Sealed Can with an Unexpected Pressure Reading
A technician heats a sealed rigid steel can of volume 0.500 L. At 22.0 °C the gauge reads 1.20 bar. The can is heated to 85.0 °C and the same gauge reads 1.75 bar. Atmospheric pressure is 1.013 bar. The can was rinsed with water and not dried before sealing, although no free water is visible after heating. The gauge has not been calibrated recently. A colleague claims the can must have leaked because the pressure rise is not as large as expected; another says a leak would lower the reading, not raise it. The technician needs to decide whether the 1.75 bar reading can be explained by residual water vapour, by gauge error, or by a leak, and what test to run next. Assume the can volume is constant and water vapour behaves as an ideal gas.
1. Calculate the gauge pressure expected at 85.0 °C if the can contains dry air only. Compare with the 1.75 bar reading. Does a leak explain the difference?
Read the worked answer
Use absolute pressures. Initially P1abs = 1.20 + 1.013 = 2.213 bar, and T1 = 295.15 K. At T2 = 358.15 K, a rigid can gives P2abs = P1abs(T2/T1) = 2.213 × (358.15/295.15) = 2.685 bar. The predicted gauge pressure is therefore 2.685 − 1.013 = 1.67 bar.
The measured 1.75 bar is about 0.08 bar higher than the dry-air prediction, not lower. A leak would reduce the amount of gas and therefore reduce the pressure, so it cannot explain a reading that is already above the dry-air value. Steel expansion over this temperature range is negligible compared with an 0.08 bar discrepancy.
The colleague who says a leak would lower the reading is correct. The remaining candidates are an extra gas or vapour contribution, a gauge calibration error, or an error in the assumed temperature or volume.
2. Can residual water explain the excess? Estimate the amount of water vapour needed, and explain why the result would be much larger if free liquid water remained at 85 °C.
Read the worked answer
The excess pressure is about 0.078 bar = 7.8 × 10³ Pa. For water vapour occupying V = 5.00 × 10⁻⁴ m³ at 358 K, n = PV/(RT) = (7.8 × 10³)(5.00 × 10⁻⁴)/(8.314 × 358) = 1.3 × 10⁻³ mol. That is about 0.024 g of water, a very small trace film.
Such a trace amount is plausible if it all evaporates by 85 °C. However, if any free liquid water remains at 85 °C, the vapour partial pressure must be the saturation value, about 0.58 bar at that temperature. The dry air would contribute about 2.65 bar absolute, so the total would be roughly 3.23 bar absolute, or about 2.22 bar gauge. That is far above the observed 1.75 bar. A visible droplet is therefore too much water for this explanation: it would leave liquid water and push the gauge much higher.
So residual water can explain the excess only if the can held a trace film of roughly 0.02 g that fully evaporated. A rinsed but visibly wet can is not a comfortable explanation. Gauge calibration becomes a serious alternative.
3. Design a test to distinguish gauge error from residual water, and state the expected result for each.
Read the worked answer
First check the gauge against a calibrated reference or a known pressure source at both 22 °C and 85 °C. If the gauge reads about 0.08 bar high at both temperatures, or has a zero offset, the anomaly is instrumental rather than chemical.
Second, repeat the experiment with a thoroughly dried can and a new seal. A dry can should follow the absolute-pressure ratio and read about 1.67 bar gauge at 85 °C. If it still reads 1.75 bar, the gauge or temperature measurement is suspect. If it now reads 1.67 bar, the original can did contain a volatile component.
Third, cool the original can back to 22 °C and re-read the gauge. A dry-air-plus-trace-vapour system should return close to the starting pressure; a gauge offset will remain. Weighing the can before and after drying, or adding a desiccant capsule, gives a direct check on water mass. A cooling curve can also reveal a saturation plateau if enough water is present.