IB · B · extended response
A gas law that fails on a Celsius axis
The particulate nature of matter · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
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(a)
Extrapolate the pressure–temperature line to zero pressure and estimate the amount of gas at 0°C.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The slope is 10/27.3 kPa °C⁻¹. Extrapolating 100 kPa down to zero requires a decrease of 273°C.
- 2
Use absolute temperature T = 273 K and SI pressure 100000 Pa in pV = nRT.
- 3
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.
(b)
Predict pressure at 136.5°C and explain the microscopic mechanism. What if the container is allowed to expand at constant pressure instead?
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
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.
- 2
Higher absolute temperature means greater mean molecular kinetic energy; more vigorous momentum transfer to the walls raises pressure at fixed volume.
- 3
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.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
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