Refresh KidLearning
LESSON 02 / 16 · TOPIC 9.1

Temperature measures an average, not a total

You will be able to: Use absolute temperature to compare average translational kinetic energy and rms speed.

Official College Board Unit 9Free study resourceReview editionTeacher review pending

Is a bigger cup automatically hotter?

Two cups of water at 20 °C have the same temperature even if one cup holds twice as much water. Temperature describes a thermal state, not the total amount of energy. To calculate particle motion simply, we next use a monatomic ideal gas.

A useful starting point: Gas pressure starts with particle collisions →

Words and symbols before equations

Kelvin T
Absolute temperature: T(K)=temperature(°C)+273.15. Use kelvin in gas equations.
Boltzmann constant k_B
1.38×10⁻²³ J/K in these examples; links particle energy to temperature.
Average kinetic energy K_avg
Mean translational kinetic energy per particle, in J.
Root-mean-square speed v_rms
Square each particle speed, average, then take the square root; not the ordinary mean speed.
Average translational energy follows TK_avg (10⁻²¹ J)Absolute temperature (K)003006.2560012.590018.75120025
Read this model snapshot. At 300 K: K_avg=6.21e-21 J and v_rms=519.6 m/s. Particle mass=4.6e-26 kg. The dot marks selected T; changing mass changes speed, not this energy curve.
What this picture assumes

Classical ideal-gas translational motion; k_B=1.38×10⁻²³ J/K. Particle mass slider values are multiplied by 10⁻²⁶ kg. Average translational kinetic energy is distinct from total internal energy.

Connect the picture to the physics

For an ideal monatomic gas, K_avg=(3/2)k_B T=(1/2)m v_rms². Increasing kelvin temperature increases average translational kinetic energy. Individual particles still have different speeds and exchange energy in collisions.

Solving gives v_rms=√(3k_B T/m). At the same temperature, heavier particles have the same average translational kinetic energy as lighter particles but a smaller rms speed. Quadrupling T doubles v_rms; doubling T does not double speed.

Temperature comparisons using ratios require absolute temperature. Going from 20 °C to 40 °C is going from 293.15 K to 313.15 K, not doubling T. We use 300 K and 600 K for simple ratios.

A worked example, step by step

For particles of mass 4.6×10⁻²⁶ kg at 300 K, estimate K_avg and v_rms. Then compare with 600 K.

  1. K_avg=1.5(1.38×10⁻²³)(300)=6.21×10⁻²¹ J.
  2. v_rms=√[3(1.38×10⁻²³)(300)/(4.6×10⁻²⁶)]=√270000≈520 m/s.
  3. At 600 K, K_avg doubles to 1.242×10⁻²⁰ J.
  4. The rms speed increases by √2 to about 735 m/s.
Common mix-up

Equal temperatures mean equal average translational kinetic energies for ideal gas particles, not equal speeds or equal total internal energies.

CHECK THE IDEA

Two ideal gases have equal T but one has four times the particle mass. Which has the larger rms speed?

Compare with an explanation

The lighter gas has twice the rms speed; both have the same average translational kinetic energy.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Change temperature with particle mass fixed. Then change particle mass with temperature fixed. Compare the energy and speed readouts before explaining the different dependencies.

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

Average translational energy follows TK_avg (10⁻²¹ J)Absolute temperature (K)003006.2560012.590018.75120025

At 300 K: K_avg=6.21e-21 J and v_rms=519.6 m/s. Particle mass=4.6e-26 kg. The dot marks selected T; changing mass changes speed, not this energy curve.

Classical ideal-gas translational motion; k_B=1.38×10⁻²³ J/K. Particle mass slider values are multiplied by 10⁻²⁶ kg. Average translational kinetic energy is distinct from total internal energy.

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. At fixed particle mass, increasing T from 300 K to 1200 K changes v_rms by…

Show answer and reasoning

2×. Speed is proportional to √T, so √4=2.

2. Two samples of the same ideal gas have the same T. One has twice as many particles. The average energy per particle is…

Show answer and reasoning

The same. Average translational kinetic energy depends on T; total internal energy also depends on particle count.

Original written challenge

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

Gas A has particle mass m and gas B has particle mass 4m. Both are at 300 K. (a) Compare K_avg. (b) Find v_rms,B/v_rms,A. (c) Find the temperature required for B to match A’s original rms speed. (d) Explain why Celsius ratios would be inappropriate.

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

Compare with the answer and four-point rubric
  1. 1 point: Equal average translational kinetic energy because T is equal.
  2. 1 point: The ratio is √(m/4m)=1/2.
  3. 1 point: T_B must be 4(300)=1200 K.
  4. 1 point: The energy relation uses absolute temperature; Celsius has an offset zero.

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 1What must be in kelvin?

T in ideal gas and particle-energy equations, and temperature ratios.

RECALL 2Does every particle move at v_rms?

No. It summarizes the squared speeds of a distribution.

RECALL 3Temperature versus total energy?

Temperature sets an average energy scale; total energy also depends on how much gas is present.

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

Temperature measures an average, not a total

  • T(K)=t(°C)+273.15.
  • K_avg=(3/2)k_B T.
  • v_rms=√(3k_B T/m).

Remember: Equal temperatures mean equal average translational kinetic energies for ideal gas particles, not equal speeds or equal total internal energies.

Conditions: Classical ideal-gas translational motion; k_B=1.38×10⁻²³ J/K. Particle mass slider values are multiplied by 10⁻²⁶ kg. Average translational kinetic energy is distinct from total internal energy.

Refresh Kid · AP Physics 2 Unit 1 (official Unit 9) · Objectives 9.1.B · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 9.1, objectives 9.1.B. 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 Temperature measures an average, not a total. 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.