Gas-thermometer extrapolation: a straight line to the wrong absolute zero
A student traps dry air in a sealed 250 mL glass flask joined to a pressure sensor by 30 cm of rubber tubing, and records the pressure with the flask in a stirred water bath: 0.0 °C gives 100.0 kPa, 20.0 °C gives 107.0 kPa, 40.0 °C gives 113.9 kPa, 60.0 °C gives 120.8 kPa, 80.0 °C gives 127.7 kPa. The sensor is accurate to plus or minus 0.3 kPa and the bath to plus or minus 0.2 °C. A spreadsheet fit gives a gradient of 0.346 kPa K⁻¹ and crosses the temperature axis at −289 °C, and the student writes that this is absolute zero, in good agreement with the accepted −273 °C. The tubing and the sensor head, which together hold about 12 mL of gas, stayed at the room temperature of 22 °C throughout; only the flask was in the bath. The flask is rated 250 mL plus or minus 5 mL. The teacher asks the group to decide whether the experiment measures absolute zero at all, and what the 16 °C discrepancy is.
1. From the stated gradient and intercept, state the zero-pressure temperature with an uncertainty justified by the instrument accuracies, and comment on the student's claim of good agreement.
Read the worked answer
The intercept follows from p = p₀ + mθ set to zero: θ₀ = −100.0/0.3463 = −288.8 °C.
Uncertainty: δp = 0.3 kPa (plus 0.2 °C × 0.346 = 0.07 kPa from the bath, negligible), so δp/p₀ = 0.3 %. The gradient is fixed by the endpoints over 80 K, so δm ≈ √2 × 0.3/80 = 0.0053 kPa K⁻¹, i.e. 1.5 %. Adding in quadrature, δθ₀/θ₀ ≈ √(0.003² + 0.015²) = 1.6 %, so δθ₀ ≈ 4.6 K. The honest result is −289 ± 5 °C.
The gap from −273 °C is 16 °C, roughly three standard uncertainties, so it cannot be dismissed as scatter. It is a systematic error, and the student's 'good agreement' claim is not supported by the numbers the apparatus actually delivers.
2. Explain the origin of the 16 °C discrepancy and justify its sign - why the apparent zero is too low rather than too high.
Read the worked answer
About 12 mL of the gas sits in the tubing and sensor head, permanently at 22 °C, while the flask is heated. Counting 'volume per kelvin' units, n_total ∝ p(V_flask/T_flask + V_dead/T_room). With V_flask = 250 mL, V_dead = 12 mL and T_room = 295 K, the dead volume holds 0.0407 units against 0.9158 for the flask at 0 °C - 4.3 % of the gas is not at the bath temperature. As the flask cools, that warm fraction becomes relatively more important and keeps the pressure higher than a pure ideal-gas thermometer would give, so the p(θ) curve is concave: its gradient decreases as θ rises.
A straight line fitted to a concave curve therefore has too small a gradient relative to p(0 °C), and p₀/m is too large: θ₀ = −p₀/m is pushed to a more negative value. Substituting the numbers above reproduces the observed data (ratios 1.000 : 1.070 : 1.139 : 1.208 : 1.277) and a chord through 0 °C and 80 °C extrapolates to ≈ −289 °C, matching the student's spreadsheet exactly. The dead volume, not the gas, is the culprit.
3. A classmate proposes two explanations: the air is not a perfectly ideal gas, and the rubber tubing expands as the pressure rises. Evaluate both, and state one change that would remove the main error.
Read the worked answer
Both are real effects, but neither is large enough. At about 1 atm the compressibility factor of air departs from 1 by roughly 0.2 %, which shifts the extrapolated intercept by only a few kelvin at most, and it is a smooth correction that does not scale with the 4 % dead-volume fraction. Rubber tubing does stretch, but the pressure changes by only 28 % across the run, and to mimic a 16 °C shift the dead volume would have to change by something like half its own size during the experiment - far more than an elastic tube does, and in the wrong pattern to be mistaken for the observed curvature.
The decisive test is that the discrepancy scales with the ratio V_dead/V_flask and with the temperature offset of the dead volume, and the model reproduces the actual measured pressures to better than 0.1 kPa. The fix is straightforward: mount the sensor directly on the flask with a short rigid connection, or immerse the whole assembly (sensor included) in the bath, or correct each pressure for the moles in the dead volume before fitting. Simply waiting longer at each temperature will not help, because this is an equilibrium systematic error, not thermal lag.