Gas Expansion into a Vacuum Environment

Introduction

            This animation considers the energy and pressure of a gas that is expanding due to the motion of one of the walls of its container.  For this case I will assume that the wall is moving uniformly so as to increase the volume accessible to the gas and that the wall does not experience any retarding force because of any external gas pressure.  This is in some ways more complicated than the usual gas physics situations where, for example, the gas expands while the pressure remain constant.

 

Figures

Figure 1: Gas expansion apparatus diagram.  The moving wall (piston) starts at the location of the vertical orange line and moves to the right end of the chamber.  Energy and pressure are plotted in blue and green, respectively.

Physics and Math

            Since the wall moves away from the occupied volume, the gas atoms lose momentum and kinetic energy when hitting it.  As a result of the energy reduction they impact all of the walls with less momentum and the pressure is reduced.

 

In the reference frame of the piston, the momentum change can be written:

 

δ p p =2m( v x v p )Leadingto: p p '=m( v x v p )2m( v x v p ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqiTdq MaamiCamaaBaaaleaacaWGWbaabeaakiabg2da9iabgkHiTiaaikda caWGTbGaaiikaiaadAhadaWgaaWcbaGaamiEaaqabaGccqGHsislca WG2bWaaSbaaSqaaiaadchaaeqaaOGaaiykaiaaykW7caaMc8UaaGPa VlaacYeacaGGLbGaaiyyaiaacsgacaGGPbGaaiOBaiaacEgacaaMc8 UaaiiDaiaac+gacaGG6aaabaGaamiCamaaBaaaleaacaWGWbaabeaa kiaacEcacqGH9aqpcaWGTbGaaiikaiaadAhadaWgaaWcbaGaamiEaa qabaGccqGHsislcaWG2bWaaSbaaSqaaiaadchaaeqaaOGaaiykaiab gkHiTiaaikdacaWGTbGaaiikaiaadAhadaWgaaWcbaGaamiEaaqaba GccqGHsislcaWG2bWaaSbaaSqaaiaadchaaeqaaOGaaiykaaaaaa@67A9@  

(1.1)

where the prime on pp indicates the value of momentum after the collision.

In the laboratory reference frame:

p ' Lab =m v p m( v x v p )=m(2 v p v x ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadchacaGGNa WaaSbaaSqaaiaadYeacaWGHbGaamOyaaqabaGccqGH9aqpcaWGTbGa amODamaaBaaaleaacaWGWbaabeaakiabgkHiTiaad2gacaGGOaGaam ODamaaBaaaleaacaWG4baabeaakiabgkHiTiaadAhadaWgaaWcbaGa amiCaaqabaGccaGGPaGaeyypa0JaaiyBaiaacIcacaaIYaGaamODam aaBaaaleaacaWGWbaabeaakiabgkHiTiaadAhadaWgaaWcbaGaamiE aaqabaGccaGGPaaaaa@5043@  

(1.2)

which results in the following energy change:

E'E= m 2 [ (2 v p v x ) 2 v x 2 ]= m 2 (4 v p 2 4 v p v x ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaacweacaGGNa GaeyOeI0Iaaiyraiabg2da9maalaaabaGaamyBaaqaaiaaikdaaaWa amWaaeaacaGGOaGaaGOmaiaadAhadaWgaaWcbaGaamiCaaqabaGccq GHsislcaWG2bWaaSbaaSqaaiaadIhaaeqaaOGaaiykamaaCaaaleqa baGaaGOmaaaakiabgkHiTiaadAhadaqhaaWcbaGaamiEaaqaaiaaik daaaaakiaawUfacaGLDbaacqGH9aqpdaWcaaqaaiaad2gaaeaacaaI YaaaaiaacIcacaaI0aGaaiODamaaDaaaleaacaWGWbaabaGaaGOmaa aakiabgkHiTiaaisdacaWG2bWaaSbaaSqaaiaadchaaeqaaOGaamOD amaaBaaaleaacaWG4baabeaakiaacMcaaaa@57A8@  

(1.3)

Obviously vx>vp if the atom is to hit the piston.  When vx=vp + δvx then becomes

δE=2m v p δ v x MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjaadw eacqGH9aqpcqGHsislcaaIYaGaamyBaiaadAhadaWgaaWcbaGaamiC aaqabaGccqaH0oazcaWG2bWaaSbaaSqaaiaadIhaaeqaaaaa@41EA@  

(1.4)

regardless of the magnitude of δvx.

