Pumping the gas in the volume has a time constant of T = V/S where V is the volume and S the pumping speed. This means P = P0*exp(-t/T). For a typical 100-liter chamber and a 1 l/s speed from a mechanical pump constrained by the conductance of piping, T = 100". Pumping from 760 to 0.1 Torr will then take 15 minutes.
On the other hand, high vacuum pumps have much larger pumping speeds, say 100 l/s and T = 1". For the same system, it will take 11.5" to go from 1E-4 to 1E-9 Torr. But this does not happen so fast. The reason is that the limiting step in the removal of molecules is adsorption/desorption in the walls, not the pump itself. For Ar, N2, the residence time on metal surfaces at room temperature is ~0.1 ns, so the estimate above is right. For water, the residence time is much larger, in the millisecond range, so pump time is much larger.
When the pressure falls very slowly and there are no apparent leaks (that would be indicated by the pressence of Ar and O2 in the residual gas), one suspects a virtual leak. This is a pocket of gas in the system that can only be pumped through a small constriction. Examples are the bottom of a threaded hole that is not vented, or a cavity that is connected to the chamber through the space between two flat plates in contact. The solution is to avoid those constructions and, when this is not possible, to provide slots to assist in the pumping of these volumes. Another type of virtual leak is the presence of a significant amount of material that has a relatively high vapor pressure (dirty spots). They can be identified by localized mild heating, that produces a very strong change in pressure.
A fingerprint may be 1 cm2 in area, 1 micron thick. It contains roughly 1E18 molecules. How long does it take to remove them? R, the number of molecules removed per second is given by the product of the pressure P and the pumping speed R = P*S. One Torr-l = 3E19 molecules. At 1E-9 Torr, R = 1E-9*100*3E19 = 3E12 molec/s so it will take 3E5" or about 100 hours to remove the dirty spot. This can be speeded up by baking
The problem with pumps is that the molecule removal rate R decreases with pressure. The pumping speed S is constant but it is a volumetric value; the number of molecules in the volume decreases with pressure. To increase R one needs to increase the pressure. This is achieved when baking, but only if one can get the pressure to be high enough. For a UHV system, one typically needs the pressure to increase to about 1E-5 Torr. If the pressure is one order of magnitude lower, a tenfold increase of time will be needed. This may be prohibitive.
Copyright: Raśl Baragiola, University of Virginia, 1999
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