Chapter 5 Kinetic Theory And Vacuum

Vern Lindberg

April 26, 2012

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Contents

5.1 A Brief Introduction to Kinetic Theory . . . . . . . . . . . . . . . . . . . . 45.2 Units of Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65.3 Gauge and Absolute Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . 75.4 More Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

5.4.1 Number Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85.4.2 Sizes and spacing of atoms . . . . . . . . . . . . . . . . . . . . . . . 85.4.3 Mean free path, mean time between collisions . . . . . . . . . . . . . 95.4.4 Collision Flux, Monolayer formation time . . . . . . . . . . . . . . . 10

5.5 Vapor Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115.6 Viscosity, Turbulence, Molecular Flow . . . . . . . . . . . . . . . . . . . . . 135.7 Thermal Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.8 Typical Vacuum System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.9 Pumping Speed, Throughput . . . . . . . . . . . . . . . . . . . . . . . . . . 165.10 Ideal Pump Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.11 Conductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.12 Effect of Conductance on Pumping Speed . . . . . . . . . . . . . . . . . . . 225.13 Conductances in Parallel and Series . . . . . . . . . . . . . . . . . . . . . . 235.14 A Practical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

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4 5.1 A Brief Introduction to Kinetic Theory

5.1 A Brief Introduction toKinetic Theory

A vacuum is a region where the pressure isless than atmospheric—i.e. where the num-ber of molecules per unit volume is less thanwhat is normally expected. The purposeof vacuum technology is to create regionsof vacuum, maintain them, and measurethem. To fully appreciate the difficulties,we need to know the connection betweenthe microscopic description of a gas—massof molecules, velocities of molecules, etc.—and the macroscopic description—pressure,volume, temperature, number of moles, etc.This is the subject of The Kinetic Theoryof Gases, which will be described fully inthermodynamics texts. A simplified versionof Kinetic Theory will allow us to determinethe functional dependence of various quan-tities without evaluating integrals.

Consider a gas consisting of a large number,N , of hard molecules. These molecules willeach have mass m and speed v. They willalso be allowed to have only kinetic energy.The molecules will be constrained to movealong one of the 3 axes - i.e. i, j, or k, andinitially will collide only with the walls of acontainer. The walls will be rigid so that inany collision the molecules will collide elas-tically: they simply reverse their direction,but continue at the same speed.

Consider the box shown in Figure 5.1. Whatis the average force exerted by the moleculeson the wall? The number of moleculesmoving toward the wall will be 1

6N . Themolecules will be spread out uniformly inthe box and will strike the walls at dif-ferent times. We can say that all of the

Figure 5.1: Box containing an ideal gas. In asimplified model, all molecules move at samespeed and along one of the axes.

16N molecules will hit the wall in a time∆t = d/v . As each molecule bounces off,its momentum (a vector) will change.

∆~p = mv(−i)−mv i = −2mv i (5.1)

This is equal to the impulse of the wall onthe molecule. In a time ∆t the total impulseon all 1

6N molecules will be

~F ∆t = ∆~p =1

6N(−2mv i

)(5.2)

where ~F is the force of the wall on themolecules. Thus

~F =1

3

N mv2

di (5.3)

is the average force of the molecules on thewall, and the average pressure, P , on thewall of area A is

P =

∣∣∣~F ∣∣∣A

=2

3N

12 mv2

Ad(5.4)

But Ad=V , the volume, and(

12 mv2

)= KE

is the kinetic energy of a molecule (for real

5.1 A Brief Introduction to Kinetic Theory 5

gas this will be the average kinetic energy.)So

P V =2

3N KE (5.5)

The model chosen is oversimplified. A realkinetic model would allow molecules to havea range of speeds, and allow motion in anydirection. Then integrals must be evaluatedto obtain the final result for the average pres-sure. Surprisingly, the detailed result is ex-actly the same as the above, with KE be-ing the average kinetic energy of a molecule.The equation above is known as the IdealGas Law, with the average kinetic energymeasured by absolute temperature, T .

P V = N k T (5.6)

with

k T =2

3KE (5.7)

Here k, sometimes written kB, is Boltz-mann’s constant, k = 1.38 × 10−23

J/K.

In chemistry we usually write

P V = nmolRT (5.8)

with R being the gas constant and nmol be-ing the number of moles of gas. You caneasily show that R = NAv k, where NAv isAvogadro’s number. We will not use molesin this text. Instead we will define n = N/Vas the number density of particles in suchunits as molecules per cm3. Do not confusethe two meanings of n! The Ideal Gas Lawcan thus be written as

P = nk T (5.9)

A real kinetic model uses Maxwell-Boltzmann distribution functions to

allow us to determine various averagespeeds.

There are different ways to compute the av-erage of the speed. The rms (root meansquare) speed is defined by

vrms =

√∑v2

N=

√3k T

m(5.10)

The average speed is:

vave = v =

∑v

N=

√8k T

πm=

√8RT

πM(5.11)

where m is the mass of one molecule in kgand M is the molecular mass (g/mol). Forair (average M = 29 g/mol) at room temper-ature, T = 298 K, the molecules have aver-age speed of 466 m/s and an rms speed of 506m/s. At the same temperature moleculeswith a smaller mass will have larger aver-age speeds. This was one of the methodsused to separate isotopes of uranium in theManhatten project.

What do we need to know about the KineticTheory that will apply to our study of vac-uums?

First: Pressure is an average quantity —close measurements would reveal fluc-tuations around an average.

Second: Pressure is dependent on temper-ature, but not on the type of molecule.Two containers filled with differentgases, but at the same temperature, canhave the same pressure. However, someof the pressure gauges will respond dif-ferently to different gases even if theyare at the same pressure.

6 5.2 Units of Pressure

5.2 Units of Pressure

The most direct units of pressure are basedon its definition, Table 5.1:

Table 5.1: Pressure Units Based on Defini-tion

SI 1 N/m2 = 1 Pascal, 1 Pacgs 1 dyne/cm2 = 0.1 Pa

British 1 psi (pound/in2) = 6895.3 Pa

Other units are based on the atmosphericpressure that is always around us on earth.In Europe the bar is the most widespreadunit, although it is being replaced with thePascal. See Table 5.2

Table 5.2: Pressure Units Based on Atmo-sphere

1 atmosphere (atm, A.P.) = 1.013 × 105 Pa1 bar (just smaller than 1 atm)= 1 × 106 dynes/cm2 = 105 Pa = 100 kPa

The third set of units, Table 5.3, and themost common in the USA, is based on themanometer, a U-tube that shows a pres-sure difference between the two sides of thetube as a difference in height of a liquid inthe tube—usually mercury1. An open-tubemanometer is shown in Figure 5.2 From thelocal gravitational field strength, g, and thedensity of the liquid, ρ, we calculate a pres-sure difference, ∆P .