For an atomic area of xp*h, where h is the piston height, the number of piston hits per second expected is

dn dt 1 x p h (< v x > v p )h= 1 x P (< v x > v p ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaiaad6gaaeaacaWGKbGaamiDaaaacqGHijYUdaWcaaqaaiaaigda aeaacaWG4bWaaSbaaSqaaiaadchaaeqaaOGaamiAaaaacaGGOaGaey ipaWJaamODamaaBaaaleaacaWG4baabeaakiabg6da+iabgkHiTiaa dAhadaWgaaWcbaGaamiCaaqabaGccaGGPaGaamiAaiabg2da9maala aabaGaaGymaaqaaiaadIhadaWgaaWcbaGaamiuaaqabaaaaOGaaiik aiabgYda8iaadAhadaWgaaWcbaGaamiEaaqabaGccqGH+aGpcqGHsi slcaWG2bWaaSbaaSqaaiaadchaaeqaaOGaaiykaaaa@555C@  

(1.5)

where xp is the length of the cylinder and h is the height of the piston.

Thus, combining equations (1.4) and (1.5), the expected energy loss per second is

dE dt =2m v p 1 x p (< v x > v p ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaiaadweaaeaacaWGKbGaamiDaaaacqGH9aqpcqGHsislcaaIYaGa amyBaiaadAhadaWgaaWcbaGaamiCaaqabaGcdaWcaaqaaiaaigdaae aacaWG4bWaaSbaaSqaaiaadchaaeqaaaaakiaacIcacqGH8aapcaWG 2bWaaSbaaSqaaiaadIhaaeqaaOGaeyOpa4JaeyOeI0IaamODamaaBa aaleaacaWGWbaabeaakiaacMcadaahaaWcbeqaaiaaikdaaaaaaa@4BD9@  

(1.6)

When vx is about equal to vp then vx doesn't change much nor does energy.  Now we must consider another question: how to compute a meaningful value for <vx> where only

<vx> is greater than vp is meaningful.  That involves integrals over the Maxwell distributions:

< v x >= v p v x exp( v x 2 v 2 )d v x v p exp( v x 2 v 2 )d v x = < v 2 > exp( v x 2 < v 2 > )| v p < v 2 > [ 1erf( v p 2 < v 2 > ) ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgYda8iaadA hadaWgaaWcbaGaamiEaaqabaGccqGH+aGpcqGH9aqpdaWcaaqaamaa pehabaGaamODamaaBaaaleaacaWG4baabeaakiGacwgacaGG4bGaai iCamaabmaabaGaeyOeI0YaaSaaaeaacaWG2bWaa0baaSqaaiaadIha aeaacaaIYaaaaaGcbaGaamODamaaCaaaleqabaGaaGOmaaaaaaaaki aawIcacaGLPaaacaWGKbGaamODamaaBaaaleaacaWG4baabeaaaeaa caWG2bWaaSbaaWqaaiaadchaaeqaaaWcbaGaeyOhIukaniabgUIiYd aakeaadaWdXbqaaiGacwgacaGG4bGaaiiCamaabmaabaGaeyOeI0Ya aSaaaeaacaWG2bWaa0baaSqaaiaadIhaaeaacaaIYaaaaaGcbaGaam ODamaaCaaaleqabaGaaGOmaaaaaaaakiaawIcacaGLPaaacaWGKbGa amODamaaBaaaleaacaWG4baabeaaaeaacaWG2bWaaSbaaWqaaiaadc haaeqaaaWcbaGaeyOhIukaniabgUIiYdaaaOGaeyypa0ZaaSaaaeaa cqGH8aapcaWG2bWaa0baaSqaaaqaaiaaikdaaaGccqGH+aGpdaabca qaaiGacwgacaGG4bGaaiiCamaabmaabaGaeyOeI0YaaSaaaeaacaWG 2bWaa0baaSqaaiaadIhaaeaacaaIYaaaaaGcbaGaeyipaWJaamODam aaDaaaleaaaeaacaaIYaaaaOGaeyOpa4daaaGaayjkaiaawMcaaaGa ayjcSdWaaWbaaSqabeaacqGHEisPaaGcdaWgaaWcbaGaamODamaaBa aameaacaWGWbaabeaaaSqabaaakeaadaGcaaqaaiabgYda8iaadAha daahaaWcbeqaaiaaikdaaaGccqGH+aGpaSqabaGcdaWadaqaaiaaig dacqGHsislcaWGLbGaamOCaiaadAgadaqadaqaamaalaaabaGaamOD amaaDaaaleaacaWGWbaabaGaaGOmaaaaaOqaaiabgYda8iaadAhada qhaaWcbaaabaGaaGOmaaaakiabg6da+aaaaiaawIcacaGLPaaaaiaa wUfacaGLDbaaaaaaaa@8DE6@  

