The First Law of Thermodynamics
Track heat and work in the first law and interpret complete pressure-volume paths.
AP Physics 2 Topic 9.4 · College Board CED printed p. 31, 32 / PDF p. 38, 39 · See this topic in the course outline
Start with the gas as your system. It can gain or lose internal energy because of thermal transfer and because a piston does work. A PV graph tracks the path, while the initial and final temperatures determine the ideal gas’s internal-energy change.
Before you start
- Ideal gas law and kelvin temperature.
- Work and conservation of energy.
- Area of rectangles and trapezoids; reading ordered graph coordinates.
Internal energy includes microscopic motion and configuration
Internal energy U is the sum of kinetic energies of constituents relative to the system’s bulk motion and potential energies associated with their configuration. A stationary block can gain U while its center of mass stays at rest: atoms vibrate more vigorously or their arrangement changes. Internal energy is distinct from the kinetic energy of the whole block moving across a room.
In the classical ideal gas model, no appreciable conservative intermolecular force acts between collisions and atomic internal structure is omitted. Thus there is no internal potential-energy contribution in this model. For an ideal monatomic gas, only translational atomic kinetic energy contributes: . For fixed N, . A change in volume alone does not change U if the temperature is unchanged. Real solids and liquids can change microscopic arrangement and potential energy without changing center-of-mass motion; do not apply this gas formula to them.
Build an energy ledger with the AP work convention
An isolated system exchanges neither matter nor energy with its surroundings, so its total energy remains constant. Energy can move between parts inside it without changing the combined total. A closed system exchanges no matter, but heat and work may cross its boundary. For the closed gas considered here, with no changes in bulk kinetic or external potential energy, conservation gives . Q is positive for heating and negative for cooling. W_on is positive when surroundings do work on the gas and negative when the gas does work on its surroundings.
This first law tracks all transfers rather than creating energy. For example, a gas receiving 500 J of heat and doing 200 J of work on its piston has Q = +500 J, W_on = −200 J, and ΔU = +300 J. If a text uses work done by the gas, its equation is ΔU = Q − W_by, where W_by = −W_on. Use one convention throughout a calculation. Internal energy depends on state; Q and W describe a particular process and are not amounts stored in the gas.
A moving boundary does pressure-volume work
A piston of area A moves a distance Δx against constant external pressure P_ext. The force magnitude is P_extA, and the volume change is AΔx. For the gas, . During expansion ΔV > 0, so the surroundings’ force opposes the outward displacement and W_on < 0. During compression ΔV < 0 and W_on > 0. If the external pressure varies, use its appropriate average over the volume interval: W_on = −P_ext,avgΔV. A volume change from 1 L to 3 L is +0.002 m³, not 2 m³.
A pressure–volume diagram uses V horizontally and P vertically, with a directional path between equilibrium states. For a quasistatic process the gas pressure matches external pressure closely, so the work magnitude is the area underneath its P-versus-V path. Rectangles give PΔV and a straight sloping segment gives the trapezoid area. Expansion makes W_on the negative of the area; compression makes it positive. For a strongly irreversible process the gas may not have one uniform pressure during the change; calculate work using external pressure rather than treating an arbitrary gas curve as the work curve.
| Directed path | Work-producing segment | W_on |
|---|---|---|
| Raise P at 1 L, then expand to 3 L | 300 kPa × 2 L = 600 J | −600 J |
| Expand at 100 kPa, then raise P at 3 L | 100 kPa × 2 L = 200 J | −200 J |
Vertical segments have zero work because ΔV = 0. The table is a numerical alternative to the two right-angle PV paths: both have the same endpoints but different areas, showing that work depends on path. Here 1 kPa·L = 1 J.