∆P = ρ g h (5.12)

Since ρ and g are constant at the earth’s

1But not always: in hospitals, water is sometimesused as the manometer fluid.

Figure 5.2: Open-tube manometer. If theliquid is mercury, the pressure difference ish Torr.

surface, we can specify the pressure by spec-ifying the height of the column of liquid,assumed to be mercury unless otherwisestated. The units are:

Table 5.3: Pressure Units Based onManometer

1 mm-of-Hg = 1 Torr1 in-of-Hg = 25.4 Torr1 µm-of-Hg = 1 µ-of-Hg = 1 µ = 1 mTorr

CAUTION: The mm-of-Hg, etc. are not tobe canceled with the units of length such asmm. Thus 1 mm-of-Hg/ 5 mm 6= 0.20

This should be clear if the “-of-Hg” is keptwith the unit. Unfortunately, people in ahurry tend to say “x mm of pressure” or“the pressure was y microns”. If you con-fuse the pressure-mm with the length-mm,you will be flogged with a piece of vacuumhose! It is best to avoid confusion by usingthe Torr.

Lest you find these units comprehensible, bewarned that people often omit the unit alto-gether and talk of “pressure of 10−6.” Usu-

5.3 Gauge and Absolute Pressure 7

ally, if a physicist from the USA or Canada istalking this means the units are Torr.

There is one convenient approximate conver-sion: 1 Torr ≈ 1 mbar. If a journal reportsa pressure of 2×10−4 mbar we can mentallyconvert to 2×10−4 Torr and know what pres-sure range we are discussing. Some conver-sions are given in Table 5.4

Table 5.4: Pressure Conversions

1 std. atm. = 760 Torr = 760 mm-of-Hg= 1.01323 bar= 1.01323× 105 Pa= 29.92 in-of-Hg= 14.69 lbs/in2

1 Torr = 1000 mTorr = 1000 µ= 0.0013 atm= 133.32 Pa= 1.34 mbar

1 millibar = 0.745 mm-of-Hg= 0.745 Torr= 100 Pa

1 Pa = 7.45× 10−3 Torr= 10 bar.

Different degrees of vacuum may be de-scribed qualitatively by the terms in Table5.5.

Table 5.5: Levels of Vacuum (in Torr)Rough vacuum 760 to 1Medium vacuum 1 to 10−3

High vacuum (HV) 10−3 to 10−8

Ultra-High vac (UHV) 10−8 and lower

In physics applications three pressureregimes are common. Sputtering, plasmaetching and other plasma processes take

place in medium vacuums of 1 mTorr to afew hundred mTorr. Evaporation of mate-rials takes place in high vacuum conditions,typically 1 × 10−6 Torr. This is a pressuretypical of low earth orbit where the spaceshuttles operate. Surface analysis equipmentand particle accelerators require UHV con-ditions of 1× 10−9 Torr or less.

As a final note, remember that 1 atmosphereof pressure is about 1 Ton/ft2. This meansthat vacuum structures are subject to largestresses!

5.3 Gauge and AbsolutePressure

Some vacuum gauges (such as the open tubemanometer) read a pressure difference. Fre-quently these gauges are marked in unitsthat show change from atmospheric pressure— this is called gauge pressure. Most phys-ical phenomena depend on absolute pres-sure. To get the absolute pressure, add at-mospheric pressure to gauge pressure. Thepressure gauges on tanks of pressurized gasare marked in gauge pressure. One abbre-viation is psig, ”pounds per square inch,gauge”

e.g. My tire pressure is 25 lbs/in2. Whatis absolute pressure?P = 25 + 14.7 = 39.7 lbs/in2.

8 5.4 More Kinetic Theory

5.4 More Kinetic The-ory

In addition to the gas law and the expres-sions for averages of speeds, 7.1.3 and 7.1.4,we need several other expressions. We willnot provide rigorous proof for the results,but will give estimates of the results followedby the more accurate equations. The fullequations involving temperature and pres-sure will be given, but frequently the tem-perature is constant at 25◦C and the gaschosen is air. These equations will also begiven.

5.4.1 Number Density

Number density, n, is the number of atomsor molecules per unit volume, usually inunits of molecules/cm3. Clearly n = N/V .We know that 1 mole of an ideal gas occu-pies 22.4 liters at STP (0◦C and 760 Torr).From this it is easy to derive

n atoms/cm3 = 9.66× 1018

(K

Torr cm3

)P

T

at 25◦C. = 3.24× 1016

(1

Torr cm3

)P

. (5.13)

The number density at standard tempera-ture and pressure, 2.5×1019 atoms per cm3,is known as Loschmidt’s Number.

Equation 5.13 was written in clear form withthe units applied to the constant as is usuallydone in physics. It is common in engineeringand other disciplines to present the equationas follows:

n = 9.66× 1018 P

T(5.14)

with P in Torr, T in Kelvin,

n in molecules/cm3.

This engineering convention will be followedin the remainder of the equations.

At room temperature and a pressure typicalof evaporation, 1 × 10−6 Torr we find thenumber density of 3.24×1010 molecules/cm3.At all pressures attainable in the laboratorywe will have huge number densities, and sothe principles arising from the ideal gas lawremain true.

5.4.2 Sizes and spacing ofatoms

Sizes of atoms and distances between them.We will make a crude estimate of the sizeof atoms as follows. In the solid and liquidstates, molecules basically touch each other.If we know the density and molecular weightwe can find the size of the molecule.

Consider water with a density of ρ = 1.0g/cm3, and a molecular mass of M = 18g/mol. Using these numbers and Avogadro’snumber, NAv we find a volume per watermolecule of 3 × 10−23 cm3. If we considerthe molecules as cubes, the side of the cubeis the cube root of the volume, about 0.3 nm= 3 A.

More careful calculations show that the di-ameter, d0, of atoms and small moleculesranges between about 0.2 and 1.0 nm. Forhelium d0 is 0.22 nm and for water vapor d0

is 0.47 nm. For air we use an average valueof 3.7× 10−8 cm = 0.37 nm = 3.7 A.

