MYP Physics · topic MYP-07.04
Gas laws and absolute temperature
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
A gas law that fails on a Celsius axis
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A fixed amount of dilute gas is sealed in a rigid container. After equilibrium at each bath temperature, its absolute pressure is measured as supplied below. Assume ideal-gas behaviour and no leakage. The gas occupies 2.0×10⁻³ m³; take R = 8.31 J mol⁻¹ K⁻¹. Supplied data (Temperature / °C; Absolute pressure / kPa): (0, 100), (27.3, 110), (54.6, 120), (81.9, 130). (a) Extrapolate the pressure–temperature line to zero pressure and estimate the amount of gas at 0°C. [3 marks] (b) Predict pressure at 136.5°C and explain the microscopic mechanism. What if the container is allowed to expand at constant pressure instead? [3 marks]
Gas thermometer calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Absolute pressure / kPa
Horizontal axis (x): Temperature / °C
Read the exact data as a table
| Temperature / °C | Absolute pressure / kPa |
|---|---|
| 0 | 100 |
| 27.3 | 110 |
| 54.6 | 120 |
| 81.9 | 130 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Extrapolate the pressure–temperature line to zero pressure and estimate the amount of gas at 0°C.
Reveal the worked solution
- The slope is 10/27.3 kPa °C⁻¹. Extrapolating 100 kPa down to zero requires a decrease of 273°C.
- Use absolute temperature T = 273 K and SI pressure 100000 Pa in pV = nRT.
- n = 100000(0.002)/(8.31×273) ≈0.0882 mol. The mathematical intercept motivates an absolute scale; a real gas condenses before this extrapolation reaches zero.
- 136.5°C corresponds to 409.5 K, 1.5 times 273 K, so fixed-volume pressure rises by the same factor to 150 kPa.
- Higher absolute temperature means greater mean molecular kinetic energy; more vigorous momentum transfer to the walls raises pressure at fixed volume.
- At fixed pressure and fixed particle number, V/T is constant, so the volume increases by 1.5. Dividing by Celsius temperatures would give an invalid proportionality.
Answer (a) Zero-pressure intercept ≈−273°C; amount ≈0.0882 mol. (b) Rigid-container pressure = 150 kPa; at constant pressure the volume would rise to 3.0×10⁻³ m³.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.