effects tend to reduce internal energy, so Δu is likely negative.

Explanation

Cooling suggests heat is leaving the system (q<0) and doing work means w>0. Both effects tend to reduce internal energy, so Δu is likely negative.

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11. A gas expands at constant pressure and absorbs 2,000 J of heat while doing 800 J of work. Δu is:

2,800 J

1,400 J

1,200 J

−1,200 J

Apply Δu=q−w with q=2000 J and w=800 J. Δu=2000−800=1200 J, so internal energy increases by 1200 J.

Explanation

Apply Δu=q−w with q=2000 J and w=800 J. Δu=2000−800=1200 J, so internal energy increases by 1200 J.

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12. Which statement best matches the first law idea?

Internal energy can change only with temperature

Heat and work are two ways to change a system’s internal energy

Work is always zero in gases

Heat is a state function

Internal energy changes when energy enters or leaves as heat or when work is done. Heat and work are process quantities (energy transfer), while u is stored energy.

Explanation

Internal energy changes when energy enters or leaves as heat or when work is done. Heat and work are process quantities (energy transfer), while u is stored energy.

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13. A system is cooled while being compressed. Which combination is possible?

Q>0, w>0

Q<0, w<0

Q<0, w>0

Q>0, w<0 and cooling impossible

Cooling means heat leaves, so q<0. Compression means work done by the system is negative (w<0), so both can be negative at the same time.

Explanation

Cooling means heat leaves, so q<0. Compression means work done by the system is negative (w<0), so both can be negative at the same time.

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14. Which statements are always true in this convention Δu=q-w?

If q>0, heat enters the system

If w>0, the system does work on surroundings

If Δu>0, temperature must increase for any substance

If q=0 and w>0, then Δu<0

A and B are true by definition of the sign convention. D follows directly from Δu=q−w: if q=0 and w>0, then Δu is negative; C is not always true because Δu and temperature don’t have a universal one-to-one link for all substances and situations.

Explanation

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15. If an ideal gas undergoes an isothermal expansion and does 500 J of work, then the heat transfer q is:

0 J

+500 J

−500 J

+1,000 J

In an isothermal ideal-gas process, Δu≈0 so 0=q−w. Therefore q=w, and for an expansion with w=+500 J, heat must enter: q=+500 J.

Explanation

In an isothermal ideal-gas process, Δu≈0 so 0=q−w. Therefore q=w, and for an expansion with w=+500 J, heat must enter: q=+500 J.

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16. A gas in a piston does 0 J of work but releases 600 J of heat. Δu is:

+600 J

0 J

−600 J

+1,200 J

If w=0, then Δu=q. Releasing heat means q is negative, so Δu=-600 J.

Explanation

If w=0, then Δu=q. Releasing heat means q is negative, so Δu=-600 J.

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17. Which are good 'first steps' in a first law problem?

State the sign convention

Identify if volume changes (work or no work)

Decide if it’s insulated (adiabatic) or not

Replace every problem with a weighted average temperature formula

Declaring the sign convention prevents mistakes with plus/minus signs. Checking volume change and insulation tells you whether w or q is zero, which simplifies the first law setup.

Explanation

Declaring the sign convention prevents mistakes with plus/minus signs. Checking volume change and insulation tells you whether w or q is zero, which simplifies the first law setup.

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18. A system absorbs 300 J and has Δu=+50 J. The work done by the system is:

−250 J

−350 J

250 J

350 J

Rearrange Δu=q−w to w=q−Δu. With q=300 J and Δu=+50 J, w=300−50=250 J.

Explanation

Rearrange Δu=q−w to w=q−Δu. With q=300 J and Δu=+50 J, w=300−50=250 J.

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19. A system absorbs 900 J of heat while the surroundings do 300 J of work on it. Δu is:

600 J

1,200 J

−600 J

−1,200 J

Using Δu=q−w with compression. Absorbing heat gives q=+900 J, and 'work done on the system' corresponds to w=-300 J. So Δu=900−(−300)=1200 J, meaning internal energy rises.

Explanation

Using Δu=q−w with compression. Absorbing heat gives q=+900 J, and 'work done on the system' corresponds to w=-300 J. So Δu=900−(−300)=1200 J, meaning internal energy rises.

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