The volume of space available to eachmolecule in the gas is just the inverse of the

5.4 More Kinetic Theory 9

Table 5.6: Molecular diametersGas diameter (nm)air 0.37He 0.22

H2O 0.47

number density, V = 1/n. From this we candetermine the average distance, D in cen-timeters, between gas molecules to be

D (cm) =1

n1/3= 4.70× 10−7

(T

P

)1/3

T in K, P in Torr

= 3.14× 10−6 P−1/3

at 25◦C. (5.15)

5.4.3 Mean free path, mean timebetween collisions

Mean-free-path, mfp or λ is the average dis-tance traveled between collisions, and meantime between collisions, τ is the average timebetween collisions of molecules in the gas.They are related by the average speed ofmolecules in the gas, 〈v〉.

λ = 〈v〉 τ (5.16)

We can estimate the mean free path as fol-lows. We will use an atom diameter (air) ofd0 = 0.37 nm, and a separation of atoms atSTP of D = 3.4 nm. In order to hit, theatoms must pass within 2d0 = 0.74 nm ofeach other, as is shown in Figure 5.3.

The collision cross section is thus about4 d2

0 = (0.74)2 = 0.55 nm2 compared withthe area occupied by an atom of D2 =

Figure 5.3: Estimating mean free path. As-sume regularly spaced atoms (gray) sepa-rated by D. Each atom has size d0. If an-other atom (black) is to collide, it must bewithin the square area shown by solid line,of size 2d0. The chance of collision in oneatomic layer is the ratio of areas of the solidsquare to the dotted square.

(3.4)2 = 11.6 nm2. We estimate the prob-ability of collision as the ratio of the two ar-eas, 4d2

0/D2, and expect a 0.55/11.6 = 0.087

chance of a collision when an atom travels3.4 nm. By dividing the separation of layersby the probability of collision in one layerwe estimate the mean free path at STP tobe 3.4 nm/0.087 = 39 nm. If you put in thesymbols this gives λ = 0.25/

(nd2

0

). A more

realistic derivation results in

λ (cm) =0.225

nd20

n in cm−3, d0 in cm

=5× 10−3

Pat 25◦C. (5.17)

For air at STP the mean free path becomes66 nm, which compares well with our crudeestimate.

At a typical evaporation pressure of 1×10−6

10 5.4 More Kinetic Theory

Torr the mean free path is 5000 cm = 50 m!Recall that the number density of particlesat this pressure is huge, n = 3.24 × 1010

molecules/cm3, but even so the moleculesrarely meet each other. For air at roomtemperature the average speed is about 480m/s, so the mean time between collisionsat 1 × 10−6 Torr is τ = λ/〈v〉 = 0.1 sec-onds.

Exercise: Suppose a chamber is a cube ofside 1 m at a pressure of 1× 10−6 Torr.Estimate (order of magnitude) the num-ber of collisions that a molecule makeswith the walls in the mean time betweencollisions.

5.4.4 Collision Flux, Monolayerformation time

We may also compute the arrival rate ofmolecules, Z, also known as collision flux.This is the number of molecules per unit areaper unit time that arrive at a surface. Wecan get an estimate for this quantity as fol-lows. Consider a cube of length a. The totalarea of the cube faces is 6a2 and the volumeis a3. With a number density n moleculesmoving at average speed 〈v〉, the collisionsper second with the walls is[

na3] [〈v〉

a

]= na2 〈v〉 (5.18)

and the number of collisions per unit areaper second is this quantity divided by thearea of the cube faces,

Z =1

6n〈v〉 (5.19)

A correct derivation results in a similar ex-pression, expressed in terms of P,M , and

T ..

Z(molecules/cm2/s

)= 3.51× 1022 P√

M TP (Torr) T (K), and M (g/mol)

= 3.77× 1020P

for air at 25◦C

.(5.20)

For air we use a weighted average M =29.

At an evaporation pressure 1×10−6 Torr androom temperature, Z is 3.8× 1014 moleculesper cm2 per second. While this is a largenumber, we must also remember that thereare a large number of atoms per square cen-timeter of a solid surface.

In coating a piece of glass we want to en-sure that every atom on the piece of glass iscoated by a deposited atom. We thus willcompute the atomic collision flux, the num-ber of gas atoms per second that hit one sur-face atom. Since each atom in a solid occu-pies about d2

0 of area, the atomic collisionflux in molecules per atom per sec is justZ ·d2

0. For air at room temperature hitting asurface where the atomic spacing is the sameas solid air,

Zatoms/atom/s) = 5.16× 105P (5.21)

A related quantity is the monolayer forma-tion time. This is the time needed to forma single layer of molecules on a surface as-suming that every molecule that strikes thesurface sticks to it, and that the moleculesform a layer one atom high. While these as-sumptions are not precisely true, we do getan estimate of the actual monolayer forma-tion time. This is just the inverse of the

5.5 Vapor Pressure 11

Table 5.7: (a) Kinetic Theory Quantities. Air at 25◦CPressure n λ τ Monolayer time(Torr) (atoms/cm3) cm (sec) (sec)

760 2.5× 1019 6.7× 10−6 1.4× 10−10 2.6× 10−9

1 3.2× 1016 5.1× 10−3 1.1× 10−7 1.9× 10−6

10−3 3.2× 1013 5.1 1.1× 10−4 1.9× 10−3

10−6 3.2× 1010 5.1× 103 1.1× 10−1 0.1910−9 3.2× 107 5.1× 106 110 1.9× 103

Pressure Arrival Rate Atomic Collision Flux(Torr) (Atoms/cm2/sec) (Atoms/atom/sec)

760 2.9× 1023 3.9× 108

1 3.8× 1020 5.2× 105

10−3 3.8× 1017 5.2× 102

10−6 3.8× 1014 0.5210−9 3.8× 1011 5.2× 10−4

atomic collision flux.

τ =1.94× 10−6

P(5.22)

Values for the various quantities are shownin Table 5.7 for air at 25◦C. A summary offormulas is in Table 5.8.

5.5 Vapor Pressure

Even if we produce a perfect vacuum, we willbe hard put to contain it. Any walls on thevacuum system—solid or liquid—will vapor-ize to some extent. This occurs even thoughthe walls are not at their boiling point. TheKinetic Theory explanation of vaporizationis that some molecules in the solid or liquidgain energy (randomly) and this will be suf-ficient to let some molecules vaporize. Thusa sealed container will fill with vapor. Themolecules in the vapor randomly collide, andoccasionally some molecules will lose energy

and condense into the liquid or solid. Even-tually an dynamic equilibrium will occur,when equal fluxes of molecules vaporize andcondense. Thus an equilibrium pressure, avapor pressure, occurs.