(1.7)

 

  

                                                                       

Converting equation (1.7) to average energy terms <E>=m<vx2> we have:

d<E> dt =2m v p 1 x p [ 2<E> mπ exp( m v p 2 <E> ) 1erf( v p 2 E ) v p ] 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaiabgYda8iaadweacqGH+aGpaeaacaWGKbGaamiDaaaacqGH9aqp cqGHsislcaaIYaGaamyBaiaadAhadaWgaaWcbaGaamiCaaqabaGcda WcaaqaaiaaigdaaeaacaWG4bWaaSbaaSqaaiaadchaaeqaaaaakmaa dmaabaWaaSaaaeaadaGcaaqaamaalaaabaGaaGOmaiabgYda8iaadw eacqGH+aGpaeaacaWGTbGaeqiWdahaaaWcbeaakiGacwgacaGG4bGa aiiCamaabmaabaGaeyOeI0YaaSaaaeaacaWGTbGaamODamaaDaaale aacaWGWbaabaGaaGOmaaaaaOqaaiabgYda8iaadweacqGH+aGpaaaa caGLOaGaayzkaaaabaGaaGymaiabgkHiTiaadwgacaWGYbGaamOzam aabmaabaWaaSaaaeaacaWG2bWaa0baaSqaaiaadchaaeaacaaIYaaa aaGcbaGaamyraaaaaiaawIcacaGLPaaaaaGaeyOeI0IaamODamaaBa aaleaacaWGWbaabeaaaOGaay5waiaaw2faamaaCaaaleqabaGaaGOm aaaaaaa@6690@  

(1.8)

Now vpdt=δxp so we have the following differential equation which can only be solved iteratively:

δ<E>=2m [ 1 2 <E> πm exp( m v p 2 <E> ) v p ] 2 δ x p x p MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabgY da8iaadweacqGH+aGpcqGH9aqpcqGHsislcaaIYaGaamyBamaadmaa baWaaSaaaeaacaaIXaaabaGaaGOmaaaadaGcaaqaamaalaaabaGaey ipaWJaamyraiabg6da+aqaaiabec8aWjaad2gaaaaaleqaaOGaciyz aiaacIhacaGGWbWaaeWaaeaacqGHsisldaWcaaqaaiaad2gacaWG2b Waa0baaSqaaiaadchaaeaacaaIYaaaaaGcbaGaeyipaWJaamyraiab g6da+aaaaiaawIcacaGLPaaacqGHsislcaWG2bWaaSbaaSqaaiaadc haaeqaaaGccaGLBbGaayzxaaWaaWbaaSqabeaacaaIYaaaaOWaaSaa aeaacqaH0oazcaWG4bWaaSbaaSqaaiaadchaaeqaaaGcbaGaamiEam aaBaaaleaacaWGWbaabeaaaaaaaa@5D47@  

(1.9)

It turns out the that exponential value for the piston speeds used in the program is very close to 1.0 so the program uses the value

δE δ x b =2 v p < v x >(< v x > v p )/ x p MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaeq iTdqMaamyraaqaaiabes7aKjaadIhadaWgaaWcbaGaamOyaaqabaaa aOGaeyypa0JaaGOmaiaadAhadaWgaaWcbaGaamiCaaqabaGccqGH8a apcaWG2bWaaSbaaSqaaiaadIhaaeqaaOGaeyOpa4JaaiikaiabgYda 8iaadAhadaWgaaWcbaGaamiEaaqabaGccqGH+aGpcqGHsislcaWG2b WaaSbaaSqaaiaadchaaeqaaOGaaiykaiaac+cacaWG4bWaaSbaaSqa aiaadchaaeqaaaaa@4FC2@  

(1.10)

Appendix: Rate of wall hits

Assuming that the container dimensions are much greater that the mean free path, l, we have the following cases:

1.  If the distance of the atom from the wall is equal to l, then the probability that the atom hits the wall in l/vx seconds is 0.5. 

2.  If the distance of the atom from the wall is equal to fl, then the probability that the atom hits the wall in l/vx seconds is 0.5/f2.

 There are N(fl/xP) atoms within nl of the piston.

