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LESSON 05 / 16 · TOPIC 9.2

Use gas graphs to test a relationship

You will be able to: Interpret slopes and intercepts of ideal-gas graphs with controlled variables.

Official College Board Unit 9Free study resourceReview editionTeacher review pending

How could a graph suggest an absolute zero of temperature?

A sealed rigid container reads 100 kPa at 300 K and 120 kPa at 360 K. The pressure rose in the same ratio as temperature. Several readings can test a relationship more convincingly than one substituted number.

A useful starting point: Connect pressure, volume, amount and temperature →

Words and symbols before equations

Controlled variables
Quantities kept fixed; here gas amount n and container volume V.
Slope
Vertical change divided by horizontal change, with units.
Extrapolation
Extending a fitted trend outside the measured range.
Intercept
Where a fitted line meets an axis; not necessarily a physically reached state.
Rigid-container data and extrapolationAbsolute pressure (kPa)Celsius temperature (°C)-273.10-173.170-73.1514026.85210126.9280
Read this model snapshot. Slope=0.3324 kPa/K for 1 mol in 0.025 m³. Ideal extrapolated zero: −273.15 °C. The three dots are synthetic 200, 300 and 400 K values; the dashed extension is not measured.
What this picture assumes

Synthetic ideal-gas data, fixed V=0.025 m³. Solid segment spans 200–400 K; dashed segment extrapolates to 0 K (−273.15 °C). Real gases do not remain ideal to absolute zero. No uncertainty is inferred from these calculated points.

Connect the picture to the physics

At fixed n and V, P=(nR/V)T. Plotting absolute P against kelvin T gives a straight line through the ideal origin with slope nR/V. Plotting P against Celsius temperature shifts the horizontal origin; extending the ideal line reaches zero pressure at −273.15 °C.

This is an extrapolation, not evidence that a real gas stays ideal to absolute zero. Real substances can condense and violate the model long before then. Use moderate temperatures, allow thermal equilibrium, record absolute pressure, and never extrapolate a short noisy data range without discussing uncertainty.

At fixed n and T, P versus V is a decreasing curve, whereas P versus 1/V is straight. At fixed n and P, V versus T in kelvin is straight. A graph’s shape is meaningful only when its axes and controls are specified.

A worked example, step by step

At fixed volume, ideal readings are (300 K,100 kPa) and (360 K,120 kPa). Find the slope and predict pressure at 330 K.

  1. Slope=(120−100)/(360−300)=20/60=1/3 kPa/K.
  2. The ideal kelvin intercept is zero, so P=(1/3)T.
  3. At 330 K, P=110 kPa.
  4. On a Celsius axis the same line has its extrapolated zero at −273.15 °C, not 0 °C.
Common mix-up

A straight line requires the correct variables and controlled conditions. An extrapolated intercept is not a measurement.

CHECK THE IDEA

Why do different gas amounts have the same extrapolated temperature intercept?

Compare with an explanation

Changing n changes the slope; the kelvin-to-Celsius offset is unchanged.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Change the number of moles in the rigid 0.025 m³ container. Compare the P-versus-Celsius slopes. Both ideal lines have the same extrapolated zero; the dashed region is outside the synthetic data range.

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

Rigid-container data and extrapolationAbsolute pressure (kPa)Celsius temperature (°C)-273.10-173.170-73.1514026.85210126.9280

Slope=0.3324 kPa/K for 1 mol in 0.025 m³. Ideal extrapolated zero: −273.15 °C. The three dots are synthetic 200, 300 and 400 K values; the dashed extension is not measured.

Synthetic ideal-gas data, fixed V=0.025 m³. Solid segment spans 200–400 K; dashed segment extrapolates to 0 K (−273.15 °C). Real gases do not remain ideal to absolute zero. No uncertainty is inferred from these calculated points.

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 and V, which plot is linear?

Show answer and reasoning

P versus T in kelvin. P=(nR/V)T.

2. A fitted ideal line reaches P=0 at −273.15 °C. This is…

Show answer and reasoning

An extrapolation of the ideal model. The ideal trend is extended beyond the measured range; real gases need not remain gaseous there.

Original written challenge

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

For a sealed rigid container, synthetic pressures are 80, 100 and 120 kPa at 240, 300 and 360 K. (a) Calculate slope. (b) Predict P at 330 K. (c) Explain how doubling n affects the slope. (d) Give one limitation of extrapolating to P=0.

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Compare with the answer and four-point rubric
  1. 1 point: Slope=(120−80)/(360−240)=1/3 kPa/K.
  2. 1 point: P=110 kPa.
  3. 1 point: Slope nR/V doubles when n doubles at fixed V.
  4. 1 point: Real gases cease to follow the ideal model, and extrapolation amplifies measurement uncertainty.

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 fixed for P proportional to T?

Gas amount and volume.

RECALL 2Why graph P against 1/V?

At fixed n and T it linearizes PV=nRT.

RECALL 3What is the Celsius intercept?

The ideal extrapolated zero is −273.15 °C.

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

Use gas graphs to test a relationship

  • Fixed n,V: P=(nR/V)T; slope=nR/V.
  • Fixed n,T: P is proportional to 1/V.
  • T(K)=t(°C)+273.15.

Remember: A straight line requires the correct variables and controlled conditions. An extrapolated intercept is not a measurement.

Conditions: Synthetic ideal-gas data, fixed V=0.025 m³. Solid segment spans 200–400 K; dashed segment extrapolates to 0 K (−273.15 °C). Real gases do not remain ideal to absolute zero. No uncertainty is inferred from these calculated points.

Refresh Kid · AP Physics 2 Unit 1 (official Unit 9) · Objectives 9.2.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 9.2, objectives 9.2.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.

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