Figure 5.4: Establishing an equilibriumvapor pressure. On the left is a non-equilibrium state. At equilibrium the flux ofparticles evaporating equals the flux of par-ticles condensing.

The equilibrium vapor pressure is verytemperature dependent. The Clausius-Clapeyron equation from thermodynamicssays that the vapor pressure should fol-

12 5.5 Vapor Pressure

Table 5.8: Important formulae from kinetic theory. P in Torr, T in K, m in kg, M in amuQuantity General For Air at 25◦C

〈v〉 cm/s

√8k T

πm=

√8RT

πM46600

vrms cm/s

√3k T

m50600

n atoms/cm3 9.66× 1018P

T3.24× 1016P

d0 nm 0.37

D cm 4.70× 10−7

(T

P

)1/3

3.14× 10−6 P−1/3

λ cm (d0 in cm)0.225

nd20

5× 10−3

P

Z atom/cm2/s 14n 〈v〉 = 3.51× 1022 P√

M T3.77× 1020P

Zd20 atom/atom/s 5.16× 105P

Monolayer formation time (s)1.94× 10−6

P

low

P = P0 exp

(−∆H

k T

)(5.23)

where ∆H is the latent heat of vaporization(enthalpy change) and is nearly temperatureindependent. Plots of the vapor pressureversus temperature will be given later.

Some typical vapor pressures are in Table5.9.

From even these few data several thingsshould become clear. Water vapor, oils(such as fingerprints), and greases have rel-atively high vapor pressures at room tem-perature. If we hope to have a UHV systemwe must avoid these items. Even at mod-

erate high vacuum this outgassing will be aproblem.

Solids have very low vapor pressures at roomtemperature, or even elevated temperatures.This is true with a few exceptions such aszinc.

In a vacuum system we have two sources ofgas: the original gas in the chamber andgas that has stuck to the surfaces of thechamber and which will leave that surfaceby a process of vaporization generally calledoutgassing. Frequently the gas load on ourpumping system will be dominated by thewater vapor in the chamber and by the out-gassing from the surface.

5.6 Viscosity, Turbulence, Molecular Flow 13

Table 5.9: Vapor Pressures of Common MaterialsMaterial T (C) P (Torr) Material T (C) P (Torr)

Water 100 760 Ice -10 1.9530 32 -50 0.0300 4.6 -78 0.56× 10−3

Mercury 100 0.27 Vacuum Grease20 1.2× 10−3 20 10−6 to 10−8

-78 3× 10−9

Copper 1000 10−9

500 10−11

20 10−35

Pump OilsRoughing 40 10−4 Diffusion 100 10−4

100 10−2 200 1

5.6 Viscosity, Turbulence,Molecular Flow

Pumping on a vacuum system means thatthe gases will be in motion. If the gas flowis relatively rapid, the gas will be in turbu-lent flow, similar to the flow of water in awhite-water rapids. The criterion for turbu-lent as opposed to laminar (non-turbulent,viscous) flow depends on the dimensions ofthe pipes containing flow, the average speedof the molecules, the density of the gas, andthe viscosity of the gas. At normal pressureswe have viscous flow of liquids (molecules ex-ert forces on each other), however at very lowpressures the flow changes to molecular flow(molecules interact only with walls.)

To understand the flow of molecules we mustexamine the interaction of a molecule witha surface. When a molecule hits a surfaceit generally sticks to the surface for a shorttime. It then can be re-emitted in any direc-tion. The distribution of directions is usually

assumed to be a “cosine distribution”: thechance of emission in a direction θ from thenormal is proportional to cos θ.

This means that the molecule is most likelyto be emitted normal to the wall, and has anequal chance of going left or right at equalangles. The direction of emission has noth-ing to do with the pressure gradient. Thisis illustrated in Figure 5.5.

Figure 5.5: Cosine Emission Law for Molec-ular Flow. The probability of emission in arange of angles θ to θ+dθ is 1

2 cos θdθ, whereθ is measured from the normal.

A viscous force is a force between molecules.In the Kinetic Model mentioned previously,the molecules had only kinetic energy; since

14 5.6 Viscosity, Turbulence, Molecular Flow

there was no potential energy, there was noforce between the molecules except duringcollisions. Real molecules will have someforce and therefore some viscosity.

Figure 5.6: Definition of Viscosity. Thebottom plate is at rest while the top platemoves. If we divide the space into regions ofsize equal to the mean free path λ, then themolecules next to a plate stick to it, the nextlayer interacts with these molecules etc.

Consider two plates with the top one mov-ing relative to the bottom one. The layerof molecules next to a plate ”sticks” slightlyto the wall and the next layer of moleculessticks to that, etc. Thus the molecules nearthe bottom plate have an average velocity ofzero, while the molecules near the top platehave the same average velocity as that of thetop plate. This is shown in Figure 5.6. Thethickness of the layers is the mean free path.The result is molecule drift speeds rangingfrom zero next to the bottom plate up to alarge value near the top plate. The viscousforce arises from the transfer of momentumbetween layers in the gas. Viscosity limitsthe speed with which gas can be pumpedout of a vacuum system. At relatively highpressures the viscous force on a plate of areaA separated by a stationary plate by a dis-tance d and moving with velocity ~v is

~F = −η ~v0A

d(5.24)

The viscosity of a gas is denoted η, and hasunits

(N · s/m2

)= 10 poise (non-SI unit).

Some values are given in Table 5.10. Athigher pressures the viscosity depends onlyon the temperature and the size and mass ofthe molecules. It is independent of the pres-sure. At lower pressures, where the meanfree path is about the same size as the size ofour vacuum tube, the viscous force is linearlyproportional to pressure. This fact is used inviscometer vacuum gauges that are used tocalibrate other high vacuum gauges.

Molecular Flow refers to low-pressure re-gions. In the molecular flow region themolecules collide most often with the wallsrather than each other. Viscosity becomesunimportant, since the molecules rarely col-lide. At the same time, however, the fluctu-ations in pressure, and density, become moreof a factor. Since the emission of gas from asurface follows the cosine law, we are equallylikely to find a molecule moving to a regionof higher pressure as to one of lower pres-sure.