Four useful process constraints
A named process tells you which term or state variable is constrained. The table applies to a fixed amount of ideal monatomic gas, with pressure work as the only work mode.
| Process | Physical condition and PV shape | First-law consequence |
|---|---|---|
| Isovolumetric | Rigid container; V constant; vertical line | W_on = 0, so ΔU = Q |
| Isothermal | Slow exchange with a constant-temperature reservoir; T constant; PV constant hyperbola | ΔU = 0, so Q = −W_on |
| Isobaric | P constant, e.g. a freely moving loaded piston; horizontal line | W_on = −PΔV; Q supplies ΔU and energy delivered as work during expansion |
| Adiabatic | Insulated boundary or sufficiently rapid process; Q = 0 | ΔU = W_on; expansion doing nonzero work against external pressure cools, while compression with positive pressure work warms |
An isotherm is a line of constant temperature on a PV diagram. For fixed n, P = nRT/V, so a hotter isotherm lies above a cooler one at the same volume. A quasistatic adiabatic expansion doing nonzero work against external pressure curves downward more steeply than an isothermal expansion from the same state because its temperature also falls. An insulated free expansion into vacuum instead has W_on = 0, ΔU = 0, and unchanged ideal-gas temperature, despite an increase in occupied volume. Its exact curve formula is not needed here. Isothermal does not mean insulated, and adiabatic does not mean constant temperature. A cycle returns to its initial state, so its net ΔU is zero even though net heat and work may be nonzero.
Heating while a gas pushes a piston
An ideal monatomic gas receives 600 J while its piston moves between prescribed stops, expanding from 1.0 L to 3.0 L against a constant external pressure of 1.0 × 10⁵ Pa. It contains 0.20 mol and starts at 300 K. No quasistatic gas-pressure equality is assumed. Find W_on, ΔU, and final T.
Choose the closed gas. Q = +600 J and ΔV = +2.0 × 10⁻³ m³. W_on = −(1.0 × 10⁵)(2.0 × 10⁻³) = −200 J.
Why this step? An outward-moving piston transfers energy out of the gas by work, giving a negative AP work term.
ΔU = Q + W_on = 600 − 200 = 400 J.
Why this step? Only the energy remaining after the work transfer increases internal energy.
ΔT = 2ΔU/(3nR) = 800/[3(0.20)(8.31)] ≈ 160 K; T_f ≈ 460 K.
Why this step? For the ideal monatomic sample, internal energy is proportional to kelvin temperature at fixed n.
Answer and check
W_on = −200 J, ΔU = +400 J, T_f ≈ 460 K. Of the 600 J supplied, 200 J leaves as work and 400 J remains as internal energy.
Different constraints give different temperatures
Compare two expansions of the same ideal monatomic gas: one isothermal, one adiabatic. For each expansion, the gas does 150 J of work on its surroundings. Describe Q and ΔU.
Both have W_on = −150 J.
Why this step? Expansion transfers energy out by work regardless of the thermal constraint.
Isothermal: ΔU = 0, so Q = +150 J.
Why this step? At fixed T the ideal monatomic gas has fixed U; heat must replace energy delivered as work.
Adiabatic: Q = 0, so ΔU = −150 J and T falls.
Why this step? Insulation prevents heat transfer, so the energy delivered as work comes from internal energy.
Answer and check
The isothermal gas receives 150 J with no temperature change. The adiabatic gas loses 150 J of internal energy and cools because the prompt specifies 150 J of expansion work. With zero work, as in insulated vacuum expansion, its ideal-gas temperature would instead stay unchanged. Equal work does not imply equal heat or equal final temperature.
Internal energy can change in two microscopic stores
A block can remain at rest while its atoms move more vigorously and their relative positions change. Its internal energy includes microscopic kinetic energy measured relative to the block’s bulk motion and potential energy associated with the constituents’ configuration. Stretching bonds, changing separations, or changing structure can change that potential-energy store. The kinetic energy of the block moving across the room is a separate bulk store.
For an ideal monatomic gas, the model omits internal structure and sustained conservative interactions between atoms. Its internal energy is therefore just translational atomic kinetic energy, U = (3/2)nRT. The absence of a potential-energy term is an assumption of that gas model. A solid or liquid generally has interactions and configurational energy, so a temperature reading alone does not determine its complete internal energy.
