Unit 9 · Topic 9.4

9.4 The First Law of Thermodynamics

The first law of thermodynamics, ΔU = Q + W, is conservation of energy for a gas: its internal energy changes only through heating or through work done on it. PV diagrams turn this into a picture, where the area under a process curve gives the work. This topic carries a lot of the unit's problem solving.

Key terms

internal energy
first law of thermodynamics
PV diagram
isotherm
isobaric and isovolumetric processes
adiabatic process

Internal energy

A system's internal energy U is the kinetic energy of all its particles plus the potential energy stored in how they're arranged. Atoms in an ideal gas don't pull on each other, so there's no internal potential energy. Its internal energy is just the total kinetic energy of the atoms. For an ideal monatomic gas:

$U = \frac{3}{2}nRT = \frac{3}{2}Nk_BT$

So the internal energy of an ideal gas depends only on its temperature. If T doesn't change, U doesn't change. Internal energy is energy inside the system: it can change while the container's center of mass sits perfectly still.

The first law

$\Delta U = Q + W$

Q is the energy added by heating (negative for cooling). W is the work done on the gas. For a gas pushed by a constant (or average) pressure:

$W = -P\Delta V$

When the gas is compressed, ΔV is negative, so W is positive: the surroundings push energy in. When the gas expands, it pushes the surroundings out of the way, so W on the gas is negative and the gas loses energy. If the system is isolated, nothing crosses its boundary, so its total energy stays constant.

Heads-up: some textbooks and videos write ΔU = Q − W, where W is the work done by the gas. Both give the same physics. The AP equation sheet uses ΔU = Q + W with W done on the gas, so stick with that on the exam.

PV diagrams

A PV diagram plots pressure (vertical) against volume (horizontal). Each point is a state of the gas, and a path between points is a process. The size of the work equals the area under the path, down to the volume axis.

Path moving right (expansion): W on the gas is negative.
Path moving left (compression): W on the gas is positive.
Vertical path (no volume change): W = 0.
Isotherms are curves of constant temperature, shaped like P ∝ 1/V. Isotherms farther from the origin are hotter.
A closed loop (cycle) returns to its start, so ΔU = 0 over the cycle. The net work is the area inside the loop. Going clockwise, the gas does net work on its surroundings, so W on the gas is negative.

Four special processes

Each special process removes one piece of the first law.

Process	What stays fixed	First law becomes
Isovolumetric	volume (W = 0)	ΔU = Q
Isobaric	pressure	ΔU = Q − PΔV
Isothermal	temperature (ΔU = 0)	Q = −W
Adiabatic	no heating or cooling (Q = 0)	ΔU = W