 Converting statement 2 into an equation we have:

P(f)= 0.5 f 2 n(f)=0.5N( fl x p ) 1 f 2 =0.5N l f x p f=( x p x)/l n(x)=0.5N l 2 x p |x x p | MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaamiuai aacIcacaWGMbGaaiykaiabg2da9maalaaabaGaaGimaiaac6cacaaI 1aaabaGaamOzamaaCaaaleqabaGaaGOmaaaaaaaakeaacaWGUbGaai ikaiaadAgacaGGPaGaeyypa0JaaGimaiaac6cacaaI1aGaamOtaiaa cIcadaWcaaqaaiaadAgacaWGSbaabaGaamiEamaaBaaaleaacaWGWb aabeaaaaGccaGGPaWaaSaaaeaacaaIXaaabaGaamOzamaaCaaaleqa baGaaGOmaaaaaaGccqGH9aqpcaaIWaGaaiOlaiaaiwdacaWGobWaaS aaaeaacaWGSbaabaGaamOzaiaadIhadaWgaaWcbaGaamiCaaqabaaa aaGcbaGaamOzaiabg2da9iaacIcacaWG4bWaaSbaaSqaaiaadchaae qaaOGaeyOeI0IaamiEaiaacMcacaGGVaGaamiBaaqaaiaad6gacaGG OaGaaiiEaiaacMcacqGH9aqpcaaIWaGaaiOlaiaaiwdacaWGobWaaS aaaeaacaWGSbWaaWbaaSqabeaacaaIYaaaaaGcbaGaamiEamaaBaaa leaacaWGWbaabeaakiaacYhacaWG4bGaeyOeI0IaamiEamaaBaaale aacaWGWbaabeaakiaacYhaaaaaaaa@700F@  

(1.11)

The number of atoms at distance |xp-x| hitting the wall per second is then n(x)vx|x-xp|/vx.

n( x ) v x | x x p | =0.5N l 2 v x | x x p | 2 x p MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape GaamOBa8aadaqadaqaa8qacaWG4baapaGaayjkaiaawMcaamaalaaa baWdbiaadAhapaWaaSbaaSqaa8qacaWG4baapaqabaaakeaadaabda qaa8qacaWG4bGaeyOeI0IaamiEa8aadaWgaaWcbaWdbiaadchaa8aa beaaaOGaay5bSlaawIa7aaaacqGH9aqpcaaIWaGaaiOlaiaaiwdaca WGobWaaSaaaeaacaWGSbWaaWbaaSqabeaacaaIYaaaaOGaamODamaa BaaaleaacaWG4baabeaaaOqaamaaemaabaWdbiaadIhacqGHsislca WG4bWdamaaBaaaleaapeGaamiCaaWdaeqaaaGccaGLhWUaayjcSdWa aWbaaSqabeaacaaIYaaaaOGaamiEamaaBaaaleaacaWGWbaabeaaaa aaaa@5649@  

(1.12)

The mean free path is

l= h x p Na MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYgacqGH9a qpdaWcaaqaaiaadIgacaWG4bWaaSbaaSqaaiaadchaaeqaaaGcbaGa amOtaiaadggaaaaaaa@3CC0@  

(1.13)

where a is the radius of the hard disc atom.

Then equation (1.12) becomes:

n( x ) v x | x p x | =0.5N ( h x p Na ) 2 v x | x p x | 2 x p =0.5 h 2 x p v x N a 2 ( x p x) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape GaamOBa8aadaqadaqaa8qacaWG4baapaGaayjkaiaawMcaamaalaaa baWdbiaadAhapaWaaSbaaSqaa8qacaWG4baapaqabaaakeaadaabda qaa8qacaWG4bWaaSbaaSqaaiaadchaaeqaaOGaeyOeI0IaamiEaaWd aiaawEa7caGLiWoaaaGaeyypa0JaaGimaiaac6cacaaI1aGaamOtam aalaaabaWaaeWaaeaadaWcaaqaaiaadIgacaWG4bWaaSbaaSqaaiaa dchaaeqaaaGcbaGaamOtaiaadggaaaaacaGLOaGaayzkaaWaaWbaaS qabeaacaaIYaaaaOGaamODamaaBaaaleaacaWG4baabeaaaOqaamaa emaabaWdbiaadIhadaWgaaWcbaGaamiCaaqabaGccqGHsislcaWG4b aapaGaay5bSlaawIa7amaaCaaaleqabaGaaGOmaaaakiaadIhadaWg aaWcbaGaamiCaaqabaaaaOGaeyypa0JaaGimaiaac6cacaaI1aWaaS aaaeaacaWGObWaaWbaaSqabeaacaaIYaaaaOGaamiEamaaBaaaleaa caWGWbaabeaakiaadAhadaWgaaWcbaGaamiEaaqabaaakeaacaWGob GaamyyamaaCaaaleqabaGaaGOmaaaakiaacIcapeGaamiEa8aadaWg aaWcbaWdbiaadchaa8aabeaakiabgkHiTiaadIhacaGGPaWaaWbaaS qabeaacaaIYaaaaaaaaaa@6E40@  