An energy ledger for a stationary solid
A stationary closed solid receives 90 J by heating and 30 J by mechanical work. Its microscopic kinetic-energy store rises by 50 J. Bulk kinetic and external gravitational energies do not change. Account for the remaining internal-energy change.
- For this boundary, Q = +90 J and W_on = +30 J, so ΔU = Q + W_on = +120 J.
- The microscopic ledger is ΔU = ΔK_micro + ΔU_configuration. Thus ΔU_configuration = 120 − 50 = +70 J.
- The 70 J belongs to changed constituent configuration, not center-of-mass motion. The stipulated kinetic increase and structural change need not be inferred from temperature alone.
Answer and check
ΔU = +120 J, with +50 J microscopic kinetic and +70 J configurational potential energy; bulk kinetic-energy change is zero.
Make the model respond
Focused Physics 2 model
Constant-pressure work and energy
Predict first. During expansion, what signs should heat, work on the gas, and the internal-energy change have?
Wₒₙ = −Pₑₓₜ(Vf − Vi); ΔU = (3/2)nRΔT; ΔU = Q + Wₒₙ
- Constant-pressure path
- Selected or reference point
- Magnitude of pressure–volume work
On a narrow screen, scroll within a figure to read its labels. The table and explanation provide the same physics in text.
Expansion: Q = 1,250 J and Wₒₙ = -500 J give ΔU = 750 J.
Read the model data and comparisons
| Quantity | Value | Unit | Meaning or limit |
|---|---|---|---|
| Pressure (gas and external) | 1.000 × 10^5 | Pa | |
| Initial volume | 0.01 | m³ | |
| Final volume | 0.015 | m³ | |
| Initial temperature | 240.5 | K | |
| Final temperature | 360.8 | K | |
| Work on gas | -500 | J | |
| Heat into gas | 1,250 | J | |
| Internal-energy change | 750 | J | |
| First-law residual ΔU − Q − Wₒₙ | 5.684 × 10^−14 | J |
Start with the gas as the system. Initial and final temperatures are calculated from the same fixed amount and pressure, so they are consistent with the ideal-gas states. Work on the gas is the negative of the signed area traversed on the P–V graph. At constant pressure, Q = (5/2)PΔV for this monatomic gas. The first law accounts for energy transfer; it does not make heat a stored state variable.
Assumptions and limits. 0.500 mol of ideal monatomic gas. Quasistatic constant-pressure change with Pgas = Pexternal. Only pressure–volume work; negligible kinetic/potential-energy change of the container and no phase change. The graph is a single path, not a heat-engine cycle.
Try it from memory
Write your answer and reason before opening the worked answer. Your self-assessment is a study decision; it is not an automatically graded score.
Question 1
A closed gas loses 80 J by cooling while surroundings do 120 J of work on it. Find ΔU and explain its sign.
Worked answer and reason
ΔU = −80 + 120 = +40 J. Its internal energy rises despite cooling.
Work and heat are separate transfers. Cooling alone does not determine the net internal-energy change.
Question 2
A PV path moves upward at fixed volume. What can you infer about W_on, T, and Q for fixed n of an ideal monatomic gas?
Worked answer and reason
W_on = 0; P rises at fixed V, so T and U rise. Thus Q = ΔU > 0.
No boundary displacement means no pressure work, while PV = nRT and U = 3nRT/2 connect the graph to heating.
Question 3
Two processes take the same closed gas from the same initial state to the same final state. Must Q, W_on, and ΔU match?
Worked answer and reason
ΔU matches because the endpoints fix U. Q and W_on can differ, but each path must satisfy Q + W_on = the same ΔU.
Heat and work depend on path; internal energy is a state property.
JavaScript enables saved drafts and self-assessment. You can always work on paper.
