The Cooling Curve Reading
A sealed 500 mL container holds a gas at 300 K and 1.0 atm. A student records the pressure every minute while the container is cooled in a water bath. The readings are: 0 min → 1.00 atm; 2 min → 0.90 atm; 4 min → 0.80 atm; 6 min → 0.70 atm; 8 min → 0.70 atm; 10 min → 0.70 atm.
The student concludes: "Pressure is directly proportional to temperature, so the gas must have stopped cooling after 6 minutes." The water bath was measured at a constant 210 K throughout, and the container is rigid with fixed volume. The gas behaves ideally.
1. Evaluate the student's conclusion. Is the pressure–temperature relationship being tested correctly, and is 'the gas stopped cooling' a valid inference?
Read the worked answer
The student conflates two ideas. Boyle's/Gay-Lussac's law states P/T = constant at fixed volume for a fixed amount of gas - so P is proportional to T in kelvin, not to time. The data show P falling 1.00 → 0.90 → 0.80 → 0.70 atm, a steady drop of 0.10 atm per 2 min, then flat.
Using P/T = constant, T = P × (300 K / 1.00 atm). At P = 0.70 atm, T = 0.70 × 300 = 210 K. That equals the bath temperature, so the gas has simply reached thermal equilibrium with the 210 K bath. It has not 'stopped cooling' in a physical sense - it has finished cooling because it can go no lower than its surroundings.
The inference is invalid as stated: the plateau reflects equilibrium with the bath, not a failure of the gas law.
2. Using the same data, predict the pressure at 3 minutes (interpolating between readings) and explain why the proportionality must use kelvin, not Celsius.
Read the worked answer
Between 2 min (0.90 atm) and 4 min (0.80 atm), the drop is linear at 0.05 atm/min, so at 3 min P ≈ 0.85 atm. Check with P/T: T at 0.85 atm = 0.85 × 300 = 255 K, which sits between the 2-min (270 K) and 4-min (240 K) values - consistent.
Kelvin is required because proportionality means P/T = constant only when T is measured from absolute zero. In Celsius, 0.90 atm corresponds to −3 °C and 0.80 atm to −33 °C; the ratio 0.90/(−3) is not equal to 0.80/(−33), and negative temperatures make ratios meaningless. Using kelvin preserves the constant ratio (0.90/270 = 0.80/240 = 0.00333 atm/K).