(1.14)

Now we must integrate this expression over all the x's from 0 to xp.

0.5 h 2 x p v x N a 2 0 x p l dx ( x p x) 2 =0.5 h 2 x p v x N a 2 [ 1 x p ( x p l) 1 x p ]= 0.5 h 2 x p v x N a 2 ( 1 l 1 x p )0.5 h 2 x p v x N a 2 l =0.5 h 2 x p v x Na N a 2 h x p =0.5 h v x a x p MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaaGimai aac6cacaaI1aWaaSaaaeaacaWGObWaaWbaaSqabeaacaaIYaaaaOGa amiEamaaBaaaleaacaWGWbaabeaakiaadAhadaWgaaWcbaGaamiEaa qabaaakeaacaWGobGaamyyamaaCaaaleqabaGaaGOmaaaaaaGcdaWd XaqaamaalaaabaGaamizaiaadIhaaeaacaGGOaGaamiEamaaBaaale aacaWGWbaabeaakiabgkHiTiaadIhacaGGPaWaaWbaaSqabeaacaaI YaaaaaaaaeaacaaIWaaabaGaamiEamaaBaaameaacaWGWbaabeaali abgkHiTiaadYgaa0Gaey4kIipakiabg2da9iaaicdacaGGUaGaaGyn amaalaaabaGaamiAamaaCaaaleqabaGaaGOmaaaakiaadIhadaWgaa WcbaGaamiCaaqabaGccaWG2bWaaSbaaSqaaiaadIhaaeqaaaGcbaGa amOtaiaadggadaahaaWcbeqaaiaaikdaaaaaaOWaamWaaeaadaWcaa qaaiaaigdaaeaacaWG4bWaaSbaaSqaaiaadchaaeqaaOGaeyOeI0Ia aiikaiaadIhadaWgaaWcbaGaamiCaaqabaGccqGHsislcaWGSbGaai ykaaaacqGHsisldaWcaaqaaiaaigdaaeaacaWG4bWaaSbaaSqaaiaa dchaaeqaaaaaaOGaay5waiaaw2faaiabg2da9aqaaiaaicdacaGGUa GaaGynamaalaaabaGaamiAamaaCaaaleqabaGaaGOmaaaakiaadIha daWgaaWcbaGaamiCaaqabaGccaWG2bWaaSbaaSqaaiaadIhaaeqaaa GcbaGaamOtaiaadggadaahaaWcbeqaaiaaikdaaaaaaOWaaeWaaeaa daWcaaqaaiaaigdaaeaacaWGSbaaaiabgkHiTmaalaaabaGaaGymaa qaaiaadIhadaWgaaWcbaGaamiCaaqabaaaaaGccaGLOaGaayzkaaGa eyisISRaaGimaiaac6cacaaI1aWaaSaaaeaacaWGObWaaWbaaSqabe aacaaIYaaaaOGaamiEamaaBaaaleaacaWGWbaabeaakiaadAhadaWg aaWcbaGaamiEaaqabaaakeaacaWGobGaamyyamaaCaaaleqabaGaaG OmaaaakiaadYgaaaGaeyypa0JaaGimaiaac6cacaaI1aWaaSaaaeaa caWGObWaaWbaaSqabeaacaaIYaaaaOGaamiEamaaBaaaleaacaWGWb aabeaakiaadAhadaWgaaWcbaGaamiEaaqabaGccaWGobGaamyyaaqa aiaad6eacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaamiAaiaadIhada WgaaWcbaGaamiCaaqabaaaaOGaeyypa0JaaGimaiaac6cacaaI1aWa aSaaaeaacaWGObGaamODamaaBaaaleaacaWG4baabeaaaOqaaiaadg gacaWG4bWaaSbaaSqaaiaadchaaeqaaaaaaaaa@A6CF@  

(1.15)