Catch the tempting mistake
- An isothermal expansion has Q = 0.
- For an ideal gas ΔU = 0 when T is fixed, but expansion gives W_on < 0. Heating supplies Q = −W_on > 0.
- Work is a property stored in a state.
- Work is a transfer that depends on the path. Internal energy is a state property.
- Compression always cools a gas.
- Adiabatic compression adds energy by work and heats it. Cooling can occur if enough energy leaves thermally.
- A closed system cannot exchange energy.
- Closed means matter does not cross its boundary. Energy can cross by heat or work; an isolated system exchanges neither matter nor energy.
Put it to work
Give the physical reason as well as the calculation. The worked solutions are on a separate page.
- Problem 7 · signed energy ledger; numerical state; model comparison
- Problem 8 · PV construction; area; state comparison; symbolic isotherm
- Problem 11 · system boundary; energy ledger; qualitative statistical argument
- Problem 13 · quantitative PV cycle; signed energy table; entropy evaluation
- Problem 15 · conduction rate; symbolic first-law derivation; microscopic scaling; entropy
Explain and apply the model
Try these problems before opening the solutions. Show the physical model, your calculation or sketch, and the reason your conclusion follows.
Where does energy go inside an object?
A closed specimen is at rest before and after a controlled deformation. It receives 80 J of work and loses 20 J thermally. Measurements in the stated model show a 15 J increase in its microscopic kinetic energy. Its bulk kinetic and external gravitational energies do not change.
(a) Calculate its internal-energy change and its configurational potential-energy change. Explain how both changes can occur while its center of mass remains at rest.
(b) A student substitutes U = (3/2)nRT for the specimen. Explain why that substitution is not established, and state the microscopic assumptions that make it valid for an ideal monatomic gas.
Written working workspace
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For the teacher
- Require a signed energy ledger before substitution: Q into gas positive, W_on from compression positive.
- On each PV path, mark direction. The area alone gives a magnitude; expansion or compression gives the work sign.
- Keep U = 3nRT/2 explicitly limited to an ideal monatomic gas; it is not a formula for solids or arbitrary molecular gases.
- When using gas pressure for PV work, state slow mechanical equilibrium with the external pressure. For irreversible expansion use external pressure, not an unspecified gas pressure.
Source objectives and scope
- 9.4.A — Describe the internal energy of a system.
- 9.4.B — Describe the behavior of a system using thermodynamic processes.
Suggested skills are teaching guidance. The course applies all science practices across its objectives.
Science references for this teaching
- College Board AP Physics 2 Course and Exam Description, Unit 9 (PDF pages 34–42; printed pages 27–35) — All Unit 9 learning objectives, essential knowledge clauses, and AP scope boundaries; source equations checked in the supplied pinned PDF.
- OpenStax University Physics Volume 2, Molecular Model of an Ideal Gas — Supporting open-textbook reading for ideal-gas assumptions, absolute pressure, and state-variable graphs; real-gas extensions in the source are beyond this unit’s AP requirements.
- OpenStax University Physics Volume 2, First Law of Thermodynamics — Supporting open-textbook reading for energy accounting; authoring translates any work-by convention into the AP work-on convention.
- OpenStax University Physics Volume 2, Entropy — Optional conceptual supporting reading for qualitative entropy and reversibility; quantitative entropy calculations in the source are outside this AP treatment.
Glossary and deeper reading · 18 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Internal energy
Internal energy is energy associated with the microscopic motion and configuration of a system’s constituents. It excludes the kinetic energy of the whole system’s bulk motion and depends on the thermodynamic state. For example, a stationary solid can gain internal energy as its atomic vibrations become more energetic.
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Mechanical work
Mechanical work is energy transferred by a force through displacement of its point of application. For a constant force on a particle, it depends on the force component along the particle’s displacement, and it can be positive, negative, or zero. For example, a magnetic force perpendicular to a particle’s displacement does zero work on that particle.