The behavior of vacuum systems will dependon which of these three types of gas flow oc-cur. Two dimensionless numbers are usedto describe gas flow. At higher pressureswe consider the Reynolds Number, definedas

R =U ρd

η(5.25)

where U is the speed of gas flow, ρ is thegas density, η is the gas viscosity, and d isthe pipe diameter. Laminar flow exists forR < 1200, turbulent flow for R > 2200, andbetween these values either type of flow mayexist.

At lower pressures the Knudsen Number is

5.7 Thermal Conductivity 15

Table 5.10: Viscosities of some materialsMaterial η (N/(m s)) η cP (centiPoise)

Chocolate Syrup, 20◦C 15 15000Motor Oil SAE60 1 1000Motor Oil SAE20 0.065 65

Water 20◦C 0.001 1.0Water 99◦C 0.00028 0.28

Air STP 1.8× 10−5 0.018

defined as

Kn =λ

d(5.26)

where λ is the mean-free-path and d is thepipe diameter. For Kn < 0.01 we havelaminar flow, for Kn > 1 we have molec-ular flow, and in between we have transitionflow.

At a pressure of 1×10−6 Torr, the mean freepath for air at room temperature is about50 meters. In a pipe of diameter less than50 meters we have molecular flow. In apipe larger than 5000 meters we have viscouslaminar flow. Thus hi-vac vacuum cham-bers usually operate in the molecular flowregion.

In the roughing lines in a vacuum chamberyou have pressures on the order of 0.05 Torr.At this pressure for air at room temperaturewe have molecular flow for pipes of diametersmaller than 1 mm and viscous laminar flowin pipes larger than 10 cm. In pipes betweenthese sizes we have transition flow, which isa bit of both kinds, and complicated to dealwith.

5.7 Thermal Conductiv-ity

At high vacuum the mean free path becomeslong and we tend to have molecular flow.This means that there are very few colli-sions between molecules in the gas: insteadthe molecules bounce from one surface to an-other in straight-line motion. The moleculescannot exchange energy with one another di-rectly, and thus the thermal conductivity ofthe gas becomes small. The kinetic theorymodel for this is similar to the model for vis-cosity in the above section. Here however themolecules transport energy from one surfaceto another.

At high pressures the thermal conductiv-ity can be shown to become constant fora given temperature and type of gas. Atlower pressures the thermal conductivity de-creases linearly with pressure, and becomesvery small. At low pressures radiation heattransfer dominates. Thermocouple and Pi-rani gauges use the dependence of conduc-tivity on pressure for their operation.

In most practical systems the walls of thechamber are kept near room temperature,while a smaller hot source may be inside thechamber. Since the molecules will have a

16 5.9 Pumping Speed, Throughput

greater chance colliding with the larger areaof the cooler surface, the temperature of thegas in the system will remain near room tem-perature. We typically say that the temper-ature of a gas in a vacuum remains constantat its initial value even when there is a heatsource in the chamber such as an evapora-tion source or substrate heaters.

A second vapor species such as the evapo-rated material may be present in the sys-tem and may have a different temperaturethan that of the background gas. The av-erage kinetic energy, and thus the temper-ature of the evaporated material will bemuch higher than the average kinetic en-ergy or temperature of the background gas.This non-equilibrium state of two tempera-tures will persist for quite a long time, muchlonger than the processing time of the depo-sition.

5.8 Typical Vacuum Sys-tem

In order to get a high vacuum, we usuallyneed two types of vacuum pumps. The me-chanical roughing pump evacuates the vesselvia the roughing line down to a pressure ofabout 70 mTorr. The diffusion pump thenpumps the vessel via the hi-vac valve downto a very high vacuum.

In Figure 5.7, V are valves, with V5 beingthe hi-vac valve, TC are thermocouple pres-sure gauges, and IG is an ionization pressuregauge. The diagram on the right uses stan-dard US symbols. European symbols are dif-ferent.

During the roughing operation valves V1,

V3, V4, and V5 are closed. Valve V2 isopened. When the vacuum is about 70mTorr, valve V2 is closed, V1 is opened andthen V5 is opened. The diffusion pump thenwill reduce the pressure to low values. ValvesV3 and V4 are needed to vent the system toatmospheric pressure.

We will need to discuss the various elementsof this typical system, including:

Piping its effect on pumping speeds

Vacuum Pumps pumping speed, ultimatepressure, etc.

Gauges ranges, use, accuracy

Valves types

Leaks finding and fixing

5.9 Pumping Speed,Throughput

We would like to know how rapidly we canevacuate a chamber. The rate of evacua-tion is a complicated function of the pump,the pressure, and the piping in the system.Pump speed, S, is defined as the rate atwhich a volume of a substance is transferredout of a system, and is measured in unitssuch as liter/sec or cubic feet per minute(CFM).

S =dV

dt(5.27)

When applied to vacuums, this poses a prob-lem. As the pressure decreases, the num-ber of molecules in a given volume decreases,and so the rate of pressure change is not con-stant.

5.9 Pumping Speed, Throughput 17

Figure 5.7: A typical vacuum system using diffusion and rotary pumps. Valves are denotedV.

Suppose we initially have 1,000,000molecules in a chamber of volume 4 cubicinches and a pressure of 2.00 Torr.

We imagine a simple pump, Figure 5.8, con-sisting of a metal plate that we insert intothe chamber isolating 1/4 of the volume.This contains 1/4 of the molecules.

Figure 5.8: A simple pump. A plate is in-serted at the 1/4 point, pulled down ejectingall molecules below it. with 3N/4 moleculesthe pressure is 3P0/4

Now I pull the plate downwards and eject

all the molecules in the small portion of thechamber. I repeat this process once per sec-ond. The gas behind the plate expands touniformly fill the volume. Thus after eachcycle I end up with 3/4 of the molecules thatI began the cycle with.

The pumping speed is just 1 cubic inch persecond. The number of molecules remainingvaries as shown in Table 5.11. If the tem-perature is fixed the pressure is just propor-tional to the number of molecules.

Figure 5.9 shows two plots of these data.When plotted on regular graph paper, acurve results. If we use semi-logarithmicgraph paper the data fit a straight line, sug-gesting an exponential decrease. Semi-logpaper is useful in any case where the pres-sure varies over a wide range of powers of10.

We can get the same results analytically.

