Learning
LESSON 08 / 16 · TOPIC 9.4

Count the energy inside the system

You will be able to: Distinguish internal energy, temperature and center-of-mass motion.

Official College Board Unit 9Free study resourceReview editionTeacher review pending

Can a stationary container gain internal energy?

A sealed container sits still while its gas warms from 300 K to 400 K. The container need not move for the gas to gain energy. Its particles can move faster relative to the stationary container.

A useful starting point: Three ways energy crosses a temperature difference →

Words and symbols before equations

System
The matter chosen for analysis, here the gas.
Internal energy U
Microscopic kinetic and interaction potential energies within the system, in J.
Monatomic ideal gas
A gas of single atoms modeled with only translational kinetic internal energy.
State function
A property determined by the current equilibrium state, not the route taken to reach it.
Total internal energy of the same amountJ · same scale for all bars0At 300 K3740At selected T3740
Read this model snapshot. For 1 mol at 300 K, U=3740 J. Average particle translational energy=6.21e-21 J. Doubling n doubles U without changing this per-particle average.
What this picture assumes

Monatomic ideal gas with translational internal energy only, R=8.31 J/(mol·K), k_B=1.38×10⁻²³ J/K. Bulk center-of-mass motion excluded. Diagram compares total U at selected T and at 300 K for the same n.

Connect the picture to the physics

Internal energy is not the kinetic energy of the whole system’s center of mass. A stationary object has microscopic motion and interactions. For real materials, internal energy can include potential energy from particle arrangements.

For a fixed amount of monatomic ideal gas, U=(3/2)nRT=(3/2)Nk_B T. The ideal model neglects interparticle potential energy, and monatomic particles have no rotational or vibrational molecular modes in this model. Do not apply this formula indiscriminately to liquids, solids or all molecular gases.

U depends on state, so equal initial and final temperatures of the same fixed monatomic ideal gas imply ΔU=0 even if heat and work were exchanged along the way. In contrast, heat and work describe transfers during a process.

Temperature, internal energy and heat
QuantityMeaningDepends on
TThermal state / particle average energy scaleState
UMicroscopic energy within chosen systemState and amount
QEnergy transferred thermallyProcess, not stored heat

A worked example, step by step

Find ΔU for 2 mol of monatomic ideal gas warmed from 300 K to 400 K. Use R=8.31 J/(mol·K).

  1. Identify the model: fixed amount, monatomic and ideal.
  2. ΔT=400−300=100 K.
  3. ΔU=(3/2)(2)(8.31)(100)=2493 J.
  4. The positive sign means the gas gains internal energy; the transfer mechanism is not determined by this calculation.
Common mix-up

U=(3/2)nRT is for a monatomic ideal gas. Equal temperature alone does not imply equal U when amounts differ.

CHECK THE IDEA

Can U change while the center of mass remains at rest?

Compare with an explanation

Yes. Microscopic motion or configuration can change without motion of the whole system.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Change gas amount and temperature separately. Compare the bar for total U with the average particle energy. Explain which change doubles total energy without changing temperature.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Total internal energy of the same amountJ · same scale for all bars0At 300 K3740At selected T3740

For 1 mol at 300 K, U=3740 J. Average particle translational energy=6.21e-21 J. Doubling n doubles U without changing this per-particle average.

Monatomic ideal gas with translational internal energy only, R=8.31 J/(mol·K), k_B=1.38×10⁻²³ J/K. Bulk center-of-mass motion excluded. Diagram compares total U at selected T and at 300 K for the same n.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant particle, temperature or energy relationship to justify your prediction.

Apply the idea to a fresh problem Practice →

Show what you understand.

Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.

1. For fixed n of monatomic ideal gas, an isothermal process gives…

Show answer and reasoning

ΔU=0. U depends only on T for fixed n; heat may still transfer.

2. Doubling n at the same T makes U…

Show answer and reasoning

Double. U is proportional to amount as well as absolute temperature.

Original written challenge

4 points · self-check · not an official AP question

A fixed monatomic ideal gas has n=1 mol and returns to its original 300 K state after a cycle. (a) Find U at that state. (b) Find ΔU for the cycle. (c) Explain whether Q must be zero. (d) Identify why the formula is model-specific.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: U=1.5(8.31)(300)=3739.5 J.
  2. 1 point: ΔU=0 because the state returns to its start.
  3. 1 point: No. Heat and work may be nonzero but balance over the cycle.
  4. 1 point: The formula counts translational energy of monatomic ideal particles and neglects interactions/internal molecular modes.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1Internal versus bulk kinetic energy?

Internal energy concerns motion and interactions within the system, not its overall center-of-mass motion.

RECALL 2What fixes U for fixed monatomic ideal gas amount?

Its absolute temperature.

RECALL 3What does state function mean?

Its change depends on endpoints, not the process path.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Count the energy inside the system

  • Monatomic ideal gas: U=(3/2)nRT.
  • Fixed amount: ΔU=(3/2)nRΔT.
  • U is a state function; Q and W are process-dependent transfers.

Remember: U=(3/2)nRT is for a monatomic ideal gas. Equal temperature alone does not imply equal U when amounts differ.

Conditions: Monatomic ideal gas with translational internal energy only, R=8.31 J/(mol·K), k_B=1.38×10⁻²³ J/K. Bulk center-of-mass motion excluded. Diagram compares total U at selected T and at 300 K for the same n.

Refresh Kid · AP Physics 2 Unit 1 (official Unit 9) · Objectives 9.4.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 9.4, objectives 9.4.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the first AP Physics 2 unit; College Board numbers it Unit 9, continuing after AP Physics 1. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.

Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.

Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.

Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.

OPTIONAL LIVE SUPPORT

Want to work through this with a tutor?

Bring your question about Count the energy inside the system. Your explanation and answers remain free to access.

Request a physics tutor →Ask about this lesson on WhatsAppThe team can confirm teacher availability and next steps.