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Ideal gas law
The ideal gas law relates absolute pressure, occupied volume, amount of gas, and kelvin temperature. The forms using moles and using particle count express the same state relationship with different constants. For example, heating a fixed amount of ideal gas in a rigid sealed container raises its pressure.
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Kelvin temperature scale
The kelvin scale measures thermodynamic temperature from absolute zero. A temperature interval of one kelvin has the same size as one degree Celsius, but their zero points differ. For example, temperature ratios in the ideal gas law must use kelvin values rather than Celsius values.
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Conservation of energy
Conservation of energy means total energy is unchanged when the full isolated system is considered. Energy can transfer between objects or change form, so one selected energy store need not remain constant. For example, friction can reduce a sliding block’s kinetic energy while increasing the thermal energy of the block and floor.
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Potential energy
Potential energy describes energy associated with the configuration of interacting parts of a system. Its numerical zero is a chosen reference, while differences in potential energy determine physical energy changes. For example, separating Earth and a ball changes their gravitational potential energy even though the energy belongs to their interaction.
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Center of mass
The center of mass is the position obtained by weighting each part’s position by its mass. Its motion represents the translational motion of the whole system under the net external force. For example, adding a heavy object to one end of a light bar shifts the center of mass toward that end.
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Ideal gas
An ideal gas model treats particles as having negligible volume and no sustained intermolecular forces between brief elastic collisions. The equilibrium state obeys the ideal gas law using absolute pressure and kelvin temperature. For example, a dilute gas far from condensation often approximates this model better than a dense gas near a phase change.
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Force
A force describes an interaction that can change an object’s momentum. It is a vector measured in newtons, so both size and direction matter when forces are added. For example, a rope can pull a cart while the floor exerts friction in another direction.
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Isolated system
An isolated system exchanges neither energy nor matter with its surroundings. In a momentum problem, conservation of total momentum specifically requires zero net external impulse over the interval analyzed. For example, two colliding carts can be treated as momentum-isolated horizontally when the external horizontal impulse is negligible.
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Heat
Heat is energy transferred between systems because of a temperature difference. It describes a transfer rather than a substance or energy store contained in an object. For example, when a warmer mug transfers energy to a cooler hand, the mug loses internal energy and the hand gains it.
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Pressure
Pressure is the normal force per unit area exerted on a surface. It is a scalar measured in pascals, with one pascal equal to one newton per square meter. For example, a gas exerts pressure throughout its volume through many microscopic collisions, even when its bulk motion is zero.
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Magnitude
Magnitude is the nonnegative size of a vector quantity. It does not specify direction, and it differs from a signed component. For example, a westward velocity can have a negative x component while its speed, the velocity magnitude, stays positive.
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State variable
A state variable is determined by a system’s present equilibrium state rather than by the route used to reach it. Internal energy and entropy are state properties, whereas heat and work describe transfers along a process. For example, two gas processes with identical endpoints have the same internal-energy change but can involve different amounts of work.
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Isothermal process
An isothermal process occurs at constant temperature. For a fixed amount of ideal gas, internal energy remains constant even though heat and work can be transferred. For example, a gas expanding slowly while exchanging energy with a temperature reservoir can remain isothermal.
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Temperature
Temperature is a state quantity that determines the direction of net energy transfer through thermal contact. For a classical ideal gas it is proportional to the average translational kinetic energy per particle, rather than to the total energy of the sample. For example, a small hot metal sample can have a higher temperature but less total internal energy than a large cooler sample.
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Kinetic energy
Kinetic energy is energy associated with motion relative to a chosen reference frame. For a nonrelativistic translating object it is one half of its mass times speed squared. For example, doubling a cart’s speed quadruples its translational kinetic energy when its mass stays fixed.
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Conservative force
A conservative force does work that depends only on the starting and ending configurations. A potential-energy function can describe it, and its work around a closed path is zero. For example, near-Earth gravity does the same work between two heights whether a ball follows a vertical or a curved path.