18 5.9 Pumping Speed, Throughput

Table 5.11: Pumping With a Simple PumpTime Number of molecules Pressure(sec) of Molecules (Torr)

0 1 000 000 2.0001 750 000 1.5002 562 500 1.1253 421 875 0.8444 316 406 0.6335 237 305 0.4756 177 979 0.3567 133 484 0.2678 100 113 0.2009 75 085 0.15010 56 314 0.112

We would like to know how many moleculesper second are entering a chamber—call thisdN/dt. This is positive for molecules enter-ing the system and negative for moleculesleaving the system. By the ideal gas law,PV = NkT . So

dN

dt=d(PV/kT )

dt(5.28)

Usually, we keep T = constant. Thus

dN

dt∝ d (PV)

dt(5.29)

A measurement of PV at a fixed T is thusproportional to a measure of N . The valued(PV)/dt is defined as the throughput, Q,of the system. The temperature to be usedis room temperature, for as we have al-ready discussed this temperature remainsapproximately fixed throughout the opera-tion.

Q =d (PV)

dt(5.30)

and is measured in such units as Torr-liters/sec. A pump of speed S and inlet

Figure 5.9: Linear and Semi-logarithmicplots of the data in Table 5.11

pressure P connected to the system has athroughput of

Q = −PS (5.31)

with the negative sign indicating the removalof molecules from the system. Equation 5.30can be expanded

Q = PdV

dt+ V

dP

dt(5.32)

You may be tempted to make the first termzero, since the volume of the chamber is con-stant. However the volume is actually thevolume of gas in the system, and this can

5.9 Pumping Speed, Throughput 19

change even when the chamber volume andpressure are constant. Even while a pumpremoves molecules from the chamber, severalmechanisms allow molecules to enter the sys-tem. Some of these mechanisms are:

Outgassing of molecules stuck to the wallswhile the chamber was vented.

Real leaks connecting the chamber to theoutside atmosphere.

Virtual leaks caused by small volumes oftrapped air, in screw threads for exam-ple.

Diffusion of gases through a spongy sur-face layer: you must clean the sur-face, perhaps by periodically sandblast-ing the chamber.

Backstreaming of molecules from thepump back into the chamber.

We will refer to all of these additions toa chamber by a throughput of moleculesinto the system Qleaks. The second termrepresents the change in the number ofmolecules due to pumping. Equation 5.32becomes

Q = Qleaks + VdP

dt(5.33)

Consider two extremes for a vacuum sys-tem.

1. The pressure has reached an ultimate,steady-state value, Pult (also called basepressure) where every molecule that en-ters from Qleaks is removed by the pumpand P is constant: dP/dt = 0. ThenQ = Qleaks = |−Pult S|.

If we know Qleaks and Pult we can deter-mine the pump speed at this pressure.

2. Qleaks = 0: The pump dominates andQ = V (dP/dt).

In the second extreme, negligible leaks, wecan determine the pressure as a function oftime. Connect a vacuum chamber of volumeV to a pump of constant speed S. As wepump down, the throughput is Q = −PS,the throughput of the pump (negative sincedP/dt is negative.) Putting this into Equa-tion 5.33 we find:

VdP

dt= −PS (5.34)

ordP

P= −S

Vdt (5.35)

If the pump speed is constant and does notdepend on pressure, and if P = P0 at t = 0we integrate to get

P (t) = P0 exp

(−SVt

)(5.36)

In general S is a function of pressure aswill be discussed in the next chapter. Manypumps operate in a pressure range wheretheir speed is approximately constant andEquation 5.36 is approximately true. Non-constant speed will lead to a deviation fromexponential pressure decrease that will usu-ally increase the time needed to pump downto a value longer than that predicted. Also,the final pressure, Pult, will not be zero, butwill be limited by the value of Qleaks.

Knowledge of the leakage rate is importantin system maintenance. If the value of Pult

increases suddenly we know that there isa problem that has occurred—a leak hassprung, the system is too dirty and needscleaning, or the substrates are outgassing

20 5.10 Ideal Pump Speed

too much. Our full expression for the pres-sure as a function of time in the case of con-stant speed is2

P (t) = P0 exp

(−SVt

)+ Pult (5.37)

We can also look at the system when we closethe valve leading to the pump so that thethroughput of the pump is zero. We expectthe pressure to rise as outgassing and leaksoccur. We can equate the two terms of thethroughput to give

VdP

dt= Qleaks (5.38)

that is easily integrated to give

P (t) =Qleaks

Vt+ Pult (5.39)

The rate of rise of pressure in a vacuum sys-tem is linear with time so long as the Qleaks

is constant. Figure 5.10 shows the expectedgraphs of pressure versus time for the casesof pump down and leak up. The pump isassumed to have a constant speed. Whengraphing the pumpdown graph on semi-logpaper, it is convenient to plot (P − Pult) onthe ordinate. This allows use of more of thedata in determining the slope.

Throughput may have various units. Com-mon ones are Torr-l/s and Pa-l/s. Anothercommon unit is the standard cc per minuteor SCCM. Standard refers to standard at-mosphere, and the volume is measured incubic centimeters. Throughput has units of

2To be totally correct, the term multiplying theexponential should be (P0 − Pult)so that at t = 0,P = P0. In practice, Pult � P0 and the equation isessentially correct as presented in the text.

Figure 5.10: Pressure changes for a vacuumsystem (a) Pump down curve for an idealpumping system of constant speed. On asemi-log plot this is linear. (b) Pressure risefor an isolated system. This plot is on regu-lar graph paper. The rate of rise of pressureis constant.

power, and can be expressed in Watts. Con-versions between the units are given in Table5.12.

5.10 Ideal Pump Speed

Many different types of hi-vac pumps exist,as we will discuss later. We can compare allof them to an ideal hi-vac pump, one thatremoves every molecule that enters it. Con-

5.11 Conductance 21

Table 5.12: Throughput Conversions1 sccm = 1.68872 Pa· L/s

= 1.68872 mW= 12.667 mT · L/s

1 W = 1000 Pa· L/s

sider an ideal pump with an inlet of area A.The number of molecules per second enter-ing the pump is given by

dN

dt= Z A =

1

4n 〈v〉A (5.40)

From the ideal gas law we have N =PV/ (k T ) and n = P/kT . Assumingconstant temperature, Equation 5.40 be-comes

1

kT

d (PV )

dt=

P

kT〈v〉 A

4(5.41)

and this allows us to cancel k T and intro-duce the throughput, Q = d (PV ) /dt

Q =1

4P 〈v〉A ≡ P S (5.42)

Hence we have an expression for the maxi-mum speed of a hi-vac pump that dependsonly on the area of the pump inlet and thespeed of the molecues,

Sideal =1

4〈v〉A (5.43)

Diffusion pumps approach this ideal state.For example a Varian VHS-6 DiffusionPump has an inlet diameter of d = 7.88inches = 200.2 mm. The area of the inlet isπd2/4 = 0.03148m2 and for air at room tem-perature the average speed is about 480 m/sso the ideal speed should be 3.780 m3/sec =3780 liters/sec. The measured speed of thispump for air is 2400 liters/sec, which is 63%of ideal.

