First Law and PV Work Quiz: Constant Pressure + PV Area

11th Grade

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By Thames

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Quizzes Created: 10017 | Total Attempts: 9,652,179

| Questions: 20 | Updated: Mar 6, 2026

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1. For a gas expanding at constant pressure, the work done by the gas is:

W=Δu

W=q

W=pΔv

W=mcΔt

Concept: PV work at constant pressure. For constant pressure expansion/compression, work equals pressure times volume change. This is a special case of w=∫p dv where p is constant.

Explanation

Concept: PV work at constant pressure. For constant pressure expansion/compression, work equals pressure times volume change. This is a special case of w=∫p dv where p is constant.

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About This Quiz

First Law and Pv Work Quiz: Constant Pressure + Pv Area - Quiz

This assessment focuses on the principles of the First Law of Thermodynamics and PV work under constant pressure. It evaluates your understanding of energy conservation, work done in thermodynamic processes, and the significance of PV area in physical systems. Mastering these concepts is essential for anyone studying thermodynamics, as they... see moreare foundational to understanding energy transfer and system behavior. see less

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2. If the gas expands, Δv is ___ (positive/negative).

Concept: Volume change sign. Expansion means final volume is larger than initial volume. Therefore Δv=v_f−v_i is positive.

Explanation

Concept: Volume change sign. Expansion means final volume is larger than initial volume. Therefore Δv=v_f−v_i is positive.

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3. A gas expands from 2.0 L to 5.0 L at p=100 kPa. Δv is:

3.0 L

7.0 L

-3.0 L

0.3 L

Concept: Calculating Δv. Volume change is final minus initial: 5.0−2.0=3.0 L. A positive result matches expansion.

Explanation

Concept: Calculating Δv. Volume change is final minus initial: 5.0−2.0=3.0 L. A positive result matches expansion.

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4. Convert Δv=3.0 L to cubic meters.

3.0×10^(-6) m³

3.0×10^(-3) m³

3.0 m³

3.0×10^3 m³

Concept: Unit conversion liters to m³. 1 L equals 10^(-3) m³. So 3.0 L becomes 3.0×10^(-3) m³.

Explanation

Concept: Unit conversion liters to m³. 1 L equals 10^(-3) m³. So 3.0 L becomes 3.0×10^(-3) m³.

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5. Using q3–q4, the work done is:

0.3 J

30 J

300 J

300,000 J

Concept: Work from pΔv with SI units. Convert pressure: 100 kPa = 100,000 Pa, and Δv=3.0×10^(-3) m³. Then w=(100,000)(3.0×10^(-3))=300 J.

Explanation

Concept: Work from pΔv with SI units. Convert pressure: 100 kPa = 100,000 Pa, and Δv=3.0×10^(-3) m³. Then w=(100,000)(3.0×10^(-3))=300 J.

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6. On a PV diagram, the area under the curve represents the work done by the gas.

Concept: PV area as work. Work is w=∫p dv, which geometrically equals the area under the p–v curve. The sign depends on whether volume increases (positive) or decreases (negative).

Explanation

Concept: PV area as work. Work is w=∫p dv, which geometrically equals the area under the p–v curve. The sign depends on whether volume increases (positive) or decreases (negative).

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7. A gas absorbs 600 J of heat and does 300 J of work. Δu is:

900 J

300 J

0 J

-300 J

Concept: First law calculation. With q=+600 J and w=+300 J, Δu=q−w=600−300=300 J. Internal energy increases because heat input exceeds work output.

Explanation

Concept: First law calculation. With q=+600 J and w=+300 J, Δu=q−w=600−300=300 J. Internal energy increases because heat input exceeds work output.

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8. If a gas is compressed at constant pressure, w (by the gas) is negative.

Concept: Work sign for compression. Compression means Δv<0, so w=pΔv is negative (for positive pressure). Negative w indicates work is done on the gas, not by it.

Explanation

Concept: Work sign for compression. Compression means Δv<0, so w=pΔv is negative (for positive pressure). Negative w indicates work is done on the gas, not by it.

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9. A gas is compressed from 4.0 L to 1.0 L at 200 kPa. Work done by the gas is closest to:

+600 J

-600 J

-6,000 J

+6,000 J

Concept: pΔv with compression. Δv=-3.0 L=-3.0×10^(-3) m³ and p=200,000 Pa. w=200,000(-0.003)=-600 J, negative as expected for compression.

Explanation

Concept: pΔv with compression. Δv=-3.0 L=-3.0×10^(-3) m³ and p=200,000 Pa. w=200,000(-0.003)=-600 J, negative as expected for compression.