Reasoning with the table

In an isothermal expansion, the gas does work, so it must absorb an equal amount of energy by heating to keep its temperature steady. In an adiabatic compression, work goes in with no way out, so U and T rise; that's why a bike pump gets warm. In an adiabatic expansion, the gas cools. On a PV diagram, an adiabat is steeper than an isotherm through the same point.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Isobaric expansion
A gas at a constant pressure of 2.0 × 10⁵ Pa expands from 0.010 m³ to 0.025 m³ while absorbing 5000 J by heating. Find the work done on the gas and the change in its internal energy. Does its temperature go up or down?
Show the solutionHide the solution
1. Step 1: ΔV = 0.025 − 0.010 = 0.015 m³.
2. Step 2: Work on the gas: $W = -P\Delta V = -(2.0 \times 10^5)(0.015) = -3000$ J. Negative, because the gas expanded and pushed its surroundings.
3. Step 3: First law: ΔU = Q + W = 5000 + (−3000) = +2000 J.
4. Step 4: U went up, and for an ideal gas U depends only on T, so the temperature rose.
Answer: W = −3000 J, ΔU = +2000 J, and the temperature increases
Example 2Calculator allowed
A rectangular cycle
A gas goes around a cycle on a PV diagram: A (1.0 × 10⁵ Pa, 0.010 m³) → B (3.0 × 10⁵ Pa, 0.010 m³) → C (3.0 × 10⁵ Pa, 0.030 m³) → D (1.0 × 10⁵ Pa, 0.030 m³) → back to A. Find the net work done on the gas and the net energy added by heating over one cycle.
Show the solutionHide the solution
1. Step 1: The path goes up, right, down, then left, which is clockwise on the diagram.
2. Step 2: The enclosed area is a rectangle: (3.0 × 10⁵ − 1.0 × 10⁵ Pa)(0.030 − 0.010 m³) = (2.0 × 10⁵)(0.020) = 4000 J.
3. Step 3: Clockwise means the gas does net work on its surroundings, so the net work on the gas is −4000 J. Check: on C→D (no volume change) and A→B, W = 0; on B→C, W = −(3.0 × 10⁵)(0.020) = −6000 J; on D→A, W = −(1.0 × 10⁵)(−0.020) = +2000 J. The total is −4000 J.
4. Step 4: Back at the start, ΔU = 0, so Q = −W = +4000 J.
Answer: Net W on the gas = −4000 J; net Q = +4000 J
Example 3
Adiabatic compression (classic trap)
A gas in an insulated cylinder is quickly compressed by a piston. A student says, “No energy was added by heating, so the temperature can't change.” Explain what actually happens.
Show the solutionHide the solution
1. Step 1: Insulated and quick means Q = 0, so the process is adiabatic.
2. Step 2: Compression means ΔV is negative, so W = −PΔV is positive: the piston does work on the gas.
3. Step 3: First law: ΔU = 0 + W > 0. The internal energy rises, and for an ideal gas that means the temperature rises.
4. Step 4: The student mixed up heating with temperature. Work alone can raise the temperature.
Answer: The temperature goes up, because the work done on the gas raises its internal energy even with Q = 0.

Common mistakes

Mixing sign conventions. On the AP exam, W is work done on the gas, so W = −PΔV is negative for expansion.
Saying an isothermal process has no heating. Q = −W there, so energy is transferred; only ΔU is zero.
Confusing adiabatic (Q = 0) with isothermal (ΔT = 0). In an adiabatic process the temperature usually changes.
Taking the area to the pressure axis instead of the volume axis, or forgetting that W = 0 for a vertical (constant-volume) path.

On the exam

PV diagrams are a favorite: rank temperatures at points (use PV ∝ T), find work from areas, and fill in a table of the signs of Q, W and ΔU for each step.
Explain using the first law in words, naming each term and its sign, and connect U to temperature for an ideal gas.
Energy bar charts for a gas before and after a process may appear in translation-between-representations questions.

Connected topics

Videos

AP Physics 2 - Unit 9 - Lesson 3 - Internal Energy
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Thermodynamics and P-V Diagrams
Bozeman ScienceWatch on YouTube (opens in a new tab)
AP Physics 2 - Thermodynamics and PV Diagrams
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
AP Physics 2 - Unit 9 - Lesson 4 - PV Cycles
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
The First Law of Thermodynamics: Internal Energy, Heat, and Work
Professor Dave ExplainsWatch on YouTube (opens in a new tab)
Thermodynamics: Crash Course Physics #23
CrashCourseWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 9.4 The First Law of Thermodynamics. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A gas absorbs 500 J of energy by heating while it expands and does 200 J of work on its surroundings. What is the change in the internal energy of the gas?

Question 2 of 4Calculator allowed

A gas at a constant pressure of $2.0 \times 10^5$ Pa expands from $1.0 \times 10^{-3}$ m³ to $3.0 \times 10^{-3}$ m³. How much work is done on the gas?

Question 3 of 4Calculator allowed

An ideal gas is compressed slowly while its temperature stays constant. Which statement is correct?

Question 4 of 4Calculator allowed

A gas in a well-insulated cylinder is compressed quickly by a piston, so no energy is transferred by heating. What happens to the temperature of the gas, and why?

0 of 4 answered