5.11 Conductance

The pipes, valves, and other plumbing thatconnect the pump to the vacuum chamberalso have their effect on the rate of evacua-tion. In order to evacuate the chamber, thepressure at the pump inlet, Pp, must be lessthan the pressure in the chamber, Pc. Thethroughput of the pipe, Q, will depend onthe size of the pressure difference (Pc − Pp),and on the geometry of the pipe. We writein general

Q = U (Pc − Pp) (5.44)

The quantity U is the conductance ofthe pipe, with units such as (liters/sec).It depends on the geometry of the pipe,and is rather difficult to compute theoreti-cally.

You may well ask why the pipe affects thethroughput. In the viscous flow region, thegas sticks to the side of the container some-what, reducing the rate of flow. In themolecular flow region, the pipe separates thechamber from the pump. In order to be re-moved, a molecule must travel a long way inone direction to get to the pump. In eitherregion, the length of the pipe is a hindranceto the evacuation.

Formulae for conductances have been devel-oped, notably by Dushman3. These are notalways correct, as is discussed by O’Hanlon4,but do give a feeling for the effect of variousparameters on the conductance.

3S Dushman, The Scientific Foundations of Vac-uum Techniques, 2nd edition, John Wiley, 1962

4J.F. O’Hanlon, “A User’s Guide to VacuumTechnology”, (Third edition), John Wiley & Sons,2003.

22 5.12 Effect of Conductance on Pumping Speed

As an example of the formulae consider acylindrical pipe of length ` and diameter D(both in cm), with `� D. For viscous con-ductance of air at 20◦C, where the averagepressure in the pipe is 〈P 〉,

U(L/s) = 182D4

`〈P 〉 (5.45)

P in Torr, D and ` in cm

For molecular conductance of air at20◦C,

U(L/s) = 12.1D3

`(5.46)

D and ` in cm

Viscous conductance will be present athigher pressures, and molecular flow at lowerpressures. The switch occurs when D is ap-proximately the mean free path λ, and is inthe range of a mTorr (alias µ) for typicalpipes of D = 5 cm. (See the earlier discus-sion of the Knudsen number.)

eg. 1 A hose 30 inches long and 1 inch indiameter connects a vacuum pump toa vacuum vessel. What is the conduc-tance in the molecular flow region?`= 30 in = 76.2 cm D = 1 in = 2.54cmU = 12.1D3/` = 2.60L/s

eg. 2 Suppose I choose a hose of the samelength, but 1 1/4 inches in diameter. Bywhat factor will conductance change?U ∝ D3 (for ` = constant)U2 = U1(D2/D1)3 = U1(1.25/1)3 =1.95 U1

Notice that a 25% change in the di-ameter almost doubles the conductance!

In the high vacuum regions of my vac-uum system where the mean-free-pathis large, in order to make the conduc-tance of my pipes large I will want thediameters to be large.

eg. 3 If I reduce the length to 1/3 ofthe original, how does the conductancechange?U ∝ 1/` (for D = constant)U2 = U1(`1/`2) = U1(1/.33) = 3U1The conductance is inversely propor-tional to the length.

For pipes of other shapes, the same qual-itative effects hold true: conductance isstrongly dependent on the cross-sectionalarea ( ∼ A3/2) and inversely proportionalto the length. For a given area, a circularcross-section will be best.

5.12 Effect of Conductanceon Pumping Speed

Given the connection of Figure 5.11, we wantto know what the effective speed of the pumpis at the chamber when the pump of speedSp is connected to the chamber through apipe of conductance U . Call this Seff . Wecan compute the throughput of the pipe ateither end in terms of the speeds and pres-sures.

At the chamber: Q = SeffPc

At the pump: Q = SpPp

These are the same, since the number ofmolecules entering the pump equals thenumber leaving. Also from the definition ofconductance,

5.13 Conductances in Parallel and Series 23

Figure 5.11: Pipe conductance U has an ef-fect on the pressures and effective speeds atvarious places in the vacuum system. Insteady-tate, the throughput is the same ev-erywhere.

Q = U (Pc − Pp) (5.47)

Combining these (eliminating the pres-sures)

Q = U

(Q

Seff− Q

Sp

)(5.48)

1

Seff=

1

Sp+

1

U(5.49)

We can also write this as

Seff =Sp

1 + Sp/U(5.50)

This equation is one of the most importantequations used in designing a vacuum sys-tem. Consider the following examples.

eg. 4 A pump of speed 300 L/s operatesin the molecular flow region. A tube6 inches long and 2 inches in diame-ter connects the pump to the chamber.

What is the effective pumping speed atthe chamber?` = 6 in = 15.2 cm D = 5.08 cmU = 12.1D3/` = 104 L/sThen Seff = Sp/(1 + Sp/U) = 300/(1 +300/104) L/s = 77 L/sThus only a small fraction of the pumpcapabilities are being used. The tubeshould be made larger.

eg. 5 A new tube is used in the above ex-ample that is 4 inches in diameter (twicethe previous diameter). What is thenew effective pumping speed?U ∝ D3 So doubling D, octuples U.U = 832 L/s andSeff = S/(1 + S/U) = 220 L/sThis is a great improvement!

Looking at Equation 5.50 we can make somegeneralizations.

Low Conductance ] If U � Sp, Seff ≈Sp/(Sp/U) = U and the size of thepump is immaterial. In order to useat least half of the capabilities of thepump, we must have U = Sp.

High Conductance For U � S, Seff ≈Sp.

The rate of evacuation is exponentially re-lated to the speed. The larger the effectivespeed, the quicker we will pump down a sys-tem.

