Pressure decay leak rate
A pressure change test measures the whole system at once: seal it at a known pressure, hold, and see how much pressure is lost. For a fixed volume the leak rate is the pV throughput Q = ΔP·V/Δt, so a 40 mbar loss from a 50 litre system over 10 minutes is 40 × 50 / 600 = 3.3 mbar·L/s. The method is cheap and needs no tracer gas, but it is blind to leak location and it is entirely at the mercy of temperature.
Temperature is the reason most pressure decay tests are wrong. Gas in a rigid volume obeys P/T = constant, so a fall of just 1 K from 293 K drops the pressure by about 0.34% — on a 6 bar absolute system that is 20 mbar of apparent loss with no leak at all. The correction is to refer the final pressure back to the starting temperature, P₂' = P₂ · T₁/T₂, using absolute pressures and absolute temperatures, and to take the leak-induced drop as P₁ − P₂'. Gauge pressures cannot be used in this correction. Where the correction is a large fraction of the raw drop, the test is not really measuring a leak; extend the hold, insulate the item, or move to a tracer-gas method.
The floor on sensitivity is set by the instrument, not the physics: the smallest leak the test can see is roughly the pressure resolution times the volume divided by the hold time. Buying sensitivity therefore means holding longer or measuring pressure better — increasing the volume makes it worse. Allow a thermal stabilisation period before starting the clock, since compression during pressurisation warms the gas and the subsequent cool-down looks exactly like a leak.
Worked example
| Test volume | 50 L |
| Start pressure (absolute) | 6000 mbar |
| End pressure (absolute) | 5940 mbar |
| Hold time | 600 s |
| Gas temperature at start | 20 C |
| Gas temperature at end | 19 C |
| Pressure gauge resolution | 1 mbar |
| Allowable leak rate | 1 mbar.L/s |
| End pressure corrected to start temperature | 5960.33 mbar |
| Leak-induced pressure drop | 39.668 mbar |
| Drop explained by temperature | 20.33 mbar |
| Leak rate | 3.306 mbar.L/s |
| Uncorrected leak rate | 5 mbar.L/s |
| Smallest detectable leak rate | 0.0833 mbar.L/s |
T1 = 293.15 K, T2 = 292.15 K. P2' = 5940 x 293.15/292.15 = 5960.332 mbar, so 20.33 mbar of the 60 mbar drop was only the gas cooling. The leak-induced drop is 6000 − 5960.332 = 39.668 mbar, giving Q = 39.668 x 50 / 600 = 3.306 mbar·L/s against 5.000 mbar·L/s if temperature were ignored. Detection floor = 1 x 50/600 = 0.0833 mbar·L/s.