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10. In Δu=q-w, if w is negative (compression), then -w is ___ (positive/negative).

Concept: Subtracting a negative. If w is negative, then -w is positive. This means compression tends to increase Δu for a given q because work is effectively adding energy to the system.

Explanation

Concept: Subtracting a negative. If w is negative, then -w is positive. This means compression tends to increase Δu for a given q because work is effectively adding energy to the system.

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11. Which statements are correct? (Select multiple answers)

Expansion at positive pressure gives positive work by the gas

Compression at positive pressure gives negative work by the gas

Work equals heat for all processes

Work depends on the process path

Concept: Work sign and path dependence. The sign of work follows the sign of Δv for positive pressure. Work depends on the path because the pressure during the volume change can vary between processes.

Explanation

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12. A gas does 150 J of work while releasing 40 J of heat. Δu equals:

+190 J

+110 J

-190 J

-110 J

Concept: Using signs for q and w. Releasing heat means q=-40, and doing work means w=+150. Δu=q−w=-40-150=-190 J, so internal energy decreases.

Explanation

Concept: Using signs for q and w. Releasing heat means q=-40, and doing work means w=+150. Δu=q−w=-40-150=-190 J, so internal energy decreases.

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13. If Δu is negative, internal energy decreased during the process.

Concept: Meaning of Δu sign. Δu<0 means the final internal energy is lower than the initial internal energy. This typically occurs when the system does more work than the heat it absorbs, or when it releases heat.

Explanation

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14. A process has Δu=+80 J and w=+50 J. Heat q is:

30 J

130 J

-30 J

-130 J

Concept: Rearranging the first law. From Δu=q−w, solve for q: q=Δu+w. q=80+50=130 J, meaning net heat enters the system.

Explanation

Concept: Rearranging the first law. From Δu=q−w, solve for q: q=Δu+w. q=80+50=130 J, meaning net heat enters the system.

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15. If a process is cyclic (returns to initial state), Δu=___.

Concept: State function over a cycle. A cycle returns the system to its initial state, so any state function has zero net change. Since u is a state function, Δu=0 for a complete cycle.

Explanation

Concept: State function over a cycle. A cycle returns the system to its initial state, so any state function has zero net change. Since u is a state function, Δu=0 for a complete cycle.

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16. In a full cycle, the first law becomes:

Net q is always zero

Net w is always zero

Net q equals net w

Net Δu is nonzero

Concept: First law for cycles. Over a cycle Δu=0, so 0=q_net−w_net. Therefore q_net = w_net, meaning net heat in equals net work out (using this convention).

Explanation

Concept: First law for cycles. Over a cycle Δu=0, so 0=q_net−w_net. Therefore q_net = w_net, meaning net heat in equals net work out (using this convention).

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17. Work done by the system can be nonzero even if the system returns to its initial temperature.

Concept: Cycles and net work. A system can return to its original state (including temperature) and still do net work if it also has net heat transfer. That’s the basic idea behind heat engines.

Explanation

Concept: Cycles and net work. A system can return to its original state (including temperature) and still do net work if it also has net heat transfer. That’s the basic idea behind heat engines.

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18. Two expansions go from the same initial volume to the same final volume. Process A stays at higher pressure than Process B during expansion. The work done by the gas is:

Same in A and B

Greater in A than in B

Greater in B than in A

Zero in both

Concept: PV area and work. Work is the area under the PV curve between the same volumes. Higher pressure at each volume means a larger area, so process A produces greater work.

Explanation

Concept: PV area and work. Work is the area under the PV curve between the same volumes. Higher pressure at each volume means a larger area, so process A produces greater work.

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19. A gas expands against zero external pressure (free expansion). The work done by the gas is w=___.

Concept: Work requires opposing pressure. Mechanical work on the surroundings requires a force (pressure) opposing expansion. If external pressure is zero, there’s nothing to push against, so w=0.

Explanation

Concept: Work requires opposing pressure. Mechanical work on the surroundings requires a force (pressure) opposing expansion. If external pressure is zero, there’s nothing to push against, so w=0.

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20. A gas expands freely into a vacuum in an insulated container. Which is correct (using Δu=q-w)?

Q>0, w>0, so Δu>0

Q<0, w<0, so Δu<0

Q=0, w=0, so Δu=0

Q=0, w>0, so Δu<0

Concept: Free expansion + adiabatic condition. Insulated means q=0, and expansion into a vacuum means no work is done (w=0). Therefore Δu=0, so internal energy does not change.

Explanation

Concept: Free expansion + adiabatic condition. Insulated means q=0, and expansion into a vacuum means no work is done (w=0). Therefore Δu=0, so internal energy does not change.

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