5.13 Conductances in Paral-lel and Series

Suppose we have several pipes, valves, etc.,and we know the conductance of each —

24 5.14 A Practical Example

U1, U2, U3, . . . If we place the sections in par-allel, the total combined conductance is ap-proximately

Uparallel = U1 + U2 + U3 + . . . (5.51)

This makes intuitive sense if we think of theparallel sections effectively increasing the di-ameter in Equations 5.45 and 5.46 .

For sections connected in series, the effectivelength is increased, so the conduction shoulddecrease. In fact approximately

1

Useries=

1

U1+

1

U2+

1

U3+ . . . (5.52)

The reciprocal of conductance (in a pipe) isdefined as impedance of a pipe: W = 1/U inunits of s/L. The rules for pipe impedancesin series and parallel are the same as therules for electrical impedances in series andparallel. These rules, unlike those in theelectrical case, are only approximate in themolecular flow regime.

A series connection of elements is the mostcommon, so how careful must we be inchoosing elements to combine. One of theelements will have the minimum conduc-tance, Umin, and for this 1/Umin is a max-imum. Clearly, none of the other elementscan increase my total conductance—theyonly make it smaller. To improve the over-all conductance I must improve the conduc-tance of the element having the minimumconductance.

eg 5 Consider a three element seriessystem: a valve of conductance 20 L/sand two pipes of conductance 50 L/sand 100 L/s. What is the conductanceof the system? What should we do to

improve the conductance of the system?

The total conductance is 1/Useries =1/50 + 1/20 + 1/100 = 0.08Useries = 12.5 L/s

If I want to increase the conductance, Ishould spend most of my efforts on thevalve that has minimum conductance,and less time (and money) on the pip-ing.

5.14 A Practical Exam-ple

How does this all come together? Here let uslook at the type of calculations that are pos-sible for a vacuum system. Assume that wehave a vacuum chamber of volume V = 530cubic feet which is connected to a diffusionpump of speed 5300 L/s. Between this pumpand the chamber are a series of a valve ofconductance 5000 L/s, a pipe of conductance6000 L/s, and a cold trap of conductance4200 L/s. The diffusion pump is backed upwith a roughing pump of speed 120 cubicfeet/minute. We expect a base pressure of8.0×10−7 Torr. These are plausible numbersfor a 10 inch diffusion pump system.

1. (a) How long will it take to pump thechamber from atmosphere to 70mTorr using the rotary pump?

(b) What is the total conductance be-tween the diffusion pump and thechamber?

(c) What is the effective speed of thediffusion pump as measured at thechamber?

5.14 A Practical Example 25

(d) How long will it take to pump thechamber from 70 mTorr to 2×10−6

Torr using the diffusion pump?

Next consider the moment whenthe chamber is at a pressure of 1mTorr.

(e) When the pressure in the chamberis 1 mTorr, what is the pressure inthe diffusion pump?

(f) When the pressure in the chamberis 1 mTorr, what is the pressure inthe rotary pump?

Finally, consider the chamberwhen it is at its base pressure,8.0× 10−7 Torr.

(g) What is the leakage throughput,Qleaks, into the chamber?

(h) If the hi-vac valve were suddenlyclosed, how long would it take forthe pressure in the system to riseto 8.0× 10−6 Torr?

Let’s first define symbols and convert to met-ric values.

Volume V = 530 cubic feet×28.3168 L/cubicfeet = 15 000 L

Diffusion pump speed SDP = 5300 L/s

Conductances U1 = 5000 L/s U2 =6000 L/s U3 = 4200 L/s

Roughing pump speed SRP = 120 cfm×0.47195 L/s / cfm = 56.63 L/s

Base pressure Pult = 8.0× 10−7 Torr

Solution

1. (a) How long will it take to pumpthe chamber from atmosphere to

70 mTorr using the rotary pump?Since both pressures are well abovethe ultimate pressure we can useEquation 5.37.

The final pressure is 70 mTorr =0.070 Torr and we start at 760Torr.

t = (V/S) ln (P0/P )= (15000/56.63) ln (760/.070)= 2461 sec = 41 minutes.

(b) What is the total conductance be-tween the diffusion pump and thechamber? We combine the con-ductances to get the total conduc-tance Utot according to Equation5.52.

1/Utot = 1/5000 + 1/6000 +1/4200 = 0.0006046

Utot = 1/0.0006046 = 1654 L/s.

(c) What is the effective speed of thediffusion pump as measured at thechamber? We use Equation 5.50to get the effective speed, Seff .

Seff = 5300/[1 + 5300/1654]= 1261 L/s

(d) How long will it take to pumpthe chamber from 70 mTorr to2 × 10−6 Torr using the diffusionpump? Now the final pressure isclose to ultimate so we use Equa-tion 5.37.

P − Pult = 2 × 10−6 − 8 × 10−7 =1.2× 10−6 Torr.

We use the effective speed at thechamber that we just calculated.

26 5.14 A Practical Example

t =15000

1261ln

(70× 10−3

1.2× 10−6

)= 131 sec = 2 minutes 10 sec.

In practice the time would be quitea bit longer due to outgassing.

(e) When the pressure in the chamberis 1 mTorr, what is the pressurein the diffusion pump? Here weuse the fact that the throughput,Q = PS, is the same at all pointsin the chamber. Thus:

Q = PchamberSeff = PDPSDP(1× 10−3 Torr

)(1261 L/sec)

= PDP (5300 L/sec)

PDP = 0.24× 10−3 Torr

(f) When the pressure in the chamberis 1 mTorr, what is the pressure inthe rotary pump? Again use thethroughput.(1× 10−3

)(1261) Torr·L/sec

= PRP (56.4 L/sec)

PRP = 22.4 × 10−3 Torr = 22.4mTorr

(g) What is the leakage throughput,Qleaks, into the chamber? Wecompute throughput at the ulti-mate pressure.

Qleaks =(8× 10−7 Torr

)(1261 L/sec)

= 1.01× 10−3 Torr · L/sec= 0.079 sccm.

(h) If the hi-vac valve were suddenlyclosed, how long would it take forthe pressure in the system to rise

to 8.0× 10−6 Torr? Finally we useEquation 5.39 for the rate of rise.

P = (Qleaks/V )t+ Pult.

8 × 10−6 Torr = 8 × 10−7 Torr +[(1.01× 10−3 Torr · L/s

)/15000 L

]t.

7.2× 10−6 Torr=(6.73× 10−8 Torr/sec

)t

t= 107 sec